Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: In the year $2001$ , the United States will host the International Mathematical Olympiad .  Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ .  What is the largest possible value of the sum $I + M + O$
$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671$

Solution: The sum is the highest if two factors are the lowest.
So, $1 \cdot 3 \cdot 667 = 2001$ and $1+3+667=671 \Longrightarrow \boxed{671}$

Answer: E 671
Problem: In the year $2001$ , the United States will host the International Mathematical Olympiad .  Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$ .  What is the largest possible value of the sum $I + M + O$
$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671$

Solution: We see since $2 + 0 + 0 + 1$ is divisible by $3$ , we can eliminate all of the first $4$ answer choices because they are way too small and get $\boxed{671}$ as our final answer.

Answer: E 671
Problem: Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?
$\textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50  \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75$

Solution: We can begin by labeling the number of initial jellybeans $x$ . If she ate $20\%$ of the jellybeans, then $80\%$ is remaining. Hence, after day 1, there are: $0.8 * x$
After day 2, there are: $0.8 * 0.8 * x$ or $0.64x$ jellybeans. $0.64x = 32$ , so $x = \boxed{50}$

Answer: B 50
Problem: Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?
$\textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50  \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75$

Solution: Testing the answers choices out, we see that the answer is $\boxed{50}$

Answer: B 50
Problem: Points $M$ and $N$ are the midpoints of sides $PA$ and $PB$ of $\triangle PAB$ . As $P$ moves along a line that is parallel to side $AB$ , how many of the four quantities listed below change?
(a) the length of the segment $MN$
(b) the perimeter of $\triangle PAB$
(c) the area of $\triangle PAB$
(d) the area of trapezoid $ABNM$
[asy] draw((2,0)--(8,0)--(6,4)--cycle); draw((4,2)--(7,2)); draw((1,4)--(9,4),Arrows); label("$A$",(2,0),SW); label("$B$",(8,0),SE); label("$M$",(4,2),W); label("$N$",(7,2),E); label("$P$",(6,4),N); [/asy]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution: (a) Triangles $ABP$ and $MNP$ are similar, and since $PM=\frac{1}{2}AP$ $MN=\frac{1}{2}AB$
(b) We see the perimeter changes. For example, imagine if P was extremely far to the left.
(c) The area clearly doesn't change, as both the base $AB$ and its corresponding height remain the same.
(d) The bases $AB$ and $MN$ do not change, and neither does the height, so the area of the trapezoid remains the same.
Only $1$ quantity changes, so the correct answer is $\boxed{1}$

Answer: B 1
Problem: The Fibonacci sequence $1,1,2,3,5,8,13,21,\ldots$ starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?
$\textbf{(A)} \ 0  \qquad \textbf{(B)} \ 4  \qquad \textbf{(C)} \ 6  \qquad \textbf{(D)} \ 7  \qquad \textbf{(E)} \ 9$

Solution: Note that any digits other than the units digit will not affect the answer.  So to make computation quicker, we can just look at the Fibonacci sequence in $\bmod{10}$
$1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$
The last digit to appear in the units position of a number in the Fibonacci sequence is $6 \Longrightarrow \boxed{6}$

Answer: C 6
Problem: The sides of a triangle with positive area have lengths $4$ $6$ , and $x$ . The sides of a second triangle with positive area have lengths $4$ $6$ , and $y$ . What is the smallest positive number that is not a possible value of $|x-y|$
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 10$

Solution: Since $6$ and $4$ are fixed sides, the smallest possible side has to be larger than $6-4=2$ and the largest possible side has to be smaller than $6+4=10$ . This gives us the triangle inequality $2<x<10$ and $2<y<10$ $7$ can be attained by letting $x=9.1$ and $y=2.1$ . However, $8=10-2$ cannot be attained. Thus, the answer is $\boxed{8}$

Answer: D 8
Problem: Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
$\textbf{(A)}\ 22 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 119 \qquad\textbf{(D)}\ 180 \qquad\textbf{(E)}\ 231$

Solution: Any two prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate A, B, and D. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is $(13)(17)-(13+17) = 221 - 30 = 191$ . Thus, we can eliminate E. So, the answer must be $\boxed{119}$

Answer: C 119
Problem: Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
$\textbf{(A)}\ 22 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 119 \qquad\textbf{(D)}\ 180 \qquad\textbf{(E)}\ 231$

