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https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_16

Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers? $\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 28$

Note that the sum of the first two numbers is $21\cdot2=42,$ the sum of the middle two numbers is $26\cdot2=52,$ and the sum of the last two numbers is $30\cdot2=60.$ It follows that the sum of the four numbers is $42+60=102,$ so the sum of the first and last numbers is $102-52=50.$ Therefore, the average of the first and last numbers is $50\div2=\boxed{25}.$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_16

Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers? $\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 28$

Let $a,b,c,$ and $d$ be the four numbers in that order. We are given that \begin{align*} \frac{a+b}{2} &= 21, &(1) \\ \frac{b+c}{2} &= 26, &(2) \\ \frac{c+d}{2} &= 30, &(3) \end{align*} and we wish to find $\frac{a+d}{2}.$ We add $(1)$ and $(3),$ then subtract $(2)$ from the result: \[\frac{a+d}{2}=21+30-26=\boxed{25}.\] ~MRENTHUSIASM

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_16

Four numbers are written in a row. The average of the first two is $21,$ the average of the middle two is $26,$ and the average of the last two is $30.$ What is the average of the first and last of the numbers? $\textbf{(A) } 24 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 26 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 28$

We can just assume some of the numbers. For example, let the first two numbers both be $21.$ It follows that the third number is $31,$ and the fourth number is $29.$ Therefore, the average of the first and last numbers is $\dfrac{21+29}2=\dfrac{50}2=\boxed{25}.$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_17

If $n$ is an even positive integer, the $\emph{double factorial}$ notation $n!!$ represents the product of all the even integers from $2$ to $n$ . For example, $8!! = 2 \cdot 4 \cdot 6 \cdot 8$ . What is the units digit of the following sum? \[2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

Notice that once $n>8,$ the units digit of $n!!$ will be $0$ because there will be a factor of $10.$ Thus, we only need to calculate the units digit of \[2!!+4!!+6!!+8!! = 2+8+48+48\cdot8.\] We only care about units digits, so we have $2+8+8+8\cdot8,$ which has the same units digit as $2+8+8+4.$ The answer is $\boxed{2}.$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_18

The midpoints of the four sides of a rectangle are $(-3,0), (2,0), (5,4),$ and $(0,4).$ What is the area of the rectangle? $\textbf{(A) } 20 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 40 \qquad \textbf{(D) } 50 \qquad \textbf{(E) } 80$

The midpoints of the four sides of every rectangle are the vertices of a rhombus whose area is half the area of the rectangle: Note that the diagonals of the rhombus have the same lengths as the sides of the rectangle. Let $A=(-3,0), B=(2,0), C=(5,4),$ and $D=(0,4).$ Note that $A,B,C,$ and $D$ are the vertices of a rhombus whose diagonals have lengths $AC=4\sqrt{5}$ and $BD=2\sqrt{5}.$ It follows that the dimensions of the rectangle are $4\sqrt{5}$ and $2\sqrt{5},$ so the area of the rectangle is $4\sqrt{5}\cdot2\sqrt{5}=\boxed{40}.$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_18

The midpoints of the four sides of a rectangle are $(-3,0), (2,0), (5,4),$ and $(0,4).$ What is the area of the rectangle? $\textbf{(A) } 20 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 40 \qquad \textbf{(D) } 50 \qquad \textbf{(E) } 80$

If a rectangle has area $K,$ then the area of the quadrilateral formed by its midpoints is $\frac{K}{2}.$ Define points $A,B,C,$ and $D$ as Solution 1 does. Since $A,B,C,$ and $D$ are the midpoints of the rectangle, the rectangle's area is $2[ABCD].$ Now, note that $ABCD$ is a parallelogram since $AB=CD$ and $\overline{AB}\parallel\overline{CD}.$ As the parallelogram's height from $D$ to $\overline{AB}$ is $4$ and $AB=5,$ its area is $4\cdot5=20.$ Therefore, the area of the rectangle is $20\cdot2=\boxed{40}.$ ~Fruitz

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_19

Mr. Ramos gave a test to his class of $20$ students. The dot plot below shows the distribution of test scores. [asy] //diagram by pog . give me 1,000,000,000 dollars for this diagram size(5cm); defaultpen(0.7); dot((0.5,1)); dot((0.5,1.5)); dot((1.5,1)); dot((1.5,1.5)); dot((2.5,1)); dot((2.5,1.5)); dot((2.5,2)); dot((2.5,2.5)); dot((3.5,1)); dot((3.5,1.5)); dot((3.5,2)); dot((3.5,2.5)); dot((3.5,3)); dot((4.5,1)); dot((4.5,1.5)); dot((5.5,1)); dot((5.5,1.5)); dot((5.5,2)); dot((6.5,1)); dot((7.5,1)); draw((0,0.5)--(8,0.5),linewidth(0.7)); defaultpen(fontsize(10.5pt)); label("$65$", (0.5,-0.1)); label("$70$", (1.5,-0.1)); label("$75$", (2.5,-0.1)); label("$80$", (3.5,-0.1)); label("$85$", (4.5,-0.1)); label("$90$", (5.5,-0.1)); label("$95$", (6.5,-0.1)); label("$100$", (7.5,-0.1)); [/asy] Later Mr. Ramos discovered that there was a scoring error on one of the questions. He regraded the tests, awarding some of the students $5$ extra points, which increased the median test score to $85$ . What is the minimum number of students who received extra points? (Note that the median test score equals the average of the $2$ scores in the middle if the $20$ test scores are arranged in increasing order.) $\textbf{(A)} ~2\qquad\textbf{(B)} ~3\qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~5\qquad\textbf{(E)} ~6\qquad$

We set up our cases as solution 1 showed, realizing that only the second case is possible. We notice that $13$ students have scores under $85$ currently and only $5$ have scores over $85$ . We find the median of these two numbers, getting: \[13-5=8\] \[\frac{8}{2}=4\] \[13-4=9\] Thus, we realize that $4$ students must have their score increased by $5$ So, the correct answer is $\boxed{4}$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); [/asy] $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$

The sum of the numbers in each row is $12$ . Consider the second row. In order for the sum of the numbers in this row to equal $12$ , the two shaded numbers must add up to $13$ [asy] unitsize(0.5cm); fill((-3,1)--(1,1)--(1,-1)--(-3,-1)--cycle,mediumgray); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); [/asy] If two numbers add up to $13$ , one of them must be at least $7$ : If both shaded numbers are no more than $6$ , their sum can be at most $12$ . Therefore, for $x$ to be larger than the three missing numbers, $x$ must be at least $8$ . We can construct a working scenario where $x=8$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$8$"); label((0,-2),"$-4$"); label((-2,0),"$6$"); label((0,0),"$7$"); [/asy] So, our answer is $\boxed{8}$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); [/asy] $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$

The sum of the numbers in each row is $-2+9+5=12,$ and the sum of the numbers in each column is $5+(-1)+8=12.$ Let $y$ be the number in the lower middle. It follows that $x+y+8=12,$ or $x+y=4.$ We express the other two missing numbers in terms of $x$ and $y,$ as shown below: [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); label((0,-2),"$y$",red+fontsize(11)); label((-2,0),"$y{+}10$",red+fontsize(11)); label((0,0),"$x{-}1$",red+fontsize(11)); [/asy] We have $x>x-1, x>y+10,$ and $x>y.$ Note that the first inequality is true for all values of $x.$ We only need to solve the second inequality so that the third inequality is true for all values of $x.$ By substitution, we get $x>(4-x)+10,$ from which $x>7.$ Therefore, the smallest possible value of $x$ is $\boxed{8}.$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); [/asy] $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$

This is based on the Solution 2 above and it is perhaps a little simpler than that. Let $y$ be the number in the lower middle. Applying summation to first two columns yields the following. [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); label((0,-2),"$y$",red+fontsize(11)); label((-2,0),"$14{-}x$",red+fontsize(11)); label((0,0),"$3{-}y$",red+fontsize(11)); [/asy] Since $x$ is greater than the other three, we have $x>14-x,$ or $x>7.$ Therefore, the smallest possible value of $x$ is $\boxed{8}.$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); [/asy] $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$

Note that the sum of the rows and columns must be $8+5-1=12$ . We proceed to test the answer choices. Testing $\textbf{(A)}$ , when $x = -1$ , the number above $x$ must be $15$ , which contradicts the precondition that the numbers surrounding $x$ is less than $x$ Testing $\textbf{(B)}$ , the number above $x$ is $9$ , which does not work. Testing $\textbf{(C)}$ , the number above $x$ is $8$ , which does not work. Testing $\textbf{(D)}$ , the number above $x$ is $6$ , which does work. Hence, the answer is $\boxed{8}$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_20

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); [/asy] $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$

The sum of the numbers in each column and row should be $5+(-1)+8=12$ . If we look at the $1^{\text{st}}$ column, the gray squares (shown below) sum to $12-(-2)=14$ [asy] draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); filldraw((-3,-3)--(-1,-3)--(-1,-1)--(-3,-1)--cycle, lightgray, black+linewidth(1)); filldraw((-1,-1)--(-3,-1)--(-3,1)--(-1,1)--cycle, lightgray, black+linewidth(1)); label(scale(1)*"All credits for original unedited asymptote for the problem go to whoever made the asymptote in the 'Problem' section.", (-0,-5), S); [/asy] If square $x$ has to be greater than or equal to the three blank squares, then the least $x$ can be is half the sum of the value of the gray squares, which is $14\div2=7$ . But square $x$ has to be greater than and not greater than or equal to the three blank squares, so the least $x$ can be is $7+1=8$ . Testing for the other rows and columns (it might be smaller than the other two squares!), we find that the smallest $x$ can be is indeed $8$ ; the other two squares are less than $8$ . Therefore, the answer is $\boxed{8}$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_21

Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first? [asy] size(7cm); draw((-8,27)--(72,27)); draw((16,0)--(16,35)); draw((40,0)--(40,35)); label("12", (28,3)); draw((25,6.5)--(25,12)--(31,12)--(31,6.5)--cycle); draw((25,5.5)--(31,5.5)); label("18", (56,3)); draw((53,6.5)--(53,12)--(59,12)--(59,6.5)--cycle); draw((53,5.5)--(59,5.5)); draw((53,5.5)--(59,5.5)); label("20", (28,18)); label("15", (28,24)); draw((25,21)--(31,21)); label("10", (56,18)); label("10", (56,24)); draw((53,21)--(59,21)); label("First Half", (28,31)); label("Second Half", (56,31)); label("Candace", (2.35,6)); label("Steph", (0,21)); [/asy] $\textbf{(A) } 7\qquad\textbf{(B) } 8\qquad\textbf{(C) } 9\qquad\textbf{(D) } 10\qquad\textbf{(E) } 11$

Let $x$ be the number of shots that Candace made in the first half, and let $y$ be the number of shots Candace made in the second half. Since Candace and Steph took the same number of attempts, with an equal percentage of baskets scored, we have $x+y=10+15=25.$ In addition, we have the following inequalities: \[\frac{x}{12}<\frac{15}{20} \implies x<9,\] and \[\frac{y}{18}<\frac{10}{10} \implies y<18.\] Pairing this up with $x+y=25$ we see the only possible solution is $(x,y)=(8,17),$ for an answer of $17-8 = \boxed{9}.$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_21

Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first? [asy] size(7cm); draw((-8,27)--(72,27)); draw((16,0)--(16,35)); draw((40,0)--(40,35)); label("12", (28,3)); draw((25,6.5)--(25,12)--(31,12)--(31,6.5)--cycle); draw((25,5.5)--(31,5.5)); label("18", (56,3)); draw((53,6.5)--(53,12)--(59,12)--(59,6.5)--cycle); draw((53,5.5)--(59,5.5)); draw((53,5.5)--(59,5.5)); label("20", (28,18)); label("15", (28,24)); draw((25,21)--(31,21)); label("10", (56,18)); label("10", (56,24)); draw((53,21)--(59,21)); label("First Half", (28,31)); label("Second Half", (56,31)); label("Candace", (2.35,6)); label("Steph", (0,21)); [/asy] $\textbf{(A) } 7\qquad\textbf{(B) } 8\qquad\textbf{(C) } 9\qquad\textbf{(D) } 10\qquad\textbf{(E) } 11$

Clearly, Steph made $15 + 10 = 25$ shots in total. Also, due to parity reasons, the difference between the amount of shots Candace made in the first and second half must be odd. Thus, we can just test 7, 9, and 11, and after doing so we find that the answer is $\boxed{9}.$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_21

Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first? [asy] size(7cm); draw((-8,27)--(72,27)); draw((16,0)--(16,35)); draw((40,0)--(40,35)); label("12", (28,3)); draw((25,6.5)--(25,12)--(31,12)--(31,6.5)--cycle); draw((25,5.5)--(31,5.5)); label("18", (56,3)); draw((53,6.5)--(53,12)--(59,12)--(59,6.5)--cycle); draw((53,5.5)--(59,5.5)); draw((53,5.5)--(59,5.5)); label("20", (28,18)); label("15", (28,24)); draw((25,21)--(31,21)); label("10", (56,18)); label("10", (56,24)); draw((53,21)--(59,21)); label("First Half", (28,31)); label("Second Half", (56,31)); label("Candace", (2.35,6)); label("Steph", (0,21)); [/asy] $\textbf{(A) } 7\qquad\textbf{(B) } 8\qquad\textbf{(C) } 9\qquad\textbf{(D) } 10\qquad\textbf{(E) } 11$

Steph made 75 percent of his shots in the first half. He makes all of his shots in the second half. The most baskets Candace could have made in the first half is 8 baskets. The most she could have made in the second half is 17 baskets. Steph makes 25 and misses 5 baskets and the only way for Candace to make 25 shots is to make 8 in the first half and 17 in the second. Thus, $17 - 8 = \boxed{9}.$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_22

A bus takes $2$ minutes to drive from one stop to the next, and waits $1$ minute at each stop to let passengers board. Zia takes $5$ minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwise she will start walking toward the next stop. Suppose the bus and Zia start at the same time toward the library, with the bus $3$ stops behind. After how many minutes will Zia board the bus? $\textbf{(A) } 17 \qquad \textbf{(B) } 19 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 23$

Initially, suppose that the bus is at Stop $0$ (starting point) and Zia is at Stop $3.$ We construct the following table of $5$ -minute intervals: \[\begin{array}{c||c|c} & & \\ [-2.5ex] \textbf{Time} & \textbf{Bus's Location} & \textbf{Zia's Location} \\ [0.5ex] \hline & & \\ [-2ex] \boldsymbol{5} \ \textbf{Minutes} & \text{Stop} \ 2 \ \text{(Waiting)} & \text{Stop} \ 4 \\ \boldsymbol{10} \ \textbf{Minutes} & \text{Midpoint of Stops} \ 3 \ \text{and} \ 4 & \text{Stop} \ 5 \\ \boldsymbol{15} \ \textbf{Minutes} & \text{Stop} \ 5 \ \text{(Leaving)} & \text{Stop} \ 6 \end{array}\] Note that Zia will wait for the bus after $15$ minutes, and the bus will arrive $2$ minutes later. Therefore, the answer is $15+2=\boxed{17}.$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_22

A bus takes $2$ minutes to drive from one stop to the next, and waits $1$ minute at each stop to let passengers board. Zia takes $5$ minutes to walk from one bus stop to the next. As Zia reaches a bus stop, if the bus is at the previous stop or has already left the previous stop, then she will wait for the bus. Otherwise she will start walking toward the next stop. Suppose the bus and Zia start at the same time toward the library, with the bus $3$ stops behind. After how many minutes will Zia board the bus? $\textbf{(A) } 17 \qquad \textbf{(B) } 19 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 23$

Since Zia will wait for the bus if the bus is at the previous stop, we can create an equation to solve for when the bus is at the previous stop. The bus travels $\frac{1}{3}$ of a stop per minute, and Zia travels $\frac{1}{5}$ of a stop per minute. Now we create the equation, $\frac{1}{3}m = \frac{1}{5}m + 3 - 1$ (the $-1$ accounts for us wanting to find when the bus reaches the stop before Zia's). Solving, we find that $m=15.$ Now Zia has to wait $2$ minutes for the bus to reach her, so our answer is $15+2=\boxed{17}.$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_23

$\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$ -by- $3$ grid. Shown below is a sample configuration with three $\triangle$ s in a line. [asy] //diagram size(5cm); defaultpen(linewidth(1.5)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,1)--(3,1)--(3,2)--(0,2)--cycle); draw((1,0)--(1,3)--(2,3)--(2,0)--cycle); draw(circle((3/2,5/2),1/3)); draw(circle((5/2,1/2),1/3)); draw(circle((3/2,3/2),1/3)); draw(shift(0.5,0.38) * equi); draw(shift(1.5,0.38) * equi); draw(shift(0.5,1.38) * equi); draw(shift(2.5,1.38) * equi); draw(shift(0.5,2.38) * equi); draw(shift(2.5,2.38) * equi); [/asy] How many configurations will have three $\triangle$ s in a line and three $\bigcirc$ s in a line? $\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96$

Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end. We take casework: Case 1: 3 lines : In this case, the lines would need to be $2$ of one shape and $1$ of another, so there are $\frac{3!}{2} = 3$ ways to arrange the lines and $2$ ways to pick which shape has only one line. In total, this is $3\cdot 2 = 6.$ Case 2: 2 lines : In this case, the lines would be one line of triangles, one line of circles, and the last one can be anything that includes both shapes. There are $3! = 6$ ways to arrange the lines and $2^3-2 = 6$ ways to choose the last line. (We subtract $2$ from the last line because one arrangement of the last line is all triangles and the other arrangement of the last line is all circles, which causes Case 2 to overlap with Case 1 and further complicating the solution.) In total, this is $6\cdot 6 = 36.$ Finally, we add and multiply: $2(36+6)=2(42)=\boxed{84}$

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_23

$\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$ -by- $3$ grid. Shown below is a sample configuration with three $\triangle$ s in a line. [asy] //diagram size(5cm); defaultpen(linewidth(1.5)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,1)--(3,1)--(3,2)--(0,2)--cycle); draw((1,0)--(1,3)--(2,3)--(2,0)--cycle); draw(circle((3/2,5/2),1/3)); draw(circle((5/2,1/2),1/3)); draw(circle((3/2,3/2),1/3)); draw(shift(0.5,0.38) * equi); draw(shift(1.5,0.38) * equi); draw(shift(0.5,1.38) * equi); draw(shift(2.5,1.38) * equi); draw(shift(0.5,2.38) * equi); draw(shift(2.5,2.38) * equi); [/asy] How many configurations will have three $\triangle$ s in a line and three $\bigcirc$ s in a line? $\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96$

We will only consider cases where the three identical symbols are the same column, but at the end we shall double our answer as the same holds true for rows. There are $3$ ways to choose a column with all $\bigcirc$ 's and $2$ ways to choose a column with all $\triangle$ 's. The third column can be filled in $2^3=8$ ways. Therefore, we have a total of $3\cdot2\cdot8=48$ cases. However, we overcounted the cases with $2$ complete columns of with one symbol and $1$ complete column with another symbol. This happens in $2\cdot3=6$ cases. $48-6=42$ . However, we have to remember to double our answer, giving us $\boxed{84}$ ways to complete the grid.

