2019 AIME I Problems/Problem 7

The 2019 AIME I takes place on March 13, 2019.

Problem 7

There are positive integers $x$ and $y$ that satisfy the system of equations $\log_{10} x + 2 \log_{10} (\gcd(x,y)) = 60$ $\log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) = 570.$ Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$ .

Solution

One immediately sees that $x = 10^{20}$ and $y = 10^{190}$ , so the answer is $3 * 40 + 2 * 380 = \boxed{880}$ because $10^{20} = 2^{20} * 5^{20}$ and similarly for $10^{190}$ .

2019 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2019 AIME I Problems/Problem 7

Problem 7

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