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

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

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 12 Problems and Solutions

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2003 AMC 12B Problems/Problem 18

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