Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

Solution: We can rewrite the expression as \[222,222-(22,222+2,222+222+22+2).\]
We note that the units digit of the addition is $0$ because all the units digits of the five numbers are $2$ and $5*2=10$ , which has a units digit of $0$
Now, we have something with a units digit of $0$ subtracted from $222,222$ . The units digit of this expression is obviously $2$ , and we get $\boxed{2}$ as our answer.

Answer: B 2
Problem: What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

Solution: 222,222-22,222 = 200,000
200,000 - 2,222 = 197778
197778 - 222 = 197556
197556 - 22 = 197534
197534 - 2 = 1957532
So our answer is $\boxed{2}$

Answer: B 2
Problem: What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

Solution: We only care about the unit's digits.
Thus, $2-2$ ends in $0$ $0-2$ ends in $8$ $8-2$ ends in $6$ $6-2$ ends in $4$ , and $4-2$ ends in $\boxed{2}$

Answer: B 2
Problem: What is the ones digit of \[222,222-22,222-2,222-222-22-2?\] $\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

Solution: We just take the units digit of each and subtract, or you can do it this way by adding an extra ten to the first number (so we don't get a negative number): \[12-2-(2+2+2+2)=10-8=2\] Thus, we get the answer $\boxed{2}$

Answer: B 2
Problem: What is the value of this expression in decimal form? \[\frac{44}{11} + \frac{110}{44} + \frac{44}{1100}\] $\textbf{(A) } 6.4\qquad\textbf{(B) } 6.504\qquad\textbf{(C) } 6.54\qquad\textbf{(D) } 6.9\qquad\textbf{(E) } 6.94$

Solution: We see that $\frac{44}{11}$ is $4$ $\frac{110}{44}$ simplifies to $\frac{5}{2}$ , which is $2.5$
and $\frac{44}{1100}$ simplifies to $\frac{1}{25}$ , which is $0.04$
$4+2.5+0.04$ reveals \[\frac{44}{11} + \frac{110}{44} + \frac{44}{1100}\] is $\boxed{6.54}$ . 
~ le_petit_chouetteur from TSMV

Answer: C 6.54
Problem: Four squares of side length $4, 7, 9,$ and $10$ are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units? [asy]  size(150); filldraw((0,0)--(10,0)--(10,10)--(0,10)--cycle,gray(0.7),linewidth(1)); filldraw((0,0)--(9,0)--(9,9)--(0,9)--cycle,white,linewidth(1)); filldraw((0,0)--(7,0)--(7,7)--(0,7)--cycle,gray(0.7),linewidth(1)); filldraw((0,0)--(4,0)--(4,4)--(0,4)--cycle,white,linewidth(1)); draw((11,0)--(11,4),linewidth(1)); draw((11,6)--(11,10),linewidth(1)); label("$10$",(11,5),fontsize(14pt)); draw((10.75,0)--(11.25,0),linewidth(1)); draw((10.75,10)--(11.25,10),linewidth(1)); draw((0,11)--(3,11),linewidth(1)); draw((5,11)--(9,11),linewidth(1)); draw((0,11.25)--(0,10.75),linewidth(1)); draw((9,11.25)--(9,10.75),linewidth(1)); label("$9$",(4,11),fontsize(14pt)); draw((-1,0)--(-1,1),linewidth(1)); draw((-1,3)--(-1,7),linewidth(1)); draw((-1.25,0)--(-0.75,0),linewidth(1)); draw((-1.25,7)--(-0.75,7),linewidth(1)); label("$7$",(-1,2),fontsize(14pt)); draw((0,-1)--(1,-1),linewidth(1)); draw((3,-1)--(4,-1),linewidth(1)); draw((0,-1.25)--(0,-.75),linewidth(1)); draw((4,-1.25)--(4,-.75),linewidth(1)); label("$4$",(2,-1),fontsize(14pt)); [/asy] $\textbf{(A)}\ 42 \qquad \textbf{(B)}\ 45\qquad \textbf{(C)}\ 49\qquad \textbf{(D)}\ 50\qquad \textbf{(E)}\ 52$

