Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: If $f(x) = ax+b$ and $f^{-1}(x) = bx+a$ with $a$ and $b$ real, what is the value of $a+b$
$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1 \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ 1 \qquad\mathrm{(E)}\ 2$

Solution: Since $f(f^{-1}(x))=x$ , it follows that $a(bx+a)+b=x$ , which implies $abx + a^2 +b = x$ . This equation holds for all values of $x$ only if $ab=1$ and $a^2+b=0$
Then $b = -a^2$ . Substituting into the equation $ab = 1$ , we get $-a^3 = 1$ . Then $a = -1$ , so $b = -1$ , and \[f(x)=-x-1.\] Likewise \[f^{-1}(x)=-x-1.\] These are inverses to one another since \[f(f^{-1}(x))=-(-x-1)-1=x+1-1=x.\] \[f^{-1}(f(x))=-(-x-1)-1=x+1-1=x.\] Therefore $a+b=\boxed{2}$

Answer:  2
Problem: A truncated cone has horizontal bases with radii $18$ and $2$ . A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?
$\mathrm{(A)}\ 6 \qquad\mathrm{(B)}\ 4\sqrt{5} \qquad\mathrm{(C)}\ 9 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 6\sqrt{3}$

Solution: Consider a trapezoid (label it $ABCD$ as follows) cross-section of the truncate cone along a diameter of the bases:
Above, $E,F,$ and $G$ are points of tangency . By the Two Tangent Theorem, $BF = BE = 18$ and $CF = CG = 2$ , so $BC = 20$ . We draw $H$ such that it is the foot of the altitude $\overline{HD}$ to $\overline{AB}$
By the Pythagorean Theorem \[r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12\] Therefore, the answer is $\boxed{6}.$

Answer: A 6
Problem: A truncated cone has horizontal bases with radii $18$ and $2$ . A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?
$\mathrm{(A)}\ 6 \qquad\mathrm{(B)}\ 4\sqrt{5} \qquad\mathrm{(C)}\ 9 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 6\sqrt{3}$

Solution: Create a trapezoid with inscribed circle $O$ exactly like in Solution #1, and extend lines $\overline{AD}$ and $\overline{BC}$ from the solution above and label the point at where they meet $H$ . Because $\frac{\overline{GC}}{\overline{BE}}$ $\frac{1}{9}$ $\frac{\overline{HG}}{\overline{HE}}$ $\frac{1}{9}$ . Let $\overline{HG} = x$ and $\overline{GE} = 8x$
Because these are radii, $\overline{GO} = \overline{OE} = \overline{OF} = 4x$ $\overline{OF} \perp \overline{BH}$ so $\overline{OF}^2 + \overline{FH}^2 = \overline{OH}^2$ . Plugging in, we get $4x^2 + \overline{FH}^2 = 5x^2$ so $\overline{FH} = 3x$ .Triangles $OFH$ and $BEH$ are similar so $\frac{\overline{OF}}{\overline{BE}}  = \frac{\overline{FH}}{\overline{EH}}$ which gives us $\frac{4x}{18} = \frac{3x}{9x}$ . Solving for x, we get \[x = 1.5\] and \[4x =6\] . Thus, the answer is $\boxed{6}$

Answer: A 6
Problem: The square
is a multiplicative magic square. That is, the product of the numbers in each row, column, and diagonal is the same. If all the entries are positive integers, what is the sum of the possible values of $g$
$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 136$

Solution: All the unknown entries can be expressed in terms of $b$ .
Since $100e = beh = ceg = def$ , it follows that $h = \frac{100}{b}, g = \frac{100}{c}$ ,
and $f = \frac{100}{d}$ . Comparing rows $1$ and $3$ then gives $50bc = 2 \cdot \frac{100}{b} \cdot \frac{100}{c}$ ,
from which $c = \frac{20}{b}$ .
Comparing columns $1$ and $3$ gives $50d \cdot \frac{100}{c}= 2c \cdot \frac{100}{d}$ ,
from which $d = \frac{c}{5} = \frac{4}{b}$ .
Finally, $f = 25b, g = 5b$ , and $e = 10$ . All the entries are positive integers
if and only if $b = 1, 2,$ or $4$ . The corresponding values for $g$ are $5, 10,$ and $20$ , and their sum is $\boxed{35}$

Answer: C 35
Problem: The square
is a multiplicative magic square. That is, the product of the numbers in each row, column, and diagonal is the same. If all the entries are positive integers, what is the sum of the possible values of $g$
$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 136$