https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_24

The figure below shows a polygon $ABCDEFGH$ , consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that $AH = EF = 8$ and $GH = 14$ . What is the volume of the prism? [asy] usepackage("mathptmx"); size(275); defaultpen(linewidth(0.8)); real r = 2, s = 2.5, theta = 14; pair G = (0,0), F = (r,0), C = (r,s), B = (0,s), M = (C+F)/2, I = M + s/2 * dir(-theta); pair N = (B+G)/2, J = N + s/2 * dir(180+theta); pair E = F + r * dir(- 45 - theta/2), D = I+E-F; pair H = J + r * dir(135 + theta/2), A = B+H-J; draw(A--B--C--I--D--E--F--G--J--H--cycle^^rightanglemark(F,I,C)^^rightanglemark(G,J,B)); draw(J--B--G^^C--F--I,linetype ("4 4")); dot("$A$",A,N); dot("$B$",B,1.2*N); dot("$C$",C,N); dot("$D$",D,dir(0)); dot("$E$",E,S); dot("$F$",F,1.5*dir(-100)); dot("$G$",G,S); dot("$H$",H,W); dot("$I$",I,NE); dot("$J$",J,1.5*S); [/asy] $\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288$

While imagining the folding, $\overline{AB}$ goes on $\overline{BC},$ $\overline{AH}$ goes on $\overline{CI},$ and $\overline{EF}$ goes on $\overline{FG}.$ So, $BJ=CI=8$ and $FG=BC=8.$ Also, $\overline{HJ}$ becomes an edge parallel to $\overline{FG},$ so that means $HJ=8.$ Since $GH=14,$ then $JG=14-8=6.$ So, the area of $\triangle BJG$ is $\frac{8\cdot6}{2}=24.$ If we let $\triangle BJG$ be the base, then the height is $FG=8.$ So, the volume is $24\cdot8=\boxed{192}.$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_1

Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need? $\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

We have $\text{water} : \text{sugar} : \text{lemon juice} = 4\cdot 2 : 2 : 1 = 8 : 2 : 1,$ so Luka needs $3 \cdot 8 = \boxed{24}$ cups.

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_1

Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need? $\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Since the amount of sugar is twice the amount of lemon juice, Luka uses $3\cdot2=6$ cups of sugar. Since the amount of water is $4$ times the amount of sugar, he uses $6\cdot4=\boxed{24}$ cups of water.

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_1

Luka is making lemonade to sell at a school fundraiser. His recipe requires $4$ times as much water as sugar and twice as much sugar as lemon juice. He uses $3$ cups of lemon juice. How many cups of water does he need? $\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

The ratio is $\text{Water}:\text{Sugar}:\text{Lemon Juice},$ or $8:2:1.$ Since we know that Luka used 3 cups of lemon juice, he needs $3\cdot2=6$ cups of sugar. Because the amount of water is $4$ times the amount of sugar Luka needs, he will need $6\cdot4=\boxed{24}$ cups of water.

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_2

Four friends do yardwork for their neighbors over the weekend, earning $$15, $20, $25,$ and $$40,$ respectively. They decide to split their earnings equally among themselves. In total, how much will the friend who earned $$40$ give to the others? $\textbf{(A) }$5 \qquad \textbf{(B) }$10 \qquad \textbf{(C) }$15 \qquad \textbf{(D) }$20 \qquad \textbf{(E) }$25$

The friends earn $$\left(15+20+25+40\right)=$100$ in total. Since they decided to split their earnings equally, it follows that each person will get $$\left(\frac{100}{4}\right)=$25$ . Since the friend who earned $$40$ will need to leave with $$25$ , he will have to give $$\left(40-25\right)=\boxed{15}$ to the others.

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_3

Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest? $\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840$

Note that the unit of the answer is strawberries , which is the product of By conversion factors, we have \[\left(6 \ \color{red}\cancel{\mathrm{ft}}\color{black}\cdot8 \ \color{red}\cancel{\mathrm{ft}}\color{black}\right)\cdot\left(4 \ \frac{\color{green}\cancel{\mathrm{plants}}}{\color{red}\cancel{\mathrm{ft}^2}}\right)\cdot\left(10 \ \frac{\mathrm{strawberries}}{\color{green}\cancel{\mathrm{plant}}}\right)=6\cdot8\cdot4\cdot10 \ \mathrm{strawberries}=\boxed{1920}.\]

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_3

Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest? $\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840$

The area of the garden is $6 \cdot 8 = 48$ square feet. Since Carrie plants $4$ strawberry plants per square foot, there are a total of $48 \cdot 4=192$ strawberry plants, each of which produces $10$ strawberries on average. Accordingly, she can expect to harvest $192 \cdot 10 = \boxed{1920}$ strawberries.

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4

Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon? [asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); for (int i=0; i<7; ++i) { A[i] = rotate(60*i)*(1,0);} path hex = A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--cycle; fill(p,grey1); draw(scale(1.25)*hex,black+linewidth(1.25)); pair S = 6A[0]+2A[1]; fill(shift(S)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(S+2*A[i])*p,grey2);} draw(shift(S)*scale(3.25)*hex,black+linewidth(1.25)); pair T = 16A[0]+4A[1]; fill(shift(T)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(T+2*A[i])*p,grey2); fill(shift(T+4*A[i])*p,grey1); fill(shift(T+2*A[i]+2*A[i+1])*p,grey1); } draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25)); [/asy] $\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49$

Looking at the rows of each hexagon, we see that the first hexagon has $1$ dot, the second has $2+3+2$ dots, and the third has $3+4+5+4+3$ dots. Given the way the hexagons are constructed, it is clear that this pattern continues. Hence, the fourth hexagon has $4+5+6+7+6+5+4=\boxed{37}$ dots.

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4

Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon? [asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); for (int i=0; i<7; ++i) { A[i] = rotate(60*i)*(1,0);} path hex = A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--cycle; fill(p,grey1); draw(scale(1.25)*hex,black+linewidth(1.25)); pair S = 6A[0]+2A[1]; fill(shift(S)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(S+2*A[i])*p,grey2);} draw(shift(S)*scale(3.25)*hex,black+linewidth(1.25)); pair T = 16A[0]+4A[1]; fill(shift(T)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(T+2*A[i])*p,grey2); fill(shift(T+4*A[i])*p,grey1); fill(shift(T+2*A[i]+2*A[i+1])*p,grey1); } draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25)); [/asy] $\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49$

The dots in the next hexagon have four bands. From innermost to outermost: Together, the answer is $1+6+12+18=\boxed{37}.$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_4

Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon? [asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); for (int i=0; i<7; ++i) { A[i] = rotate(60*i)*(1,0);} path hex = A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--cycle; fill(p,grey1); draw(scale(1.25)*hex,black+linewidth(1.25)); pair S = 6A[0]+2A[1]; fill(shift(S)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(S+2*A[i])*p,grey2);} draw(shift(S)*scale(3.25)*hex,black+linewidth(1.25)); pair T = 16A[0]+4A[1]; fill(shift(T)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(T+2*A[i])*p,grey2); fill(shift(T+4*A[i])*p,grey1); fill(shift(T+2*A[i]+2*A[i+1])*p,grey1); } draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25)); [/asy] $\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49$

The first hexagon has $1$ dot, the second hexagon has $1+6$ dots, the third hexagon has $1+6+12$ dots, and so on. The pattern continues since to go from hexagon $n$ to hexagon $(n+1),$ we add a new band of dots around the outside of the existing ones, with each side of the band having side length $(n+1).$ Thus, the number of dots added is $6(n+1)-6 = 6n$ (we subtract $6$ as each of the corner hexagons in the band is counted as part of two sides.). We therefore predict that the fourth hexagon has $1+6+12+18=\boxed{37}$ dots.

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_5

Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive? $\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25$

Each cup is filled with $\frac{3}{4} \cdot \frac{1}{5} = /frac{3}{20}$ of the amount of juice in the pitcher, so the percentage is $\frac{3}{20} \cdot 100 = \boxed{15}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_5

Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive? $\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25$

The pitcher is $\frac{3}{4}$ full, i.e. $75\%$ full. Therefore each cup receives $\frac{75}{5}=\boxed{15}$ percent of the total capacity.

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_5

Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive? $\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25$

Assume that the pitcher has a total capacity of $100$ ounces. Since it is filled three fourths with pineapple juice, it contains $75$ ounces of pineapple juice, which means that each cup will contain $\frac{75}{5}=15$ ounces of pineapple juice. Since the total capacity of the pitcher was $100$ ounces, it follows that each cup received $15\%$ of the total capacity of the pitcher, yielding $\boxed{15}$ as the answer.