Solution: We work inwards. The area of the outer shaded square is the area of the whole square minus the area of the second largest square. The area of the inner shaded region is the area of the third largest square minus the area of the smallest square. The sum of these areas is \[10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{52}\]

Answer: E 52
Problem: Four squares of side length $4, 7, 9,$ and $10$ are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units? [asy]  size(150); filldraw((0,0)--(10,0)--(10,10)--(0,10)--cycle,gray(0.7),linewidth(1)); filldraw((0,0)--(9,0)--(9,9)--(0,9)--cycle,white,linewidth(1)); filldraw((0,0)--(7,0)--(7,7)--(0,7)--cycle,gray(0.7),linewidth(1)); filldraw((0,0)--(4,0)--(4,4)--(0,4)--cycle,white,linewidth(1)); draw((11,0)--(11,4),linewidth(1)); draw((11,6)--(11,10),linewidth(1)); label("$10$",(11,5),fontsize(14pt)); draw((10.75,0)--(11.25,0),linewidth(1)); draw((10.75,10)--(11.25,10),linewidth(1)); draw((0,11)--(3,11),linewidth(1)); draw((5,11)--(9,11),linewidth(1)); draw((0,11.25)--(0,10.75),linewidth(1)); draw((9,11.25)--(9,10.75),linewidth(1)); label("$9$",(4,11),fontsize(14pt)); draw((-1,0)--(-1,1),linewidth(1)); draw((-1,3)--(-1,7),linewidth(1)); draw((-1.25,0)--(-0.75,0),linewidth(1)); draw((-1.25,7)--(-0.75,7),linewidth(1)); label("$7$",(-1,2),fontsize(14pt)); draw((0,-1)--(1,-1),linewidth(1)); draw((3,-1)--(4,-1),linewidth(1)); draw((0,-1.25)--(0,-.75),linewidth(1)); draw((4,-1.25)--(4,-.75),linewidth(1)); label("$4$",(2,-1),fontsize(14pt)); [/asy] $\textbf{(A)}\ 42 \qquad \textbf{(B)}\ 45\qquad \textbf{(C)}\ 49\qquad \textbf{(D)}\ 50\qquad \textbf{(E)}\ 52$

Solution: We can calculate it as the sum of the areas of $2$ smaller trapezoids and $2$ larger trapezoids. \[2\left(\cfrac{(7+4)(7-4)}{2}\right)+2\left(\cfrac{(10+9)(10-9)}{2}\right)=10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = \boxed{52}\]

Answer: E 52
Problem: When Yunji added all the integers from $1$ to $9$ , she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?
$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$

Solution: The sum of the digits from $1-9$ are $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$ . Note that one of the answer choices (that is equal to Yunju's digits) subtracted from her sum of $45$ must equal a square.
Note that $6^2 = 36$ is a very close square to the sum of 45. Checking, we see that $45 - 9 = 36 = 6^2$ works.
Therefore, the missing number is $\boxed{9}$

Answer: E 9
Problem: Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of $6$ . Which of the following integers cannot be the sum of the two numbers?
$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$

Solution: First, figure out all pairs of numbers whose product is 6. Then, using the process of elimination, we can find the following:
$\textbf{(A)}$ is possible: $2\times 3$
$\textbf{(C)}$ is possible: $1\times 6$
$\textbf{(D)}$ is possible: $2\times 6$
$\textbf{(E)}$ is possible: $3\times 6$
The only integer that cannot be the sum is $\boxed{6}.$

Answer: B 6
Problem: $3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?
$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$

Solution: We can eliminate B, C, and D, because they are not $21$ subtracted by any multiple of $4$ . Finally, we see that there is no way to have A, so the solution is $\boxed{5}$

Answer: E 5
Problem: $3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?
$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$

Solution: Let $x$ be the number of $1x1$ tiles. There are $21$ squares and each $2x2$ or $1x4$ tile takes up 4 squares, so $x \equiv 1 \pmod{4}$ , so it is either $1$ or $5$ . Color the columns, starting with red, then blue, and alternating colors, ending with a red column. There are $12$ red squares and $9$ blue squares, but each $2x2$ and $1x4$ shape takes up an equal number of blue and red squares, so there must be $3$ more $1x1$ tiles on red squares than on blue squares, which is impossible if there is just one, so the answer is $\boxed{5}$ , which can easily be confirmed to work.