Solution: We know because this is a multiplicative magic square that each of the following are equal to each other: $100e=ceg=50dg=beh=2cf=50bc=def=2gh$
From this we know that $50dg=2hg$ , thus $h=25d$ .
Thus $beh=be(25d)$ and $be(25d)=100e$ . Thus $b=\frac{4}{d}$ From this we know that $50bc=(50)(\frac{4}{d})(c)=50dg$ . Thus $c=\frac{d^2g}{4}$ .
Now we know from the very beginning that $100e=ceg$ or $100=cg$ or $100=\frac{d^2g}{4}(g)$ or $\frac{d^2g^2}{4}$ . Rearranging the equation $100=\frac{d^2g^2}{4}$ we have $(d^2)(g^2)=400$ or $dg=20$ due to $d$ and $g$ both being positive. Now that $dg=20$ we find all pairs of positive integers that multiply to $20$ . There is $(d,g)= (20,1);(10,2);(5,4);(4,5);(2,10);(1,20)$ . Now we know that $b=\frac{4}{d}$ and b has to be a positive integer. Thus $d$ can only be $1$ $2$ , or $4$ . Thus $g$ can only be $20$ $10$ ,or $5$ . Thus sum of $20+10+5$ $35$ . The answer is $\boxed{35}$

Answer: C 35
Problem: The polynomial $x^3 - 2004 x^2 + mx + n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two.  How many values of $n$ are possible?
$\mathrm{(A)}\ 250,\!000 \qquad\mathrm{(B)}\ 250,\!250 \qquad\mathrm{(C)}\ 250,\!500 \qquad\mathrm{(D)}\ 250,\!750 \qquad\mathrm{(E)}\ 251,\!000$

Solution: Let the roots be $r,s,r + s$ , and let $t = rs$ . Then
and by matching coefficients, $2(r + s) = 2004 \Longrightarrow r + s = 1002$ . Then our polynomial looks like \[x^3 - 2004x^2 + (t + 1002^2)x - 1002t = 0\] and we need the number of possible products $t = rs = r(1002 - r)$ . Because $m=t+1002^2$ is an integer, we also note that $t$ must be an integer.
Since $r > 0$ and $t > 0$ , it follows that $0 < t = r(1002-r) < 501^2 = 251001$ , with the endpoints not achievable because the roots must be distinct and positive. Because neither $r$ nor $1002-r$ can be an integer, there are $251,000 - 500 = \boxed{250,500}$ possible values of $n = -1002t$

Answer: C 250,500
Problem: The polynomial $x^3 - 2004 x^2 + mx + n$ has integer coefficients and three distinct positive zeros. Exactly one of these is an integer, and it is the sum of the other two.  How many values of $n$ are possible?
$\mathrm{(A)}\ 250,\!000 \qquad\mathrm{(B)}\ 250,\!250 \qquad\mathrm{(C)}\ 250,\!500 \qquad\mathrm{(D)}\ 250,\!750 \qquad\mathrm{(E)}\ 251,\!000$

Solution: Letting the roots be $r$ $s$ , and $t$ , where $t = r+s$ , we see that by Vieta's Formula's, $2004 = r+s+t = t + t = 2t$ , and so $t = 1002$ . Therefore, $x-1002$ is a factor of $x^3 - 2004x^2 + mx + n$ . Letting $x = 0$ gives that $1002 \mid n$ because $x - 1002 \mid x^3 - 2004x^2 + mx + n$ . Letting $n = -1002a$ and noting that $x^3 - 2004x^2 + mx + n = (x-1002)(x^2 - bx + a)$ for some $b$ , we see that $b$ is the sum of the roots of $x^2 - bx + a$ $r$ and $s$ , and so $b = 1002$ . Now, we have that $x^2 - 1002x + a$ has roots $r$ and $s$ , and we wish to find the number of possible values of $a$ . By the quadratic formula, we see that \[\frac{1002 \pm \sqrt{1002^2 - 4a}}{2} = 501 \pm \sqrt{501^2 - a}\] are the two values of noninteger positive real numbers $r$ and $s$ , neither of which is equal to $1002$ . This information gives us that $0 < 501^2 - a < 501^2$ , and so since $501^2 - a$ is evidently not a square, we have $501^2 - 1 - 500 = 251,001 - 500 - 1 = \boxed{250,500}$ possible values of $n = 1002a$