Solution: Let the two primes be $p$ and $q$ . We wish to obtain the value of $pq-(p+q)$ , or $pq-p-q$ . Using Simon's Favorite Factoring Trick , we can rewrite this expression as $(1-p)(1-q) -1$ or $(p-1)(q-1) -1$ . Noticing that $(13-1)(11-1) - 1 = 120-1 = 119$ , we see that the answer is $\boxed{119}$

Answer: C 119
Problem: Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$

Solution: Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$ .  Thus, the original sequence can be generated from a quadratic function.
If $f(n) = an^2 + bn + c$ , and $f(0) = 1$ $f(1) = 5$ , and $f(2) = 13$ , we get a system of three equations in three variables:
$f(0) = 1$ gives $c = 1$
$f(1) = 5$ gives $a + b + c = 5$
$f(2) = 13$ gives $4a + 2b + c = 13$
Plugging in $c=1$ into the last two equations gives
$a + b = 4$
$4a + 2b = 12$
Dividing the second equation by 2 gives the system:
$a + b = 4$
$2a + b = 6$
Subtracting the first equation from the second gives $a = 2$ , and hence $b = 2$ .  Thus, our quadratic function is:
$f(n) = 2n^2 + 2n + 1$
Calculating the answer to our problem, $f(100) = 20000 + 200 + 1 = 20201$ , which is choice $\boxed{20201}$

Answer: C 20201
Problem: Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$

Solution: We can see that each figure $n$ has a central box and 4 columns of $n$ boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are $\sum_{n=1}^{100} n = 5050$ squares. $4 \cdot 5050 = 20200$ . Adding in the original center box we have $20200 + 1 = \boxed{20201}$

Answer: C 20201
Problem: Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$

Solution: Let $a_n$ be the number of squares in figure $n$ .  We can easily see that \[a_0=4\cdot 0+1\] \[a_1=4\cdot 1+1\] \[a_2=4\cdot 3+1\] \[a_3=4\cdot 6+1.\] See that we multiply the number we are on to the next consecutive number. 
Note that in $a_n$ , the number multiplied by the 4 is the $n$ th triangular number.  Hence, $a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{20201}$ .
~ edited by mathlover66

Answer: C 20201
Problem: Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$

Solution: Let $f_n$ denote the number of unit cubes in a figure. We have \[f_0=1\] \[f_1=5\] \[f_2=13\] \[f_3=25\] \[f_4=41\] \[...\]
Computing the difference between the number of cubes in each figure yields \[4,8,12,16,...\] It is easy to notice that this is an arithmetic sequence, with the first term being $4$ and the difference being $4$ . Let this sequence be $a_n$
From $f_0$ to $f_{100}$ , the sequence will have $100$ terms. Using the arithmetic sum formula yields
\[S_{100}=\frac{100[2\cdot 4+(100-1)4]}{2}\] \[=50(2\cdot 4+99\cdot 4)\] \[=50(101\cdot 4)\] \[=200\cdot 101\] \[=20200\]
So $f_{100}=1+20200=\boxed{20201}$ unit cubes.

Answer: C 20201
Problem: Figures $0$ $1$ $2$ , and $3$ consist of $1$ $5$ $13$ , and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\textbf{(E)}\ 40801$

Solution: Newton’s little formula states that $a_n = A \binom{n-1}{0} + B \binom{n-1}{1} + C \binom{n-1}{2} + \cdots + K \binom{n-1}{m}$ if first term is $a_1$ and $A =$ first difference, $B =$ second difference, and so on. Hence we apply the formula (because we start at term 0, term 100 is $a_{101}$ ): $a_{101} = 5 \binom{100}{0} + 8 \binom{100}{1} + 4 \binom{100}{2} = \boxed{20201}.$ ~Peelybonehead
~clarification by LeonidasTheConquerer

Answer: C 20201
Problem: Let $A$ $M$ , and $C$ be nonnegative integers such that $A+M+C=10$ .  What is the maximum value of $A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$
$\textbf{(A)}\ 49 \qquad\textbf{(B)}\ 59 \qquad\textbf{(C)}\ 69 \qquad\textbf{(D)}\ 79 \qquad\textbf{(E)}\ 89$

