# 2003 AMC 12B Problems/Problem 18

## 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 1

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 $a^cb^d = 11^{8} \cdot 7^{5}$. $a + b + c + d = 11 + 7 + 8 + 5 = \boxed{\textbf{(B)}\ 31}$

## Solution 2

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, $a+b+c+d=7+5+11+8=31$. $\boxed{\textbf{(B)}\ 31}$ is correct.

## See Also

 2003 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

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