Answer: C 250,500
Problem: Given that $2^{2004}$ is a $604$ digit number whose first digit is $1$ , how many elements of the set $S = \{2^0,2^1,2^2,\ldots ,2^{2003}\}$ have a first digit of $4$
$\mathrm{(A)}\ 194 \qquad \mathrm{(B)}\ 195 \qquad \mathrm{(C)}\ 196 \qquad \mathrm{(D)}\ 197 \qquad \mathrm{(E)}\ 198$

Solution: Given $n$ digits, there must be exactly one power of $2$ with $n$ digits such that the first digit is $1$ . Thus $S$ contains $603$ elements with a first digit of $1$ . For each number in the form of $2^k$ such that its first digit is $1$ , then $2^{k+1}$ must either have a first digit of $2$ or $3$ , and $2^{k+2}$ must have a first digit of $4,5,6,7$ . Thus there are also $603$ numbers with first digit $\{2,3\}$ and $603$ numbers with first digit $\{4,5,6,7\}$ . By using complementary counting , there are $2004 - 3 \times 603 = 195$ elements of $S$ with a first digit of $\{8,9\}$ . Now, $2^k$ has a first digit of $\{8,9\}$ if and only if the first digit of $2^{k-1}$ is $4$ , so there are $\boxed{195}$ elements of $S$ with a first digit of $4$

Answer: B 195
Problem: A set $S$ of points in the $xy$ -plane is symmetric about the origin, both coordinate axes, and the line $y=x$ . If $(2,3)$ is in $S$ , what is the smallest number of points in $S$
$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 16$

Solution: If $(2,3)$ is in $S$ , then $(3,2)$ is also, and quickly we see that every point of the form $(\pm 2, \pm 3)$ or $(\pm 3, \pm 2)$ must be in $S$ . Now note that these $8$ points satisfy all of the symmetry conditions. Thus the answer is $\boxed{8}$

Answer: D 8
Problem: Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$ . For how many values of $n$ is $q+r$ divisible by $11$
$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$

Solution: When a $5$ -digit number is divided by $100$ , the first $3$ digits become the quotient, $q$ , and the last $2$ digits become the remainder, $r$
Therefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive.
For each of the $9\cdot10\cdot10=900$ possible values of $q$ , there are at least $\left\lfloor \frac{100}{11} \right\rfloor = 9$ possible values of $r$ such that $q+r \equiv 0\pmod{11}$
Since there is $1$ "extra" possible value of $r$ that is congruent to $0\pmod{11}$ , each of the $\left\lfloor \frac{900}{11} \right\rfloor = 81$ values of $q$ that are congruent to $0\pmod{11}$ have $1$ more possible value of $r$ such that $q+r \equiv 0\pmod{11}$
Therefore, the number of possible values of $n$ such that $q+r \equiv 0\pmod{11}$ is $900\cdot9+81\cdot1=8181 \Rightarrow\boxed{8181}$

Answer: B 8181
Problem: Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$ . For how many values of $n$ is $q+r$ divisible by $11$
$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$

Solution: Notice that $q+r\equiv0\pmod{11}\Rightarrow100q+r\equiv0\pmod{11}$ . This means that any number whose quotient and remainder sum is divisible by 11 must also be divisible by 11. Therefore, there are $\frac{99990-10010}{11}+1=8181$ possible values. The answer is $\boxed{8181}$

Answer: B 8181
Problem: Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$ . For how many values of $n$ is $q+r$ divisible by $11$
$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$

Solution: Let $abcde$ be the five digits of $n$ . Then $q = abc$ and $r = de$ . By the divisibility rules of $11$ $q = a - b + c \pmod{11}$ and $r = -d + e \pmod{11}$ , so $q + r = a - b + c - d + e = abcde = n \pmod{11}$ . Thus, $n$ must be divisble by $11$ . There are $\frac{99990 - 10010}{11} + 1 = 8181$ five-digit multiples of $11$ , so the answer is $\boxed{8181}$

Answer: B 8181
Problem: Objects $A$ and $B$ move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object $A$ starts at $(0,0)$ and each of its steps is either right or up, both equally likely. Object $B$ starts at $(5,7)$ and each of its steps is either to the left or down, both equally likely. Which of the following is closest to the probability that the objects meet?
$\mathrm{(A)} \ 0.10 \qquad \mathrm{(B)} \ 0.15 \qquad \mathrm{(C)} \ 0.20 \qquad \mathrm{(D)} \ 0.25 \qquad \mathrm{(E)} \ 0.30 \qquad$