Answer: E 5
Problem: $3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?
$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$

Solution: Suppose there are $a$ different $2\times 2$ tiles, $b$ different $4\times 1$ tiles and $c$ different $1\times 1$ tiles. Since the areas of these tiles must total up to $21$ (area of the whole grid), we have \[4a + 4b + c = 21.\] Reducing modulo $4$ gives $c\equiv 1\pmod{4}$ , or $c = 1$ or $c = 5$
If $c = 1$ , then $a + b = 5$ . After some testing, there is no valid pair $(a, b)$ that works, so the answer must be $\boxed{5}$ , which can be constructed in many ways.

Answer: E 5
Problem: On Monday Taye has $$2$ . Every day, he either gains $$3$ or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, $3$ days later?
$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$

Solution: How many values could be on the first day? Only $2$ dollars. The second day, you can either add $3$ dollars, or double, so you can have $5$ dollars, or $4$ . For each of these values, you have $2$ values for each. For $5$ dollars, you have $10$ dollars or $8$ , and for $4$ dollars, you have $8$ dollars or $ $7$ . Now, you have $2$ values for each of these. For $10$ dollars, you have $13$ dollars or $20$ , for $8$ dollars, you have $16$ dollars or $11$ , for $8$ dollars, you have $16$ dollars or $11$ , and for $7$ dollars, you have $14$ dollars or $10$
On the final day, there are 11, 11, 16, and 16 repeating, leaving you with $8-2 = \boxed{6}$ different values.

Answer: D 6
Problem: On Monday Taye has $$2$ . Every day, he either gains $$3$ or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, $3$ days later?
$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$

Solution: Continue as in Solution 1 to get $7$ $8$ , or $10$ dollars by the 2nd day. The only way to get the same dollar amount occurring twice by branching (multiply by $2$ or adding $3$ ) from here is if $7+3=10\cdot 2$ or $7+3=8\cdot 2$ which both aren't true. Hence our answer is $3\cdot2=\boxed{6}$

Answer: D 6
Problem: All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?
$\textbf{(A) } 24\qquad\textbf{(B) } 25\qquad\textbf{(C) } 26\qquad\textbf{(D) } 27\qquad\textbf{(E) } 28$

Solution: Since she has half as many red marbles as green, we can call the number of red marbles $x$ , and the number of green marbles $2x$ .
Since she has half as many green marbles as blue, we can call the number of blue marbles $4x$ . 
Adding them up, we have $7x$ marbles. The number of marbles therefore must be a multiple of $7$ . The only possible answer is $\boxed{28}.$

Answer: E 28
Problem: All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?
$\textbf{(A) } 24\qquad\textbf{(B) } 25\qquad\textbf{(C) } 26\qquad\textbf{(D) } 27\qquad\textbf{(E) } 28$

Solution: Suppose Maria has $g$ green marbles and let $t$ be the total number of marbles. She then has $\frac{g}{2}$ red marbles and $2g$ blue marbles. Altogether, Maria has \[g + \frac{g}{2} + 2g = \frac{7g}{2} = t\] marbles, implying that $g = \dfrac{2t}{7},$ so $t$ must be a multiple of $7$ . The only multiple of $7$ is $\boxed{28}.$