Solution: The trick is to realize that the sum $AMC+AM+MC+CA$ is similar to the product $(A+1)(M+1)(C+1)$ . If we multiply $(A+1)(M+1)(C+1)$ , we get \[(A+1)(M+1)(C+1) = AMC + AM + AC + MC + A + M + C + 1.\] We know that $A+M+C=10$ , therefore $(A+1)(M+1)(C+1) = (AMC + AM + MC + CA) + 11$ and \[AMC + AM + MC + CA = (A+1)(M+1)(C+1) - 11.\] Now consider the maximal value of this expression. Suppose that some two of $A$ $M$ , and $C$ differ by at least $2$ .  Then this triple $(A,M,C)$ is not optimal. (To see this, WLOG let $A\geq C+2.$ We can then increase the value of $(A+1)(M+1)(C+1)$ by changing $A \to A-1$ and $C \to C+1$ .)
Therefore the maximum is achieved when $(A,M,C)$ is a rotation of $(3,3,4)$ . The value of $(A+1)(M+1)(C+1)$ in this case is $4\cdot 4\cdot 5=80,$ and thus the maximum of $AMC + AM + MC + CA$ is $80-11 = \boxed{69}.$

Answer: C 69
Problem: Let $A$ $M$ , and $C$ be nonnegative integers such that $A+M+C=10$ .  What is the maximum value of $A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$
$\textbf{(A)}\ 49 \qquad\textbf{(B)}\ 59 \qquad\textbf{(C)}\ 69 \qquad\textbf{(D)}\ 79 \qquad\textbf{(E)}\ 89$

Solution: Notice that if we want to maximize $AMC + AM + MC + AC$ , we want A, M, and C to be as close as possible. For example, if $A = 7, B = 2,$ and $C=1,$ then the expression would have a much smaller value than if we were to substitute $A = 4, B = 5$ , and $C = 1$ . So to make A, B, and C as close together as possible, we divide $\frac{10}{3}$ to get $3$ . Therefore, A must be 3, M must be 3, and C must be 4. $AMC + AM + MC + AC = 36 + 12 + 12 + 9 = 69$ . So the answer is $\boxed{69.}$

Answer: C 69.
Problem: Let $A$ $M$ , and $C$ be nonnegative integers such that $A+M+C=10$ .  What is the maximum value of $A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$
$\textbf{(A)}\ 49 \qquad\textbf{(B)}\ 59 \qquad\textbf{(C)}\ 69 \qquad\textbf{(D)}\ 79 \qquad\textbf{(E)}\ 89$

Solution: According to our knowledge in AM-GM, the closer numbers are, the value of their product is bigger. Assume that $A=B<C$ , we can get the set $A=3,B=3,C=4$ which the answer is $\boxed{69.}$ ~bluesoul

Answer: C 69.
Problem: One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
$\text {(A)}\ 3 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 5 \qquad \text {(D)}\ 6 \qquad \text {(E)}\ 7$

Solution: If there were 4 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be too much coffee. If there were 6 people in the family, and each of them drank exactly the same amount of coffee and milk as Angela, there would be not enough milk. Thus, it has to be $\boxed{5}$

Answer:  5
Problem: One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
$\text {(A)}\ 3 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 5 \qquad \text {(D)}\ 6 \qquad \text {(E)}\ 7$

Solution: The number of ounces that Angela's family drank has to be a multiple of 8, so we can find the right answer by guessing random values for the number of ounces of coffee and milk Angela drank. With Angela drinking 4 ounces of milk and 4 ounces of coffee, we get 40 total ounces Angela's family drank. Dividing that by 8(each person drank 8 ounces) we get 5 members who drank the coffee with milk, or $\boxed{5}$ ~ Mathyguy88

Answer: C 5
Problem: Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\]

Solution: Let $y = \frac{x}{3}$ ; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$ . Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$ , and $z = -\frac{1}{3}, \frac{2}{9}$ . These sum up to $\boxed{19}$

Answer: B 19
Problem: Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\]

Solution: This is quite trivially solved, as $3x = \dfrac{9x}{3}$ , so $P(3x) = P(9x/3) = 81x^2 + 9x + 1 = 7$ $81x^2+9x-6 = 0$ has solutions $-\frac{1}{3}$ and $\frac{2}{9}$ . Adding these yields a solution of $\boxed{19}$

Answer: B 19
Problem: Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\]

Solution: Similar to Solution 1, we have $=81z^2+9z-6=0.$ The answer is the sum of the roots, which by Vieta's Formulas is $-\frac{b}{a}=-\frac{9}{81}=\boxed{19}$

Answer: B 19
Problem: Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\]

Solution: Set $f\left(\frac{x}{3} \right) = x^2+x+1=7$ to get $x^2+x-6=0.$ From either finding the roots (-3 and 2), or using Vieta's formulas, we find the sum of these roots to be $-1.$ Each root of this equation is $9$ times greater than a corresponding root of $f(3z) = 7$ (because $\frac{x}{3} = 3z$ gives $x = 9z$ ), thus the sum of the roots in the equation $f(3z)=7$ is $-\frac{1}{9}$ or $\boxed{19}$

