# 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.

## Solution 3

Another way to solve this problem solve for x. First, we can divide both sides by 7 to get: $$x^5 = \frac{11y^{13}}{7}$$

Next, we take the fifth root on both sides, which gets us: $$x = \sqrt{\frac{11y^{13}}{7}}$$ $$x = y^2 \cdot \sqrt{\frac{11y^{3}}{7}}$$

Since we know x is a positive integer that we are trying to minimize, we can let y equal the smallest number that will make x an integer. In this case, we let $y = 11^3 \cdot 7^2$(Make sure you see why this makes x the smallest integer possible!), which when plugged in, results in: $$x = (11^3 \cdot 7^2)^2 \cdot \sqrt{\frac{11}{7}\cdot (11^3 \cdot 7^2)^3}$$ $$x = (11^3 \cdot 7^2)^2 \cdot \sqrt{\frac{11}{7}\cdot 11^9 \cdot 7^6}$$ $$x = (11^3 \cdot 7^2)^2 \cdot \sqrt{11^{10} \cdot 7^5}$$ $$x = (11^3 \cdot 7^2)^2 \cdot 11^2 \cdot 7^1$$ $$x = 7^5 \cdot 11^ 8$$

This gets us $a^cb^d = 7^5\cdot 11^8$, so $a + b + c + d = 7 + 5 + 11 + 8 = \boxed{\textbf{(B)}\ 31}$ ~lucaswujc, $\LaTeX$ help from Technodoggo

## Solution 4 (easy)

According to the problem, we have that $x^5$ and $y^{13}$ must be a multiple of both $7$ and $11$. Thus, in their prime factorisation, there must be $7$ and $11$. Thus, we have $a=7$ and $b=11$. Now, let $x=7^c11^d\implies x^5=7^{5c}11^{5d}\implies7x^5=7^{5c+1}11^{5d}$. We can now divide both sides by 11 in our original equation $7x^5=11y^{13}$ to get $y^{13}=7^{5c+1}11^{5d-1}$. As we are only considering integers, we have $$5c+1\equiv0~(\text{mod}~13)\implies5c\equiv12~(\text{mod}~13)$$ and $$5d-1\equiv0~(\text{mod}~13)\implies5d\equiv1~(\text{mod}~13).$$

We can apply brute force to solve for $c$ and $d$ as the numbers aren't big. For the first congruence, we find that $25$ is the smallest number that satisfies, thus $c=5$. For the second congruence, we find that $40$ is the smallest number that satisfies, thus $d=8$. Summarising, we get $a+b+c+d=7+11+5+8=\boxed{\textbf{(D)}~31}$.

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