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_7

How many integers between $2020$ and $2400$ have four distinct digits arranged in increasing order? (For example, $2347$ is one integer.) $\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}$

Firstly, observe that the second digit of such a number cannot be $1$ or $2$ , because the digits must be distinct and increasing. The second digit also cannot be $4$ as the number must be less than $2400$ , so it must be $3$ . It remains to choose the latter two digits, which must be $2$ distinct digits from $\left\{4,5,6,7,8,9\right\}$ . That can be done in $\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15$ ways; there is then only $1$ way to order the digits, namely in increasing order. This means the answer is $\boxed{15}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_7

How many integers between $2020$ and $2400$ have four distinct digits arranged in increasing order? (For example, $2347$ is one integer.) $\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}$

As in Solution 1, we find that the first two digits must be $23$ , and the third digit must be at least $4$ . If it is $4$ , then there are $5$ choices for the last digit, namely $5$ $6$ $7$ $8$ , or $9$ . Similarly, if the third digit is $5$ , there are $4$ choices for the last digit, namely $6$ $7$ $8$ , and $9$ ; if $6$ , there are $3$ choices; if $7$ , there are $2$ choices; and if $8$ , there is $1$ choice. It follows that the total number of such integers is $5+4+3+2+1=\boxed{15}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_8

Ricardo has $2020$ coins, some of which are pennies ( $1$ -cent coins) and the rest of which are nickels ( $5$ -cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least amounts of money that Ricardo can have? $\textbf{(A) }\text{806} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}$

Clearly, the amount of money Ricardo has will be maximized when he has the maximum number of nickels. Since he must have at least one penny, the greatest number of nickels he can have is $2019$ , giving a total of $(2019\cdot 5 + 1)$ cents. Analogously, the amount of money he has will be least when he has the greatest number of pennies; as he must have at least one nickel, the greatest number of pennies he can have is also $2019$ , giving him a total of $(2019\cdot 1 + 5)$ cents. Hence the required difference is \[(2019\cdot 5 + 1)-(2019\cdot 1 + 5)=2019\cdot 4-4=4\cdot 2018=\boxed{8072}\]

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_8

Ricardo has $2020$ coins, some of which are pennies ( $1$ -cent coins) and the rest of which are nickels ( $5$ -cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least amounts of money that Ricardo can have? $\textbf{(A) }\text{806} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}$

Suppose Ricardo has $p$ pennies, so then he has $(2020-p)$ nickels. In order to have at least one penny and at least one nickel, we require $p \geq 1$ and $2020 - p \geq 1$ , i.e. $1 \leq p \leq 2019$ . The number of cents he has is $p+5(2020-p) = 10100-4p$ , so the maximum is $10100-4 \cdot 1$ and the minimum is $10100 - 4 \cdot 2019$ , and the difference is therefore \[(10100 - 4\cdot 1) - (10100 - 4\cdot 2019) = 4\cdot 2019 - 4 = 4\cdot 2018 = \boxed{8072}\]

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10

Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this? $\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Let the Aggie, Bumblebee, Steelie, and Tiger, be referred to by $A,B,S,$ and $T$ , respectively. If we ignore the constraint that $S$ and $T$ cannot be next to each other, we get a total of $4!=24$ ways to arrange the 4 marbles. We now simply have to subtract out the number of ways that $S$ and $T$ can be next to each other. If we place $S$ and $T$ next to each other in that order, then there are three places that we can place them, namely in the first two slots, in the second two slots, or in the last two slots (i.e. $ST\square\square, \square ST\square, \square\square ST$ ). However, we could also have placed $S$ and $T$ in the opposite order (i.e. $TS\square\square, \square TS\square, \square\square TS$ ). Thus there are 6 ways of placing $S$ and $T$ directly next to each other. Next, notice that for each of these placements, we have two open slots for placing $A$ and $B$ . Specifically, we can place $A$ in the first open slot and $B$ in the second open slot or switch their order and place $B$ in the first open slot and $A$ in the second open slot. This gives us a total of $6\times 2=12$ ways to place $S$ and $T$ next to each other. Subtracting this from the total number of arrangements gives us $24-12=12$ total arrangements $\implies\boxed{12}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10

Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this? $\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

Let's try complementary counting. There $4!$ ways to arrange the 4 marbles. However, there are $2\cdot3!$ arrangements where Steelie and Tiger are next to each other. (Think about permutations of the element ST, A, and B or TS, A, and B). Thus, \[4!-2\cdot3!=\boxed{12}\]

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10

Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this? $\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

We use complementary counting: we will count the numbers of ways where Steelie and Tiger are together and subtract that from the total count. Treat the Steelie and the Tiger as a "super marble." There are $2!$ ways to arrange Steelie and Tiger within this "super marble." Then there are $3!$ ways to arrange the "super marble" and Zara's two other marbles in a row. Since there are $4!$ ways to arrange the marbles without any restrictions, the answer is given by $4!-2!\cdot 3!=\boxed{12}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_10

Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this? $\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

We will use the following $\textbf{Georgeooga-Harryooga Theorem:}$ The Georgeooga-Harryooga Theorem states that if you have $a$ distinguishable objects and $b$ of them cannot be together, then there are $\frac{(a-b)!(a-b+1)!}{b!}$ ways to arrange the objects. $\textit{Proof. (Created by AoPS user RedFireTruck)}$ Let our group of $a$ objects be represented like so $1$ $2$ $3$ , ..., $a-1$ $a$ . Let the last $b$ objects be the ones we can't have together. Then we can organize our objects like so $\square1\square2\square3\square...\square a-b-1\square a-b\square$ We have $(a-b)!$ ways to arrange the objects in that list. Now we have $a-b+1$ blanks and $b$ other objects so we have $_{a-b+1}P_{b}=\frac{(a-b+1)!}{(a-2b+1)!}$ ways to arrange the objects we can't put together. By fundamental counting principle our answer is $\frac{(a-b)!(a-b+1)!}{(a-2b+1)!}$ Proof by RedFireTruck talk ) 12:09, 1 February 2021 (EST) Back to the problem. By the Georgeooga-Harryooga Theorem , our answer is $\frac{(4-2)!(4-2+1)!}{(4-2\cdot2+1)!}=\boxed{12}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_11

After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds? [asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7}); for (int i = 0; i < 6; ++i) { for (int j = 0; j < 6; ++j) { draw((i,0)--(i,6), grey); draw((0,j)--(6,j), grey); } } for (int i = 1; i <= 6; ++i) { draw((-0.1,i)--(0.1,i),linewidth(1.25)); draw((i,-0.1)--(i,0.1),linewidth(1.25)); label(string(5*i), (i,0), 2*S); label(string(i), (0, i), 2*W); } draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25)); label(rotate(90) * "Distance (miles)", (-0.5,3), W); label("Time (minutes)", (3,-0.5), S); dot("Naomi", (2,6), 3*dir(305)); dot((6,6)); label("Maya", (4.45,3.5)); draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy] $\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24$

Naomi travels $6$ miles in a time of $10$ minutes, which is equivalent to $\dfrac{1}{6}$ of an hour. Since $\text{speed} = \frac{\text{distance}}{\text{time}}$ , her speed is $\frac{6}{\left(\frac{1}{6}\right)} = 36$ mph. By a similar calculation, Maya's speed is $12$ mph, so the answer is $36-12 = \boxed{24}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_11

After school, Maya and Naomi headed to the beach, $6$ miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds? [asy] unitsize(1.25cm); dotfactor = 10; pen shortdashed=linetype(new real[] {2.7,2.7}); for (int i = 0; i < 6; ++i) { for (int j = 0; j < 6; ++j) { draw((i,0)--(i,6), grey); draw((0,j)--(6,j), grey); } } for (int i = 1; i <= 6; ++i) { draw((-0.1,i)--(0.1,i),linewidth(1.25)); draw((i,-0.1)--(i,0.1),linewidth(1.25)); label(string(5*i), (i,0), 2*S); label(string(i), (0, i), 2*W); } draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25)); label(rotate(90) * "Distance (miles)", (-0.5,3), W); label("Time (minutes)", (3,-0.5), S); dot("Naomi", (2,6), 3*dir(305)); dot((6,6)); label("Maya", (4.45,3.5)); draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35)); draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed); [/asy] $\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24$

Naomi's speed of $6$ miles in $10$ minutes is equivalent to $6 \cdot 6 = 36$ miles per hour, while Maya's speed of $6$ miles in $30$ minutes (i.e. half an hour) is equivalent to $6 \cdot 2 = 12$ miles per hour. The difference is consequently $36-12=\boxed{24}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_12

For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\] $\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$

We have $5! = 2 \cdot 3 \cdot 4 \cdot 5$ , and $2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!$ . Therefore, the equation becomes $3 \cdot 4 \cdot 10! = 12 \cdot N!$ , and so $12 \cdot 10! = 12 \cdot N!$ . Cancelling the $12$ s, it is clear that $N=\boxed{10}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_12

For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\] $\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$

Since $5! = 120$ , we obtain $120\cdot 9!=12\cdot N!$ , which becomes $12\cdot 10\cdot 9!=12\cdot N!$ and thus $12 \cdot 10!=12\cdot N!$ . We therefore deduce $N=\boxed{10}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_12

For a positive integer $n$ , the factorial notation $n!$ represents the product of the integers from $n$ to $1$ . What value of $N$ satisfies the following equation? \[5!\cdot 9!=12\cdot N!\] $\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$

We notice that $5! \cdot 9! = (5!)^2 \cdot (9 \cdot 8 \cdot 7 \cdot 6).$ We know that $5! = 120,$ so we have $120(5! \cdot 9 \cdot 8 \cdot 7 \cdot 6) = 12 \cdot N!$ Isolating $N!$ we have $N! = 10 \cdot 5! \cdot 9 \cdot 8 \cdot 7 \cdot 6 \Rightarrow N! = 10! \Rightarrow N = \boxed{10}.$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_13

Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add? $\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

After Jamal adds $x$ purple socks, he has $(18+x)$ purple socks and $6+18+12+x=(36+x)$ total socks. This means the probability of drawing a purple sock is $\frac{18+x}{36+x}$ , so we obtain \[\frac{18+x}{36+x}=\frac{3}{5}\] Since $\frac{18+9}{36+9}=\frac{27}{45}=\frac{3}{5}$ , the answer is $\boxed{9}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_13

Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add? $\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

As in Solution 1, we have the equation $\frac{18+x}{36+x}=\frac{3}{5}$ . Cross-multiplying yields $90+5x=108+3x \Rightarrow 2x=18 \Rightarrow x=9$ . Thus, Jamal added $\boxed{9}$ purple socks.