Answer: B 19
Problem: Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$ . Find the sum of all values of $z$ for which $f(3z) = 7$
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\]

Solution: Since we have $f(x/3)$ $f(3z)$ occurs at $x=9z.$ Thus, $f(9z/3) = f(3z) = (9z)^2 + 9z + 1$ . We set this equal to 7:
$81z^2 + 9z +1 = 7 \Longrightarrow 81z^2 + 9z - 6 = 0$ . For any quadratic $ax^2 + bx +c = 0$ , the sum of the roots is $-\frac{b}{a}$ . Thus, the sum of the roots of this equation is $-\frac{9}{81} = \boxed{19}$

Answer: B 19
Problem: Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution: Because $y^3=x^2$ , set $x=a^3$ $y=a^2$ $a\neq 0$ ). Put them in $(y-x)^2=4y^2$ we get $(a^2(a-1))^2=4a^4$ which implies $a^2-2a+1=4$ . Solve the equation to get $a=3$ or $-1$ . Since $x$ and $y$ are positive, $a=3$ and $x+y=3^3+3^2=\boxed{36}$

Answer: D 36
Problem: Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution: Let's take the second equation and square root both sides. This will obtain $y-x = \pm2y$ . Solving the case where $y-x=+2y$ , we'd find that $x=-y$ . This is known to be false because both $x$ and $y$ have to be positive, and $x=-y$ implies that at least one of the variables is not positive. So we instead solve the case where $y-x=-2y$ . This means that $x=3y$ . Inputting this value into the first equation, we find: \[y^3 = (3y)^2\] \[y^3 = 9y^2\] \[y=9\] This means that $x=3y=3(9)=27$ . Therefore, $x+y=9+27=\boxed{36}$

Answer:  36
Problem: Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution: first expand
$(y-x)^2 = 4y^2$
$y^2-2xy+x^2 = 4y^2$
$y^2-2yx+x^2 = 4y^2$
$x^2-2xy-3y^2 = 0$
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
consider 
a=1
b=-2y
c=-3y^2
$x=\frac{2y\pm\sqrt{(-2y)^2-4(1)(-3y^2)}}{2a}$
$x=\frac{2y\pm\sqrt{16y^2}}{2a}$
$\frac{2y+4y^{2}}{2}$ or $\frac{2y-4y^{2}}{2}$
$x=y+2y$ and $x=y-2y$
$x=3y$ and $x=-y$
we can see both x and y will be postive in $x=3y$
now do same as solution 2 \[y^3 = (3y)^2\] \[y^3 = 9y^2\] \[y=9\] This means that $x=3y=3(9)=27$ . Therefore, $x+y=9+27=\boxed{36}$

Answer:  36
Problem: Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution: Since $a^2 = |a|^2$ , we can rewrite the second equation as $(x-y)^2=4y^2$
Let $u=x+y$ . The second equation becomes
\[(u-2y)^2 = 4y^2\] \[u^2 - 4uy = 0\] \[u = 4y\] \[x+y = 4y\] \[x = 3y.\]
Substituting this into the first equation, we have
\[y^3 = (3y)^2,\] so $x = 9$
Hence $x = 27$ and $x + y = \boxed{36}.$

Answer: D 36
Problem: Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$ . What is $x+y$ $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution: We will use the difference of squares in the second equation.
\[(y-x)^2=4y^2\] \[(y-x)^2-(2y)^2=0\] \[(y-x-2y)(y-x+2y)=0\] \[-(x+y)(3y-x)=0\]
Since x and y are positive, x+y is non-zero. Thus, \[3y=x\]
Substituting into the first equation:
\[y^3=x^2\] \[y^3=9y^2\] \[y=9, x=27 \rightarrow x+y=\boxed{36}\]

Answer: D 36
Problem: What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$
$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$

Solution: Remind that $\text{slope}=\dfrac{\Delta y}{\Delta x}=\tan \theta$ where $\theta$ is the angle between the slope and $x$ -axis. $k_1=2=\tan \alpha$ $k_2=\dfrac{1}{3}=\tan \beta$ . The angle formed by the two lines is $\alpha-\beta$ $\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{2-1/3}{1+2\cdot 1/3}=1$ . Therefore, $\alpha-\beta=\boxed{45}$

Answer: C 45