Answer: E 28
Problem: In January $1980$ the Mauna Loa Observatory recorded carbon dioxide $(CO2)$ levels of $338$ ppm (parts per million). Over the years the average $CO2$ reading has increased by about $1.515$ ppm each year. What is the expected $CO2$ level in ppm in January $2030$ ? Round your answer to the nearest integer.
$\textbf{(A)}\ 399\qquad \textbf{(B)}\ 414\qquad \textbf{(C)}\ 420\qquad \textbf{(D)}\ 444\qquad \textbf{(E)}\ 459$

Solution: This is a time period of $50$ years, so we can expect the ppm to increase by $50*1.515=75.75~76$ $76+338=\boxed{414}$

Answer: B 414
Problem: In January $1980$ the Mauna Loa Observatory recorded carbon dioxide $(CO2)$ levels of $338$ ppm (parts per million). Over the years the average $CO2$ reading has increased by about $1.515$ ppm each year. What is the expected $CO2$ level in ppm in January $2030$ ? Round your answer to the nearest integer.
$\textbf{(A)}\ 399\qquad \textbf{(B)}\ 414\qquad \textbf{(C)}\ 420\qquad \textbf{(D)}\ 444\qquad \textbf{(E)}\ 459$

Solution: For each year that has passed, the ppm will increase by $1.515$ . In $2030$ , the CO2 would have increased by $50\cdot 1.515 \approx. 76,$ so the total ppm of CO2 will be $76 + 338 = \boxed{414}.$

Answer: B 414
Problem: In January $1980$ the Mauna Loa Observatory recorded carbon dioxide $(CO2)$ levels of $338$ ppm (parts per million). Over the years the average $CO2$ reading has increased by about $1.515$ ppm each year. What is the expected $CO2$ level in ppm in January $2030$ ? Round your answer to the nearest integer.
$\textbf{(A)}\ 399\qquad \textbf{(B)}\ 414\qquad \textbf{(C)}\ 420\qquad \textbf{(D)}\ 444\qquad \textbf{(E)}\ 459$

Solution: 2030 - 1980 = 50 years.
338 + 50 * 1.515 = 338 + 75.75 = 413.75 for 2030 ppm level $=\boxed{414}$

Answer: B 414
Problem: The coordinates of $\triangle ABC$ are $A(5,7)$ $B(11,7)$ , and $C(3,y)$ , with $y>7$ . The area of $\triangle ABC$ is 12. What is the value of $y$
[asy]  draw((3,11)--(11,7)--(5,7)--(3,11));  dot((5,7)); label("$A(5,7)$",(5,7),S);  dot((11,7)); label("$B(11,7)$",(11,7),S);  dot((3,11)); label("$C(3,y)$",(3,11),NW);  [/asy]
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$

Solution: The triangle has base $6,$ which means its height satisfies \[\dfrac{6h}{2}=3h=12.\] This means that $h=4,$ so the answer is $7+4=\boxed{11}$

Answer: D 11
Problem: The coordinates of $\triangle ABC$ are $A(5,7)$ $B(11,7)$ , and $C(3,y)$ , with $y>7$ . The area of $\triangle ABC$ is 12. What is the value of $y$
[asy]  draw((3,11)--(11,7)--(5,7)--(3,11));  dot((5,7)); label("$A(5,7)$",(5,7),S);  dot((11,7)); label("$B(11,7)$",(11,7),S);  dot((3,11)); label("$C(3,y)$",(3,11),NW);  [/asy]
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$

Solution: [asy] size(10cm); draw((5,7)--(11,7)--(3,11)--cycle); draw((3,11)--(3,7)--(5,7),red); draw((3,7.5)--(3.5,7.5)--(3.5,7)); label("$A(5,7)$", (5,7),S); label("$B(11,7)$", (11,7),S); label("$C(3,y)$", (3,11),W); label("$D(3,7)$", (3,7),SW); [/asy] Label point $D(3,7)$ as the point at which $CD\perp DA$ . We now have $[\triangle ABC] = [\triangle BCD] - [\triangle ACD]$ , where the brackets denote areas. On the right hand side, both of these triangles are right, so we can just compute the two sides of each triangle. The two side lengths of $\triangle ACD$ are $y-7$ and $5-3=2$ . The two side lengths of $\triangle BCD$ are $y-7$ and $11-3 = 8.$ Now,
\[[\triangle ABC] = 12  = \frac{1}{2}\cdot (y-7)\cdot 8 - \frac{1}{2}\cdot (y-7)\cdot 2  = 3(y-7)\]
Dividing by $3$ gives $y -7 = 4,$ so $y = \boxed{11}.$