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_13

Jamal has a drawer containing $6$ green socks, $18$ purple socks, and $12$ orange socks. After adding more purple socks, Jamal noticed that there is now a $60\%$ chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add? $\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24$

$6$ green socks and $12$ orange socks together should be $100\%-60\% = 40\%$ of the new total number of socks, so that new total must be $\frac{6+12}{0.4}= 45$ . Therefore, $45-6-18-12=\boxed{9}$ purple socks were added.

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_14

There are $20$ cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all $20$ cities? [asy] // made by SirCalcsALot size(300); pen shortdashed=linetype(new real[] {6,6}); for (int i = 2000; i < 9000; i = i + 2000) { draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey); label(string(i), (0,i), W); } for (int i = 500; i < 9300; i=i+500) { draw((0,i)--(150,i),linewidth(1.25)); if (i % 2000 == 0) { draw((0,i)--(250,i),linewidth(1.25)); } } int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750}; int data_length = 20; int r = 550; for (int i = 0; i < data_length; ++i) { fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey); draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)); } draw((0,4750)--(11450,4750),shortdashed); label("Cities", (11450*0.5,0), S); label(rotate(90)*"Population", (0,9000*0.5), 10*W); // axis draw((0,0)--(0,9300), linewidth(1.25)); draw((0,0)--(11550,0), linewidth(1.25)); [/asy] $\textbf{(A) }65000 \qquad \textbf{(B) }75000 \qquad \textbf{(C) }85000 \qquad \textbf{(D) }95000 \qquad \textbf{(E) }105000$

We can see that the dotted line is exactly halfway between $4500$ and $5000$ , so it is at $4750$ . As this is the average population of all $20$ cities, the total population is simply $4750 \cdot 20 = \boxed{95000}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15

Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$ $\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

Since $20\% = \frac{1}{5}$ , multiplying the given condition by $5$ shows that $y$ is $15 \cdot 5 = \boxed{75}$ percent of $x$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15

Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$ $\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

Letting $x=100$ (without loss of generality), the condition becomes $0.15\cdot 100 = 0.2\cdot y \Rightarrow 15 = \frac{y}{5} \Rightarrow y=75$ . Clearly, it follows that $y$ is $75\%$ of $x$ , so the answer is $\boxed{75}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15

Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$ $\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

We have $15\%=\frac{3}{20}$ and $20\%=\frac{1}{5}$ , so $\frac{3}{20}x=\frac{1}{5}y$ . Solving for $y$ , we multiply by $5$ to give $y = \frac{15}{20}x = \frac{3}{4}x$ , so the answer is $\boxed{75}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15

Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$ $\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

We are given $0.15x = 0.20y$ , so we may assume without loss of generality that $x=20$ and $y=15$ . This means $\frac{y}{x}=\frac{15}{20}=\frac{75}{100}$ , and thus the answer is $\boxed{75}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15

Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$ $\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

$15\%$ of $x$ is $0.15x$ , and $20\%$ of $y$ is $0.20y$ . We put $0.15x$ and $0.20y$ into an equation, creating $0.15x = 0.20y$ because $0.15x$ equals $0.20y$ . Solving for $y$ , dividing $0.2$ to both sides, we get $y = \frac{15}{20}x = \frac{3}{4}x$ , so the answer is $\boxed{75}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_15

Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$ $\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$

$15\%$ of $x$ can be written as $\frac{15}{100}x$ , or $\frac{15x}{100}$ $20\%$ of $y$ can similarly be written as $\frac{20}{100}y$ , or $\frac{20y}{100}$ . So now, $\frac{15x}{100} = \frac{20y}{100}$ . Using cross-multiplication, we can simplify the equation as: $1500x = 2000y$ . Dividing both sides by $500$ , we get: $3x = 4y$ $\frac{3}{4}$ is the same thing as $75\%$ , so the answer is $\boxed{75}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_16

Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $B?$ [asy] size(200); dotfactor = 10; pair p1 = (-28,0); pair p2 = (-111,213); draw(p1--p2,linewidth(1)); pair p3 = (-160,0); pair p4 = (-244,213); draw(p3--p4,linewidth(1)); pair p5 = (-316,0); pair p6 = (-67,213); draw(p5--p6,linewidth(1)); pair p7 = (0, 68); pair p8 = (-350,10); draw(p7--p8,linewidth(1)); pair p9 = (0, 150); pair p10 = (-350, 62); draw(p9--p10,linewidth(1)); pair A = intersectionpoint(p1--p2, p5--p6); dot("$A$", A, 2*W); pair B = intersectionpoint(p5--p6, p3--p4); dot("$B$", B, 2*WNW); pair C = intersectionpoint(p7--p8, p5--p6); dot("$C$", C, 1.5*NW); pair D = intersectionpoint(p3--p4, p7--p8); dot("$D$", D, 2*NNE); pair EE = intersectionpoint(p1--p2, p7--p8); dot("$E$", EE, 2*NNE); pair F = intersectionpoint(p1--p2, p9--p10); dot("$F$", F, 2*NNE); [/asy] $\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5$

Following the first few steps of Solution 1, we have $2(A+C+D+E+F)+3B=47$ . Because an even number ( $2(A+C+D+E+F)$ ) subtracted from an odd number (47) is always odd, we know that $3B$ is odd, showing that $B$ is odd. Now we know that $B$ is either 1, 3, or 5. If we try $B=1$ , we get $43=47$ which is not true. Testing $B=3$ , we get $45=47$ , which is also not true. Therefore, we have $B = \boxed{5}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_17

How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$ $\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

As in Solution 1, we prime factorize $2020$ as $2^2\cdot 5\cdot 101$ , and we recall the standard formula that the number of positive factors of an integer is found by adding $1$ to each exponent in its prime factorization, and then multiplying these. Thus $2020$ has $(2+1)(1+1)(1+1) = 12$ factors. The only number which has one factor is $1$ . For a number to have exactly two factors, it must be prime, and the only prime factors of $2020$ are $2$ $5$ , and $101$ . For a number to have three factors, it must be a square of a prime (this follows from the standard formula mentioned above), and from the prime factorization, the only square of a prime that is a factor of $2020$ is $4$ . Thus, there are $5$ factors of $2020$ which themselves have $1$ $2$ , or $3$ factors (namely $1$ $2$ $4$ $5$ , and $101$ ), so the number of factors of $2020$ that have more than $3$ factors is $12-5=\boxed{7}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_17

How many positive integer factors of $2020$ have more than $3$ factors? (As an example, $12$ has $6$ factors, namely $1,2,3,4,6,$ and $12.$ $\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

Let $d(n)$ be the number of factors of n. We know by prime factorization that $d(2020) = 12$ . These $12$ numbers can be divided into unordered pairs ${a,b}$ where $ab = 2020$ . Since $d(2020) = d(a)d(b)$ , one of $d(a), d(b)$ has $3$ or less factors and the other has $4$ or more - in to total $6$ factors of $2020$ with more than $3$ factors. However, this argument has exceptions where $a$ and $b$ share a nontrivial common factor, which in this case can only be two. There are two cases - One in which $5$ and $101$ divide the same factor, WLOG assumed to be $a$ , so that $d(a) = 2^3 > 3$ and $d(b) = 2$ , as otherwise. In the other case, $a = 5\cdot2$ and $b = 101\cdot2$ , so that $d(a) = d(b) = 4$ . This adds one factor with more than $3$ factors, so the answer is $\boxed{7}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18

Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$ [asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); [/asy] $\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$

[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$O$", (0,0), 1.25*S); draw((0,0)--(-8,15));[/asy] Let $O$ be the center of the semicircle. The diameter of the semicircle is $9+16+9=34$ , so $OC = 17$ . By symmetry, $O$ is the midpoint of $DA$ , so $OD=OA=\frac{16}{2}= 8$ . By the Pythagorean theorem in right-angled triangle $ODC$ (or $OBA$ ), we have that $CD$ (or $AB$ ) is $\sqrt{17^2-8^2}=15$ . Accordingly, the area of $ABCD$ is $16\cdot 15=\boxed{240}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18

Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$ [asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); [/asy] $\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$

Let the midpoint of segment $FE$ be the origin. Evidently, point $D=(-8,0)$ and $A=(8,0)$ . Since points $C$ and $B$ share $x$ -coordinates with $D$ and $A$ respectively, it suffices to find the $y$ -coordinate of $B$ (which will be the height of the rectangle) and multiply this by $DA$ (which we know is $16$ ). The radius of the semicircle is $\frac{9+16+9}{2} = 17$ , so the whole circle has equation $x^2+y^2=289$ ; as already stated, $B$ has the same $x$ -coordinate as $A$ , i.e. $8$ , so substituting this into the equation shows that $y=\pm15$ . Since $y>0$ at $B$ , the y-coordinate of $B$ is $15$ . Therefore, the answer is $16\cdot 15 = \boxed{240}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18

Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$ [asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); [/asy] $\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$

Draw the other half of the circle as follows: [asy] draw(arc((0,0),17,360,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*SE); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*SW); dot("$E$",(17,0), 1.25*E); dot("$F$",(-17,0), 1.25*W); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); draw((-8,-15)--(-8,0)--(8,0)--(8,-15)--cycle); dot("$B'$",(8,-15), 1.25*S); dot("$C'$",(-8,-15), 1.25*S); [/asy] By the Power of a Point Theorem $FD\cdot DE = CD\cdot C'D$ . By symmetry, $CD = C'D$ . We see that $FD = 9$ and $DE = 16 + 9 = 25$ . Substituting in these values, $9\cdot 25 = CD^2$ , giving $CD^2 = 225$ and $CD = 15$ . The area of the rectangle is therefore $15\cdot 16 = \boxed{240}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_18

Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$ [asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); [/asy] $\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$

[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),S); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); dot("$G$", (0,15), SE); dot("$O$", (0,0), NE); draw((0,0)--(0, 15)); draw((-7.5,15)--(0,0)); [/asy] According to the Pythagorean Theorem and the Vertical Theorem, we can find out that $OG=\sqrt{\left(\frac{2\times9+16}{2}\right)^2 - \left(\frac{16}{2}\right)^2}=15$ . Therefore, the answer is $15\times16=\boxed{240}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_19

A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$ $\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$

After finding out that the last digit must be $5$ , the number is of the form $5\square 5\square 5$ . If the unknown digit is $x$ , we can find that one of the solutions to $x$ is $0$ , since $5+5+5$ is equal to $15$ , which is divisible by $3$ . After trying every one digit number, you'll notice that $x$ must be a multiple of $3$ , meaning that $x=0$ $3$ $6$ , or $9$ $50505$ $53535$ $56565$ , and $59595$ are the $\boxed{4}$ solutions to this question.