Solution: We know that the sum of the vertical steps must be equal to $7$ . We also know that they must take $6$ steps each. Since moving vertically or horizontally is equally likely, we can write all the possible paths as a generating function:
\[P(x)=(x+1)^{12}\]
Where we need to extract the $x^5$ coefficient. By the binomial coefficient theorem, that term is $\binom{12}{5}=792$ paths. Since there are also $2^{12}$ paths, we have:
$\frac{792}{2^{12}}=\frac{792}{4096}\approx\frac{800}{4000}\approx\boxed{0.20}$

Answer: C 0.20
Problem: How many perfect squares are divisors of the product $1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!$
$\textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008$

Solution: We want to find the number of perfect square factors in the product of all the factorials of numbers from $1 - 9$ . We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. This comes out to be equal to $2^{30} \cdot 3^{13} \cdot 5^5 \cdot 7^3$ . To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even: $2^{15} \cdot 3^{6}\cdot 5^2 \cdot 7^1$ . To find the total number of possibilities, we add $1$ to each exponent and multiply them all together. This gives us $16 \cdot 7 \cdot 3 \cdot 2 = 672$ $\Rightarrow\boxed{672}$

Answer: B 672
Problem: How many perfect squares are divisors of the product $1! \cdot 2! \cdot 3! \cdot \hdots \cdot 9!$
$\textbf{(A)}\ 504\qquad\textbf{(B)}\ 672\qquad\textbf{(C)}\ 864\qquad\textbf{(D)}\ 936\qquad\textbf{(E)}\ 1008$

Solution: (Explanation of 1)
Factorials up to 9 in product lead to prime factorization $2^{30}$ $*$ $3^{13}$ $*$ $5^5$ $*$ $7^3$ So the number of pairs possible is { $2^{0}$ $,$ $2^{2}$ $,$ ... $2^{30}$ }{ $3^{0}$ $,$ $3^{2}$ $,$ ... $3^{12}$ }{ $5^{0}$ $,$ $5^{2}$ $,$ ... $5^{4}$ }{ $7^{0}$ $,$ $7^{2}$
Which leads to resulting number of pairs = $16*7*3*2$ $=$ $672$ $\Rightarrow\boxed{672}$

Answer: B 672
Problem: If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$
$\mathrm{(A)}\ -2      \qquad \mathrm{(B)}\ 0     \qquad \mathrm{(C)}\ 2      \qquad \mathrm{(D)}\ 3      \qquad \mathrm{(E)}\ 4$

Solution: Using logarithmic rules, we see that
\[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\] \[=2-\left(\log_{a}b+\frac {1}{\log_{a}b}\right)\]
Since $a$ and $b$ are both greater than $1$ , using AM-GM gives that the term in parentheses must be at least $2$ , so the largest possible values is $2-2=0 \Rightarrow \boxed{0}.$

Answer: B 0
Problem: If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$
$\mathrm{(A)}\ -2      \qquad \mathrm{(B)}\ 0     \qquad \mathrm{(C)}\ 2      \qquad \mathrm{(D)}\ 3      \qquad \mathrm{(E)}\ 4$

Solution: Similar to the previous solution, we use our logarithmic rules and start off the same way. \[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\]
However, now, we use a different logarithmic rule stating that $\log_{a}b$ is simply equal to $\frac {\log_{10}b}{\log_{10}a}$ . With this, we can rewrite our previous equation to give us \[2-(\log_{a}b+\log_{b}a) = 2 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right)\]
We can now cross multiply to get that \[2 - \left(\frac {\log_{10}b}{\log_{10}a} + \frac {\log_{10}a}{\log_{10}b}\right) = 2 - \left(\frac {2 \cdot \log_{10}b \cdot \log_{10}a}{\log_{10}b \cdot \log_{10}a}\right)\] Finally, we cancel to get $2-2=0 \Rightarrow \boxed{0}.$

Answer: B 0
Problem: If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$
$\mathrm{(A)}\ -2      \qquad \mathrm{(B)}\ 0     \qquad \mathrm{(C)}\ 2      \qquad \mathrm{(D)}\ 3      \qquad \mathrm{(E)}\ 4$

