2019 AIME I Problem 7

Attempt Problem 7 of the 2019 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AIME I solutions, or check the answer key.

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7.

There are positive integers xx and yy that satisfy the system of equations log10x+2log10(gcd(x,y))=60\log_{10} x + 2\log_{10}(\gcd(x, y)) = 60 log10y+2log10(lcm(x,y))=570. \begin{aligned} &\log_{10} y \\ &\quad {}+ 2\log_{10}(\operatorname{lcm}(x, y)) = 570. \end{aligned} Let mm be the number of (not necessarily distinct) prime factors in the prime factorization of x,x, and let nn be the number of (not necessarily distinct) prime factors in the prime factorization of y.y. Find 3m+2n.3m + 2n.

Answer: 880
Concepts:greatest common divisorleast common multiplelogarithmprime factorization
Difficulty rating: 2460
Solution:

The equations say xgcd(x,y)2=1060x \cdot \gcd(x,y)^2 = 10^{60} and ylcm(x,y)2=10570,y \cdot \operatorname{lcm}(x,y)^2 = 10^{570}, so xx and yy are products of the primes 22 and 55 only. Fix one of these primes and let aa and bb be its exponents in xx and y.y. Since the gcd takes the smaller exponent and the lcm the larger, a+2min(a,b)=60,b+2max(a,b)=570. \begin{aligned} a + 2\min(a, b) &= 60, \\ b + 2\max(a, b) &= 570. \end{aligned}

If a>b,a \gt b, then a+2b=60a + 2b = 60 and b+2a=570;b + 2a = 570; adding gives a+b=210,a + b = 210, and subtracting gives ab=510,a - b = 510, forcing b<0,b \lt 0, impossible. So ab,a \le b, and the equations become 3a=603a = 60 and 3b=570,3b = 570, giving a=20a = 20 and b=190b = 190 for both primes.

Thus x=220520x = 2^{20} 5^{20} and y=21905190,y = 2^{190} 5^{190}, so m=40,m = 40, n=380,n = 380, and 3m+2n=120+760=880.3m + 2n = 120 + 760 = 880.

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