Answer: D 11
Problem: The coordinates of $\triangle ABC$ are $A(5,7)$ $B(11,7)$ , and $C(3,y)$ , with $y>7$ . The area of $\triangle ABC$ is 12. What is the value of $y$
[asy]  draw((3,11)--(11,7)--(5,7)--(3,11));  dot((5,7)); label("$A(5,7)$",(5,7),S);  dot((11,7)); label("$B(11,7)$",(11,7),S);  dot((3,11)); label("$C(3,y)$",(3,11),NW);  [/asy]
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$

Solution: By the Shoelace Theorem, $\triangle ABC$ has area \[\frac{1}{2}|(y \cdot 11 + 7 \cdot 5 + 7 \cdot 3) - (3 \cdot 7 + 11 \cdot 7 + 5 \cdot y)| = \frac{1}{2}|(11y + 56) - (98 + 5y)| = \frac{1}{2}|6y - 42|\] . From the problem, this is equal to $12$ . We now solve for y.
$\frac{1}{2}|6y - 42| = 12$
$|6y-42| = 24$
$6y - 42 = 24$ OR $6y - 42 = -24$
$6y = 66$ OR $6y = 18$
$y = 11$ OR $y = 3$
However, since, as stated in the problem, $y > 7$ , our only valid solution is $\boxed{11}$

Answer: D 11
Problem: The coordinates of $\triangle ABC$ are $A(5,7)$ $B(11,7)$ , and $C(3,y)$ , with $y>7$ . The area of $\triangle ABC$ is 12. What is the value of $y$
[asy]  draw((3,11)--(11,7)--(5,7)--(3,11));  dot((5,7)); label("$A(5,7)$",(5,7),S);  dot((11,7)); label("$B(11,7)$",(11,7),S);  dot((3,11)); label("$C(3,y)$",(3,11),NW);  [/asy]
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$

Solution: As in the figure, the triangle is determined by the vectors $\begin{bmatrix}-2 \\ y-7\end{bmatrix}$ and $\begin{bmatrix}6\\0\end{bmatrix}$ . Recall that the absolute value of the determinant of these vectors is the area of the parallelogram determined by those vectors; the triangle has half the area of that parallelogram. Then we must have that $\frac{1}{2}|\begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix}| = 12 \implies \begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix} = \pm 24$ . Expanding the determinants, we find that $-6(y-7) = 24$ or $-6(y-7) = -24$ . Solving each equation individually, we find that $y = 3$ or $y = 11$ . However, the problem states that $y > 7$ , so the only valid solution is $\boxed{11}$

Answer: D 11
Problem: Rohan keeps a total of 90 guppies in 4 fish tanks.
How many guppies are in the 4th tank?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$

Solution: Let $x$ denote the number of guppies in the first tank.
Then, we have the following for the number of guppies in the rest of the tanks:
The number of guppies in all of the tanks combined is 90, so we can write the equation
$x + x + 1 + x + 1 + 2 + x + 1 + 2 + 3 = 90$
Simplifying the equation gives
$4x + 10 = 90$
Solving the resulting equation gives $x = 20$ , so the number of guppies in the fourth tank is $20 + 1 + 2 + 3 = 26$
Therefore, the correct answer is $\boxed{26}$

Answer: E 26
Problem: Rohan keeps a total of 90 guppies in 4 fish tanks.
How many guppies are in the 4th tank?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$