Solution: By the logarithmic rules, we have $2-(\log_a{b}+\frac{1}{\log_a{b}})$ .
Let $x=\log_a{b}$ . Thus, the expression becomes $2-(x+\frac{1}{x})$ . We want to find the maximum of the function. To do so, we will find its derivative and let it equal to 0. We get: $\frac{d}{dx}\big(2-(x+\frac{1}{x})\big)=\frac{d}{dx}\big(2-x-\frac{1}{x})=-1+x^{-2}=0 \implies \frac{1}{x^2}=1, x^2=1, x=\pm1.$ Since $a\geq b>1, x=-1$ is not a solution. Thus, $x=1$ . Substituting it into the original expression $2-(x+\frac{1}{x})$ , we get $2-(1+\frac{1}{1})=2-2=\boxed{0}$

Answer:  0
Problem: Let $f(x)= \sqrt{ax^2+bx}$ .  For how many real values of $a$ is there at least one positive value of $b$ for which the domain of $f$ and the range of $f$ are the same set
$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{infinitely \ many} }$

Solution: If $f(x)=y$ , then squaring both sides of the given equation and subtracting $ax^2$ and $bx$ yields $y^2-ax^2-bx=0$ . Completing the square, we get $(x+\frac{b}{2a})^2-\frac{y^2}{a}=\frac{b^2}{4a^2}$ where $y\geq 0$ . Divide out by $\frac{b^2}{4a^2}$ to put the equation in the standard form of an ellipse or hyperbola (depending on the sign of $a$ ) to get $\frac{(x+\frac{b}{2a})^2}{(\frac{b}{2a})^2}-\frac{y^2}{(\frac{b}{2\sqrt {\pm a}})^2}=1$
Before continuing, it is important to note that because $f(x)=\sqrt{ax^2+bx}=\sqrt{ax(x+\frac{b}{a})}$ $f(x)$ has roots 0 and $-\frac{b}{a}$ . Now, we can use the function we deduced to figure out some of its properties when:
$a>0$ : A semi-hyperbola above or on the x-axis. Therefore, no positive value of $a$ allows the domain and range to be the same set because the range will always be $[0, \infty)$ and the domain will always be undefined on some finite range between some value and zero.
$a=0$ : We must refer back to the original function; this results in a horizontal semiparabola in the first quadrant, which satisfies that the domain and range of the function are equal. Specifically, both sets are $[0, \infty )$ $a=0$ is the only case where this happens.
$a<0$ : A semi-ellipse in quadrant one. Since its roots are 0 and $-\frac{b}{a}$ , its domain must be $[0, -\frac{b}{a}]$ . To make its domain and range equal, the maximum value of the ellipse must then be $-\frac{b}{a}$ . But we have another expression for the maximum value of the ellipse, which is $0+\frac{b}{2\sqrt {\pm a}}$ . Setting these two expressions equal to each other will find us the final value of $a$ that satisfies the question.
$\frac{b}{-a}=\frac{b}{2\sqrt {\pm a}}$
$-a=2\sqrt {\pm a}$
$a^2=\pm 4a$
$a=0,\pm 4$
We already knew 0 was a solution from earlier, so -4 is our only new solution (we already ruled out any positive value of $a$ as a solution, so 4 does not work). Thus there are $\boxed{2}$ values of $a$ that make the domain and range of $f(x)$ the same set.

Answer: C 2
Problem: Penniless Pete's piggy bank has no pennies in it, but it has 100 coins, all nickels,dimes, and quarters, whose total value is $8.35. It does not necessarily contain coins of all three types. What is the difference between the largest and smallest number of dimes that could be in the bank?
$\text {(A) } 0 \qquad \text {(B) } 13 \qquad \text {(C) } 37 \qquad \text {(D) } 64 \qquad \text {(E) } 83$

Solution: Where $a,b,c$ is the number of nickels, dimes, and quarters, respectively, we can set up two equations:
\[(1)\ 5a+10b+25c=835\ \ \ \ (2)\ a+b+c=100\]
Eliminate $a$ by subtracting $(2)$ from $(1)/5$ to get $b+4c=67$ . Of the integer solutions $(b,c)$ to this equation, the number of dimes $b$ is least in $(3,16)$ and greatest in $(67,0)$ , yielding a difference of $67-3=\boxed{64}$

Answer: D 64
Problem: Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way?
$\text {(A) } 1 \qquad \text {(B) } 2 \qquad \text {(C) } 3 \qquad \text {(D) } 4 \qquad \text {(E) } 5$