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_19

A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$ $\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$

assume the number is $ababa$ $10101a+1010b=0 (mod 15)\newline$ $6a+5b=0 (mod 15)\newline$ $a=0 (mod 5)\newline$ $5b=0 (mod 15)\newline$ $b=0 (mod 3)\newline$ Solutions: $(5,0),(5,3),(5,6),(5,9)\newline$ $\boxed{4}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_21

A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are there from $P$ to $Q?$ (The figure shows a sample path.) [asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy] $\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$

Notice that, in order to step onto any particular white square, the marker must have come from one of the $1$ or $2$ white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it(also called the Water Fall Method). To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated). [asy] int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } label("$1$", (5.5, .5)); label("$1$", (4.5, 1.5)); label("$1$", (6.5, 1.5)); label("$1$", (3.5, 2.5)); label("$1$", (7.5, 2.5)); label("$2$", (5.5, 2.5)); label("$1$", (2.5, 3.5)); label("$3$", (6.5, 3.5)); label("$3$", (4.5, 3.5)); label("$4$", (3.5, 4.5)); label("$3$", (7.5, 4.5)); label("$6$", (5.5, 4.5)); label("$10$", (4.5, 5.5)); label("$9$", (6.5, 5.5)); label("$19$", (5.5, 6.5)); label("$9$", (7.5, 6.5)); label("$28$", (6.5, 7.5)); [/asy] The answer is therefore $\boxed{28}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_21

A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are there from $P$ to $Q?$ (The figure shows a sample path.) [asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy] $\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$

Suppose we "extend" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board. [asy] int N = 7; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } draw((8,0) -- (8,8),red); label("$P$", (5.5,.5)); label("$Q$", (6.5,7.5)); label("$X$", (8.5,3.5)); label("$Y$", (8.5,5.5)); [/asy] The total number of paths from $P$ to $Q$ , including invalid paths which cross over the red line, is then the number of paths which make $4$ steps up-and-right and $3$ steps up-and-left, which is $\binom{4+3}{3} = \binom{7}{3} = 35$ . We need to subtract the number of invalid paths, i.e. the number of paths that pass through $X$ or $Y$ . To get to $X$ , the marker has to make $3$ up-and-right steps, after which it can proceed to $Q$ with $3$ steps up-and-left and $1$ step up-and-right. Thus, the number of paths from $P$ to $Q$ that pass through $X$ is $1 \cdot \binom{3+1}{3} = 4$ . Similarly, the number of paths that pass through $Y$ is $\binom{4+1}{1}\cdot 1 = 5$ . However, we have now double-counted the invalid paths which pass through both $X$ and $Y$ ; from the diagram, it is clear that there are only $2$ of these (as the marker can get from $X$ to $Y$ by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is $4+5-2=7$ , and the number of valid paths from $P$ to $Q$ is $35-7 = \boxed{28}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_21

A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are there from $P$ to $Q?$ (The figure shows a sample path.) [asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy] $\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$

On any white square, we may choose to go left or right, as long as we do not cross over the border of the board. Call the moves $L$ and $R$ respectively. Every single legal path consists of $4$ $R's$ and $3$ $L's$ , so now all we have to find is the number of ways to order $4 R's$ and $3 L's$ in any way, which is ${7 \choose 3}=35$ . However, we originally promised that we will not go over the border, and now we have to subtract the paths that do go over the border. The paths that go over the border are any paths that start with RRR(1 path), RR(5 paths) and LRRR(1 path) so our final number of paths is $35-7=\boxed{28}.$ ~PEKKA

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_22

When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below. [asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label("$N$",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label("if $N$ is even",(9.25,1.25),N); label("if $N$ is odd",(9.25,-1.25),N); label("$\frac N2$",(12,1.25)); label("$3N+1$",(12.6,-1.25)); [/asy] For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$ \[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\] When the same $6$ -step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$ \[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\] $\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$

We start with final output of $1$ and work backward, taking cares to consider all possible inputs that could have resulted in any particular output. This produces following set of possibilities each stage: \[\{1\}\rightarrow\{2\}\rightarrow\{4\}\rightarrow\{1,8\}\rightarrow\{2,16\}\rightarrow\{4,5,32\}\rightarrow\{1,8,10,64\}\] where, for example, $2$ must come from $4$ (as there is no integer $n$ satisfying $3n+1=2$ ), but $16$ could come from $32$ or $5$ (as $\frac{32}{2} = 3 \cdot 5 + 1 = 16$ , and $32$ is even while $5$ is odd). By construction, last set in this sequence contains all the numbers which will lead to number $1$ to end of the $6$ -step process, and sum is $1+8+10+64=\boxed{83}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_22

When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below. [asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label("$N$",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label("if $N$ is even",(9.25,1.25),N); label("if $N$ is odd",(9.25,-1.25),N); label("$\frac N2$",(12,1.25)); label("$3N+1$",(12.6,-1.25)); [/asy] For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$ \[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\] When the same $6$ -step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$ \[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\] $\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$

As in Solution 1, we work backwards from $1$ , this time showing the possible cases in a tree diagram: [asy] // Upper branches draw((-6, 1.5)--(-5, 1)--(-3, 1)--(-2,0)--(0, 0)); draw((-6, 0.5)--(-5, 1)); // Lower branches draw((-6, -1.5)--(-5, -1.5)--(-4, -1)--(-3, -1)--(-2, 0)); draw((-6, -0.5)--(-5, -0.5)--(-4, -1)); label("$1$", (0, 0), UnFill(0.1mm)); label("$2$", (-1, 0), UnFill(0.1mm)); label("$4$", (-2, 0), UnFill(0.1mm)); // Upper branches label("$1$", (-3, 1), UnFill(0.1mm)); label("$2$", (-4, 1), UnFill(0.1mm)); label("$4$", (-5, 1), UnFill(0.1mm)); label("$\textbf{8}$", (-6, 1.5), UnFill(0.1mm)); label("$\textbf{1}$", (-6, 0.5), UnFill(0.1mm)); // Lower branches label("$8$", (-3, -1), UnFill(0.1mm)); label("$16$",(-4, -1), UnFill(0.1mm)); label("$5$", (-5, -0.5), UnFill(0.1mm)); label("$32$", (-5, -1.5), UnFill(0.1mm)); label("$\textbf{10}$", (-6, -0.5), UnFill(0.1mm)); label("$\textbf{64}$", (-6, -1.5), UnFill(0.1mm)); [/asy] The possible numbers are those at the "leaves" of the tree (the ends of the various branches), which are $1$ $8$ $64$ , and $10$ . Thus the answer is $1+8+64+10=\boxed{83}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_22

When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below. [asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label("$N$",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label("if $N$ is even",(9.25,1.25),N); label("if $N$ is odd",(9.25,-1.25),N); label("$\frac N2$",(12,1.25)); label("$3N+1$",(12.6,-1.25)); [/asy] For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$ \[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\] When the same $6$ -step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$ \[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\] $\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$

We begin by finding the inverse of the function that the machine uses. Call the input $I$ and the output $O$ . If $I$ is even, $O=\frac{I}{2}$ , and if $I$ is odd, $O=3I+1$ . We can therefore see that $I=2O$ when $I$ is even and $I=\frac{O-1}{3}$ when $I$ is odd. Therefore, starting with $1$ , if $I$ is even, $I=2$ , and if $I$ is odd, $I=0$ , but the latter is not valid since $0$ is not actually odd. This means that the 2nd-to-last number in the sequence has to be $2$ . Now, substituting $2$ into the inverse formulae, if $I$ is even, $I=4$ (which is indeed even), and if $I$ is odd, $I=\frac{1}{3}$ , which is not an integer. This means the 3rd-to-last number in the sequence has to be $4$ . Substituting in $4$ , if $I$ is even, $I=8$ , but if $I$ is odd, $I=1$ . Both of these are valid solutions, so the 4th-to-last number can be either $1$ or $8$ . If it is $1$ , then by the argument we have just made, the 5th-to-last number has to be $2$ , the 6th-to-last number has to be $4$ , and the 7th-to-last number, which is the first number, must be either $1$ or $8$ . In this way, we have ultimately found two solutions: $N=1$ and $N=8$ On the other hand, if the 4th-to-last number is $8$ , substituting $8$ into the inverse formulae shows that the 5th-to-last number is either $16$ or $\frac{7}{3}$ , but the latter is not an integer. Substituting $16$ shows that if $I$ is even, $I=32$ , but if I is odd, $I=5$ , and both of these are valid. If the 6th-to-last number is $32$ , then the first number must be $64$ , since $\frac{31}{3}$ is not an integer; if the 6th-to-last number is $5,$ then the first number has to be $10$ , as $\frac{4}{3}$ is not an integer. This means that, in total, there are $4$ solutions for $N$ , specifically, $1$ $8$ $10$ , and $64$ , which sum to $\boxed{83}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_23

Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed? $\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$

Firstly, observe that it is not possible for a single student to receive $4$ or $5$ awards because this would mean that one of the other students receives no awards. Thus, each student must receive either $1$ $2$ , or $3$ awards. If a student receives $3$ awards, then the other two students must each receive $1$ award; if a student receives $2$ awards, then another student must also receive $2$ awards and the remaining student must receive $1$ award. We consider each of these two cases in turn. If a student receives three awards, there are $3$ ways to choose which student this is, and $\binom{5}{3}$ ways to give that student $3$ out of the $5$ awards. Next, there are $2$ students left and $2$ awards to give out, with each student getting one award. There are clearly just $2$ ways to distribute these two awards out, giving $3\cdot\binom{5}{3}\cdot 2=60$ ways to distribute the awards in this case. In the other case, two students receive $2$ awards and one student recieves $1$ award . We know there are $3$ choices for which student gets $1$ award. There are $\binom{3}{1}$ ways to do this. Then, there are $\binom{5}{2}$ ways to give the first student his two awards, leaving $3$ awards yet to distribute. There are then $\binom{3}{2}$ ways to give the second student his $2$ awards. Finally, there is only $1$ student and $1$ award left, so there is only $1$ way to distribute this award. This results in $\binom{5}{2}\cdot\binom{3}{2}\cdot 1\cdot 3 =90$ ways to distribute the awards in this case. Adding the results of these two cases, we get $60+90=\boxed{150}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_23

Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed? $\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$

Upon inspection (specified in the above solution), there are two cases of the distribution of awards to the students: one student gets 3 awards and the other each get 1 award or one student gets 1 award and the other two get 2 awards. In the first case, there are $\binom{3}{1} = 3$ ways to choose the person who gets 3 awards. From here, there are $\binom{5}{3} = 10$ ways to choose the 3 awards from the 5 total awards. Now, one person has $2$ choices for awards and the other has $1$ choice for the award. Thus, the total number of ways to choose awards in this case is $3 \cdot 10 \cdot 2 \cdot 1 = 60$ In the other case, there are $\binom{3}{1} = 3$ ways to choose the person who gets 1 award, and $5$ choices for his/her award. Then, one person has $\binom{4}{2} = 6$ ways to have his/her awards and the other person has $\dbinom{2}{2} = 1$ ways to have his/her awards. This gives $3 \cdot 5 \cdot 6 \cdot 1 = 90$ ways for this case. Adding these cases together, we get $60 + 90 = 150$ ways to distribute the awards, or choice $\boxed{150}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_23

Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed? $\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240$

Without the restriction that each student receives at least one award, we could take each of the awards and choose one of the $3$ students to give it to. This would be $3^5$ ways to distribute the awards in total. Now we need to subtract the cases where at least one student doesn't receive an award. If a student doesn't receive an award, there are $3$ choices for which student that is, so $2^5$ ways of choosing a student to receive each of the awards; in total, $3\cdot32=96$ However, if $2$ students both don't receive an award, then this case would be counted twice among the $96$ , so we need to add back in these cases. Said in other words, $2$ students not receiving an award is equivalent to $1$ student receiving $5$ awards, and there are $3$ choices for whom that student would be. To finish, the total number of ways to distribute the awards is $243 - 96+3$ , or $\boxed{150}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25

Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units? [asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy] $\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$

Let the side length of each square $S_k$ be $s_k$ . Then, from the diagram, we can line up the top horizontal lengths of $S_1$ $S_2$ , and $S_3$ to cover the top side of the large rectangle, so $s_{1}+s_{2}+s_{3}=3322$ . Similarly, the short side of $R_2$ will be $s_1-s_2$ , and lining this up with the left side of $S_3$ to cover the vertical side of the large rectangle gives $s_{1}-s_{2}+s_{3}=2020$ . We subtract the second equation from the first to obtain $2s_{2}=1302$ , and thus $s_{2}=\boxed{651}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25

Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units? [asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy] $\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$

Assuming that the problem is well-posed, it should be true in the particular case where $S_1 \cong S_3$ and $R_1 \cong R_2$ . Let the sum of the side lengths of $S_1$ and $S_3$ be $x$ , and let the length of square $S_2$ be $y$ . We then have the system \[\begin{dcases}x+y =3322 \\x-y=2020\end{dcases}\] which we solve to determine $y=\boxed{651}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25

Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units? [asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy] $\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$

Since, for each pair of rectangles, the side lengths have a sum of $3322$ or $2020$ and a difference of $S_2$ , the answer must be $\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{651}$

https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_25

Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units? [asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy] $\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$

Let the side length of $S_2$ be s, and the shorter side length of $R_1$ and $R_2$ be $r$ . We have [asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); label("$r$",(5.2,5/2)); label("$r$",(3.2,1/2)); label("$s$",(3.2,3/2)); [/asy] From this diagram, it is evident that $r+s+r=2020$ . Also, the side length of $S_1$ and $S_3$ is $r+s$ . Then, $r+s+s+r+s=3322$ . Now, we have 2 systems of equations. \begin{align*}r+s+r &= 2020 \\ r+s+r+s+s &= 3322 \\ \end{align*} We can see an $r+s+r$ in the 2nd equation, so substituting that in gives us $2020+2s=3322 \Rightarrow 2s= 1302 \Rightarrow s=\boxed{651}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_1

Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can, and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how many items will they buy? $\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$

We know that the sandwiches cost $4.50$ dollars. Guessing will bring us to multiplying $4.50$ by 6, which gives us $27.00$ . Since they can spend $30.00$ they have $3$ dollars left. Since sodas cost $1.00$ dollar each, they can buy 3 sodas, which makes them spend $30.00$ Since they bought 6 sandwiches and 3 sodas, they bought a total of $9$ items. Therefore, the answer is $\boxed{9}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_1

Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can, and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how many items will they buy? $\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$

Let $s$ be the number of sandwiches and $d$ be the number of sodas. We have to satisfy the equation of \[4.50s+d=30\] In the question, it states that Ike and Mike buys as many sandwiches as possible. So, we drop the number of sodas for a while. We have: \begin{align*} 4.50s&=30 \\ s&=\frac{30}{4.5} \\ s&=6R3 \end{align*} We don't want a remainder so the maximum number of sandwiches is $6$ . The total money spent is $6\cdot 4.50=27$ . The number of dollar left to spent on sodas is $30-27=3$ dollars. $3$ dollars can buy $3$ sodas leading us to a total of $6+3=9$ items. Hence, the answer is $\boxed{9}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_2

Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle $ABCD$ [asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy] $\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$

We can see that there are $2$ rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is $5$ , so the bigger side is $10$ , if we do $5 \cdot 2 = 10$ . Now we get the sides of the big rectangle being $15$ and $10$ , so the area is $\boxed{150}$ . ~avamarora

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_4

Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$ [asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label("A",(-13,0),W); label("B",(0,5),N); label("C",(13,0),E); label("D",(0,-5),S); [/asy] $\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$

[asy] draw((-12,0)--(0,5)); draw((0,5)--(12,0)); draw((12,0)--(0,-5)); draw((0,-5)--(-12,0)); draw((0,0)--(12,0)); draw((0,0)--(0,5)); draw((0,0)--(-12,0)); draw((0,0)--(0,-5)); dot((-12,0)); dot((0,5)); dot((12,0)); dot((0,-5)); label("A",(-12,0),W); label("B",(0,5),N); label("C",(12,0),E); label("D",(0,-5),S); label("E",(0,0),SW); [/asy] A rhombus has sides of equal length. Because the perimeter of the rhombus is $52$ , each side is $\frac{52}{4}=13$ . In a rhombus, diagonals are perpendicular and bisect each other, which means $\overline{AE}$ $12$ $\overline{EC}$ Consider one of the right triangles: [asy] draw((-12,0)--(0,5)); draw((0,0)--(-12,0)); draw((0,0)--(0,5)); dot((-12,0)); dot((0,5)); label("A",(-12,0),W); label("B",(0,5),N); label("E",(0,0),SE); [/asy] $\overline{AB}$ $13$ , and $\overline{AE}$ $12$ . Using the Pythagorean theorem, we find that $\overline{BE}$ $5$ . You know the Pythagorean triple, (5, 12, 13). Thus the values of the two diagonals are $\overline{AC}$ $24$ and $\overline{BD}$ $10$ . The area of a rhombus is = $\frac{d_1\cdot{d_2}}{2}$ $\frac{24\cdot{10}}{2}$ $120$ $\boxed{120}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_4

Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$ [asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label("A",(-13,0),W); label("B",(0,5),N); label("C",(13,0),E); label("D",(0,-5),S); [/asy] $\textbf{(A) }60\qquad\textbf{(B) }90\qquad\textbf{(C) }105\qquad\textbf{(D) }120\qquad\textbf{(E) }144$

Right off the bat, we can see that the perimeter of the figure is 52. Dividing this by four, we can get that each side is equal to 13. By drawing a line perpendicular to the one given, we can split the figure into four right triangles. 12 (24/2) is equal to the height of one small right triangle, and 13 is the slanted side. Using the Pythagorean theorem we can find that 169 (13 squared) - 144 (12 squared) = 25 (five squared). With this, we can determine that each small right triangle equals 30. Multiplying that by four we can get $\boxed{120}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_7

Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests? $\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$

We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables $x$ and $y$ for the scores on the last two tests. \[\frac{76+94+87+x+y}{5} = 81,\] \[\frac{257+x+y}{5} = 81.\] We can now cross multiply to get rid of the denominator. \[257+x+y = 405,\] \[x+y = 148.\] Now that we have this equation, we will assign $y$ as the lowest score of the two other tests, and so: \[x = 100,\] \[y=48.\] Now we know that the lowest score on the two other tests is $\boxed{48}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_7

Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests? $\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$

Right now, she scored $76, 94,$ and $87$ points, for a total of $257$ points. She wants her average to be $81$ for her $5$ tests, so she needs to score $405$ points in total. This means she needs to score a total of $405-257= 148$ points in her next $2$ tests. Since the maximum score she can get on one of her $2$ tests is $100$ , the least possible score she can get is $\boxed{48}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_7

Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests? $\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }66\qquad\textbf{(D) }70\qquad\textbf{(E) }74$

We can compare each of the scores with the average of $81$ $76$ $\rightarrow$ $-5$ $94$ $\rightarrow$ $+13$ $87$ $\rightarrow$ $+6$ $100$ $\rightarrow$ $+19$ So the last one has to be $-33$ (since all the differences have to sum to $0$ ), which corresponds to $81-33 = \boxed{48}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_8

Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself? $\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$

After Gilda gives $20$ % of the marbles to Pedro, she has $80$ % of the marbles left. If she then gives $10$ % of what's left to Ebony, she has $(0.8*0.9)$ $72$ % of what she had at the beginning. Finally, she gives $25$ % of what's left to her brother, so she has $(0.75*0.72)$ $\boxed{54}$ of what she had in the beginning left.