Solution: Suppose there are no guppies in the first tank.
Then, the number of guppies in the other tanks are $1,3,$ and $6,$ or $10$ guppies in total.
We need to add $90 - 10 = 80$ guppies into $4$ tanks or $20$ guppies in each tank.
So the number of guppies in the fourth tank is $20 + 6 = \boxed{26}.$

Answer: E 26
Problem: Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground?
(For example, one sequence of hops is up-up-down-down-up-down.)
2024-AMC8-q13.png
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$

Solution: Looking at the answer choices, you see that you can list them out. 
Doing this gets you:
$UUDDUD$
$UDUDUD$
$UUUDDD$
$UDUUDD$
$UUDUDD$
Counting all the paths listed above gets you $\boxed{5}$

Answer: B 5
Problem: Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground?
(For example, one sequence of hops is up-up-down-down-up-down.)
2024-AMC8-q13.png
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$

Solution: Any combination can be written as some re-arrangement of $UUUDDD$ . Clearly we must end going down, and start going up, so we need the number of ways to insert 2 $U$ 's and 2 $D$ 's into $U\, \_ \, \_ \, \_ \, \_ \, D$ . There are ${4\choose 2}=6$ ways, but we have to remove the case $UDDUUD$ , giving us $\boxed{5}$

Answer: B 5
Problem: Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground?
(For example, one sequence of hops is up-up-down-down-up-down.)
2024-AMC8-q13.png
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$

Solution: These numbers are clearly the Catalan numbers. Since we have 6 steps, we need the third Catalan number, which is $\boxed{5}$ .
~andliu766

Answer: B 5
Problem: Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground?
(For example, one sequence of hops is up-up-down-down-up-down.)
2024-AMC8-q13.png
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$

Solution: First step is U, last step is D.
After third step we can get only positions 3 or 1.
In the first case there is only one way UUUDDD.
In the second case we have two way to get this position UDU and UUD.
Similarly, we have two way return to position 0 (UDD and DUD).
Therefore, we have $1 + 2 \cdot 2 = \boxed{5}$

Answer: B 5
Problem: Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground?
(For example, one sequence of hops is up-up-down-down-up-down.)
2024-AMC8-q13.png
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$

Solution: We can find the total cases then deduct the ones that don't work.
Let $U$ represent "Up" and $D$ represent "Down". We know that in order to land back at the bottom of the stairs, we must have an equal number of $U$ 's and $D$ 's, therefore six hops means $3$ of each.
The number of ways to arrange $3$ $U$ 's and $3$ $D$ 's is $\dfrac{6!}{(3!)^2}=\dfrac{720}{36}=20$
Case $1$ : Start with $D$
Case $2$ : Start with $UDD$
Case $3$ : Start with $UUDDD$
Case $4$ : Start with $UDUDD$
Case $1$ is asking us how many ways there are to arrange $3$ $U$ 's and $2$ $D$ 's, which is $\dfrac{5!}{3!2!}=\dfrac{120}{12}=10$
Case $2$ is asking us how many ways there are to arrange $2$ $U$ 's and $1$ $D$ , which is $\dfrac{3!}{2!1!}=\dfrac{6}{2}=3$
Case $3$ is asking us how many ways there are to arrange $1$ $U$ , which is $1$
Case $4$ is asking us the same thing as Case $3$ , giving us $1$
Therefore, deducting all cases from $20$ gives $20-10-3-1-1=\boxed{5}$

Answer: B 5
Problem: The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below(not drawn to scale).The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?
2024-AMC8-q14.png
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 30 \qquad \textbf{(D)}\ 31 \qquad \textbf{(E)}\ 32$

Solution: We can simply see that path $A \rightarrow X \rightarrow M \rightarrow Y \rightarrow C \rightarrow Z$ will give us the smallest value. Adding, $5+2+6+5+10 = \boxed{28}$ . This is nice as it’s also the smallest value, solidifying our answer.

Answer:  28