Solution: Place the first triangle. Now, we can place the second triangle either adjacent to the first, or with one side between them, for a total of $\boxed{2}$

Answer: B 2
Problem: Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way?
$\text {(A) } 1 \qquad \text {(B) } 2 \qquad \text {(C) } 3 \qquad \text {(D) } 4 \qquad \text {(E) } 5$

Solution: Take ${5 \choose 2}$ to realize there are 10 ways to choose 2 different triangles. Then divide by 5 for each vertex of a pentagon to get $\boxed{2}$

Answer: B 2
Problem: If $\log (xy^3) = 1$ and $\log (x^2y) = 1$ , what is $\log (xy)$
$\mathrm{(A)}\ -\frac 12  \qquad\mathrm{(B)}\ 0  \qquad\mathrm{(C)}\ \frac 12 \qquad\mathrm{(D)}\ \frac 35  \qquad\mathrm{(E)}\ 1$

Solution: We rewrite the logarithms in the problem. \[\log(x) + 3\log(y) = 1\] \[2\log(x) + \log(y) = 1\] \[\log(x) + \log(y) = c\] where $c$ is the desired quantity. Set $u = \log(x)$ and $v = \log(y)$ . Then we have that \[u + 3y = 1 \textbf{(1)}\] \[2u + y = 1 \textbf{(2)}\] \[u + v = c\] . Notice that \[2 \cdot \textbf{(2)} + \textbf{(1)} = 5u + 5v = 3 \implies u + v = \frac{3}{5} \implies c = \boxed{35}\]

Answer: D 35
Problem: Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$

Solution: Substitute $a^cb^d$ into $x$ . We then have $7(a^{5c}b^{5d}) = 11y^{13}$ . Divide both sides by $7$ , and it follows that:
\[(a^{5c}b^{5d}) = \frac{11y^{13}}{7}.\]
Note that because $11$ and $7$ are prime, the minimum value of $x$ must involve factors of $7$ and $11$ only. Thus, we try to look for the lowest power $p$ of $11$ such that $13p + 1 \equiv 0 \pmod{5}$ , so that we can take $11^{13p + 1}$ to the fifth root. Similarly, we want to look for the lowest power $n$ of $7$ such that $13n - 1 \equiv 0 \pmod{5}$ . Again, this allows us to take the fifth root of $7^{13n - 1}$ . Obviously, we want to add $1$ to $13p$ and subtract $1$ from $13n$ because $11^{13p}$ and $7^{13n}$ are multiplied by $11$ and divided by $7$ , respectively. With these conditions satisfied, we can simply multiply $11^{p}$ and $7^{n}$ and substitute this quantity into $y$ to attain our answer.
We can simply look for suitable values for $p$ and $n$ . We find that the lowest $p$ , in this case, would be $3$ because $13(3) + 1 \equiv 0 \pmod{5}$ . Moreover, the lowest $q$ should be $2$ because $13(2) - 1 \equiv 0 \pmod{5}$ . Hence, we can substitute the quantity $11^{3} \cdot 7^{2}$ into $y$ . Doing so gets us:
\[(a^{5c}b^{5d}) = \frac{11(11^{3} \cdot 7^{2})^{13}}{7} = 11^{40} \cdot 7^{25}.\]
Taking the fifth root of both sides, we are left with $a^cb^d = 11^{8} \cdot 7^{5}$ $a + b + c + d = 11 + 7 + 8 + 5 = \boxed{31}$

Answer: B 31
Problem: Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$

Solution: A simpler way to tackle this problem without all that modding is to keep the equation as:
\[7*a^{5b}c^{5d} = 11y^{13}\]
As stated above, $a$ and $c$ must be the factors 7 and 11 in order to keep $x$ at a minimum. Moving all the non-y terms to the left hand side of the equation, we end up with:
\[7^{5b+1}11^{5d-1}=y^{13}\]
The above equation means that $y$ must also contain only the factors 7 and 11 (again, in order to keep $x$ at a minimum), so we end up with:
\[7^{5b+1}11^{5d-1}=7^{13h}11^{13j}\]
$h$ and $j$ are arbitrary variables placed in order to show that $y$ could have more than just one 7 or one 11 as factors)
Since 7 and 11 are prime, we know that $5b+1=13h$ and $5d-1=13j$ . The smallest positive combinations that would work are $b=5,h=2$ and $d=8,j=3$ . Therefore, $a+b+c+d=7+5+11+8=31$ $\boxed{31}$ is correct.

Answer: B 31