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_8

Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself? $\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$

Suppose Gilda has 100 marbles. Then, she gives Pedro 20% of 100 = 20, she remains with 80 marbles. Out of 80 marbles, she gives 10% of 80 = 8 to Ebony. Thus, she remains with 72 marbles. Then, she gives 25% of 72 = 18 to Jimmy, finally leaving her with 54. And, $\frac{54}{100}$ =54%= $\boxed{54}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_8

Gilda has a bag of marbles. She gives $20\%$ of them to her friend Pedro. Then Gilda gives $10\%$ of what is left to another friend, Ebony. Finally, Gilda gives $25\%$ of what is now left in the bag to her brother Jimmy. What percentage of her original bag of marbles does Gilda have left for herself? $\textbf{(A) }20\qquad\textbf{(B) }33\frac{1}{3}\qquad\textbf{(C) }38\qquad\textbf{(D) }45\qquad\textbf{(E) }54$

(Only if you have lots of time do it this way) Since she gave away 20% and 10% of what is left and then another 25% of what is actually left, we can do 20+10+25 or 55%. But it is actually going to be a bit more than 55% because 10% of what is left is not 10% of the total amount. So, the only option that is greater than 100% - 55% is $\boxed{54}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_11

The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign language class? $\textbf{(A) }16\qquad\textbf{(B) }53\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$

Let $x$ be the number of students taking both a math and a foreign language class. By P-I-E, we get $70 + 54 - x$ $93$ Solving gives us $x = 31$ But we want the number of students taking only a math class, which is $70 - 31 = 39$ $\boxed{39}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_11

The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign language class? $\textbf{(A) }16\qquad\textbf{(B) }53\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$

We have $70 + 54 = 124$ people taking classes. However, we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that $31$ people took the two classes. To find the amount of people who took only math class, we subtract the people who didn't take only one math class, so we get $70 - 31 = \boxed{39}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_11

The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign language class? $\textbf{(A) }16\qquad\textbf{(B) }53\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$

[asy] draw(circle((-0.5,0),1)); draw(circle((0.5,0),1)); label("$\huge{x}$", (0, 0)); label("$70-x$", (-1, 0)); label("$54-x$", (1, 0)); [/asy] We know that the sum of all three areas is $93$ So, we have: \[93 = 70-x+x+54-x\] \[93 = 70+54-x\] \[93 = 124 - x\] \[-31=-x\] \[x=31\] We are looking for the number of students in only math. This is $70-x$ . Substituting $x$ with $31$ , our answer is $\boxed{39}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_13

palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$ $\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$

Note that the only positive 2-digit palindromes are multiples of 11, namely $11, 22, \ldots, 99$ . Since $N$ is the sum of 2-digit palindromes, $N$ is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so $N=110$ is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as $110=77+22+11$ . Then, $N = 110$ , and the sum of the digits of $N$ is $1+1+0 = \boxed{2}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_13

palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$ $\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$

We already know that two-digit palindromes can only be two-digit multiples of 11; which are: $11, 22, 33, 44, 55, 66, 77, 88,$ and $99$ . Since this is clear, we will need to find out the least multiple of 11 that is not a palindrome. Then, we start counting. $110 \ldots$ Aha! This multiple of 11, 110, not only isn’t a palindrome, but it also is the sum of three distinct two-digit palindromes, for example: 11 + 22 + 77, 22 + 33 + 55, and 44 + 11 + 55! The sum of $N$ ’s digits is $1+1+0 = \boxed{2}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_13

palindrome is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let $N$ be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of $N$ $\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$

As stated above, two-digit palindromes can only be two-digit multiples of 11. We can see that if we add anything that are multiples of 11 together, we will again get a multiple of 11. For instance, $11+22=33$ . Since we know this fact and we are finding the smallest value possible, we can start with the first three-digit multiple of 11 which is $110$ . Since this is not a palindrome and can be the sum of 3 two-digit palindromes (see above solutions for more details), $110$ fits the bill. We can see that the sum of $110$ 's digits is $1+1+0 = \boxed{2}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16

Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip? $\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$

The only option that is easily divisible by $55$ is $110$ , which gives 2 hours of travel. And, the formula is $\frac{15}{30} + \frac{110}{55} = \frac{5}{2}$ And, $\text{Average Speed}$ $\frac{\text{Total Distance}}{\text{Total Time}}$ Thus, $\frac{125}{50} = \frac{5}{2}$ Both are equal and thus our answer is $\boxed{110}.$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16

Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip? $\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$

To calculate the average speed, simply evaluate the total distance over the total time. Let the number of additional miles he has to drive be $x.$ Therefore, the total distance is $15+x$ and the total time (in hours) is \[\frac{15}{30}+\frac{x}{55}=\frac{1}{2}+\frac{x}{55}.\] We can set up the following equation: \[\frac{15+x}{\frac{1}{2}+\frac{x}{55}}=50.\] Simplifying the equation, we get \[15+x=25+\frac{10x}{11}.\] Solving the equation yields $x=110,$ so our answer is $\boxed{110}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16

Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip? $\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$

If he travels $15$ miles at a speed of $30$ miles per hour, he travels for 30 min. Average rate is total distance over total time so $(15+d)/(0.5 + t) = 50$ , where d is the distance left to travel and t is the time to travel that distance. Solve for $d$ to get $d = 10+50t$ . You also know that he has to travel $55$ miles per hour for some time, so $d=55t$ . Plug that in for d to get $55t = 10+50t$ and $t=2$ and since $d=55t$ $d = 2\cdot55 =110$ , the answer is $\boxed{110}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_16

Qiang drives $15$ miles at an average speed of $30$ miles per hour. How many additional miles will he have to drive at $55$ miles per hour to average $50$ miles per hour for the entire trip? $\textbf{(A) }45\qquad\textbf{(B) }62\qquad\textbf{(C) }90\qquad\textbf{(D) }110\qquad\textbf{(E) }135$

Let $h$ be the amount of hours Qiang drives after his first 15 miles. Average speed, which we know is $50$ mph, means total distance over total time. For 15 miles at 30 mph, the time taken is $\frac{1}{2}$ hour, so the total time for this trip would be $\frac{1}{2} + h$ hours. For the total distance, 15 miles are traveled in the first part and $55h$ miles in the second. This gives the following equation: \[\dfrac{15+55h}{\frac{1}{2}+h} = 50.\] Cross multiplying, we get that $15 + 55h = 50h + 25$ , and simple algebra gives $h=2$ . In 2 hours traveling at 55 mph, the distance traveled is $\frac{2 \hspace{0.05 in} \text{hours}}{1} \cdot \frac{55 \hspace{0.05 in} \text{miles}}{1 \hspace{0.05 in} \text{hour}} = 2 \cdot 55 \hspace{0.05 in} \text{miles} = 110 \hspace{0.05 in} \text{miles}$ , which is choice $\boxed{110}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_19

In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams? $\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$

This isn't finished to another. This gives equality, as each team wins once and loses once as well. For a win, we have $3$ points, so a team gets $3\times2=6$ points if they each win a game and lose a game. This case brings a total of $18+6=24$ points. Therefore, we use Case 2 since it brings the greater amount of points, or $\boxed{24}$

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_19

In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams? $\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$

We can name the top three teams as $A$ $B$ , and $C$ . We can see that (respective scores of) $A=B=C$ because these teams have the same points. If we look at the matches that involve the top three teams, we see that there are some duplicates: $AB$ $BC$ , and $AC$ come twice. In order to even out the scores and get the maximum score, we can say that in match $AB$ $A$ and $B$ each win once out of the two games that they play. We can say the same thing for $AC$ and $BC$ . This tells us that each team $A$ $B$ , and $C$ win and lose twice. This gives each team a total of $3 + 3 + 0 + 0 = 6$ points. Now, we need to include the other three teams. We can label these teams as $D$ $E$ , and $F$ . We can write down every match that $A, B,$ or $C$ plays in that we haven't counted yet: $AD$ $AD$ $AE$ $AE$ $AF$ $AF$ $BD$ $BD$ $BE$ $BE$ $BF$ $BF$ $CD$ $CD$ $CE$ $CE$ $CF$ , and $CF$ . We can say $A$ $B$ , and $C$ win each of these in order to obtain the maximum score that $A$ $B$ , and $C$ can have. If $A$ $B$ , and $C$ win all six of their matches, $A$ $B$ , and $C$ will have a score of $18$ $18 + 6$ results in a maximum score of $\boxed{24}$