# 2019 AIME I Problems/Problem 7

## 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

Add the two equations to get that log x+log y+2(log(gcd(x,y))+log(lcm(x,y)))=630. Then, use the theorem log a+log b=log ab to get the equation log xy+2(log(gcd(x,y))+log(lcm(x,y)))=630. Use the theorem that the product of the gcd and lcm of two numbers equals to the product of the number along with the log a+log b=log ab theorem to get the equation 3log xy=630. This can easily be simplified to log xy=210, or xy = 10^210. 10^210 can be factored into 2^210 * 5^210, and m+n equals to the sum of the exponents of 2 and 5, which is 210+210 = 420. Multiply by two to get 2m +2n, which is 840. Then, use the first equation, which is log x + 2log(gcd(x,y)) = 60, to realize that x has to have a lower degree of 2 and 5 than y, therefore making the gcd x. Then, turn the equation into 3log x = 60, yielding log x = 20, or x = 10^20. Factor this into 2^20 * 5^20, and add the two 20's, resulting in m, which is 40. Add m to 2m + 2n (which is 840) to get 40+840 = 880.

## Solution 2 (Almost same as above but cleaner)

First simplifying the first and second equations, we get that $$\log_{10}(x\cdot\text{gcd}(x,y)^2)=60$$ $$\log_{10}(y\cdot\text{lcm}(x,y)^2)=570$$

Thus, when the two equations are added, we have that $$\log_{10}(x\cdot y\cdot\text{gcd}\cdot\text{lcm}^2)=630$$ When simplified, this equals $$\log_{10}(x^3y^3)=630$$ so this means that $$x^3y^3=10^{630}$$ so $$xy=10^{210}.$$

Now, the following cannot be done on a proof contest but let's (intuitively) assume that $x and $x$ and $y$ are both powers of $10$. This means the first equation would simplify to $$x^3=10^{60}$$ and $$y^3=10^{570}.$$ Therefore, $x=10^{20}$ and $y=10^{190}$ and if we plug these values back, it works! $10^{20}$ has $20\cdot2=40$ total factors and $10^{190}$ has $190\cdot2=380$ so $$3\cdot 40 + 2\cdot 380 = \boxed{880}.$$

Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.

## See Also

 2019 AIME I (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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