2003 AMC 12B Problem 18

Below is the professionally curated solution for Problem 18 of the 2003 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12B solutions, or check the answer key.

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Concepts:prime factorization modular arithmetic

Difficulty rating: 1710

18.

Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^c b^d.$ What is $a + b + c + d?$

$30$

$31$

$32$

$33$

$34$

Solution:

For the minimum $x,$ neither $x$ nor $y$ has prime factors other than $7$ and $11.$ Write $x = 7^c 11^d,$ so $7x^5 = 7^{5c+1} 11^{5d}.$ Writing $y = 7^m 11^n,$ we need $7^{5c+1}11^{5d} = 7^{13m}11^{13n+1}.$

Matching exponents: $5c + 1 \equiv 0 \pmod{13}$ gives the least $c = 5,$ and $5d \equiv 1 \pmod{13}$ gives the least $d = 8.$ So $a = 7,$ $b = 11,$ and $a + b + c + d = 7 + 11 + 5 + 8 = 31.$

Thus, the correct answer is B.

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