# Difference between revisions of "2019 AIME I Problems/Problem 7"

## Problem 7

There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} 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 1

Add the two equations to get that $\log x+\log y+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630$. Then, we use the theorem $\log a+\log b=\log ab$ to get the equation, $\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630$. Using the theorem that $\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y$, along with the previously mentioned theorem, we can get the equation $3\log(xy)=630$. This can easily be simplified to $\log(xy)=210$, or $xy = 10^{210}$.

$10^{210}$ can be factored into $2^{210} \cdot 5^{210}$, 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 ($\log x + 2\log(\gcd(x,y)) = 60$) to show that x has to have lower degrees of 2 and 5 than y. Therefore, making the $gcd x$. Then, turn the equation into $3\log x = 60$, which yields $\log x = 20$, or $x = 10^{20}$. Factor this into $2^{20} \cdot 5^{20}$, and add the two 20's, resulting in m, which is 40. Add $m$ to $2m + 2n$ (which is 840) to get $40+840 = \boxed{880}$.

## Solution 2 (Crappier Solution)

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 $$\log_{10}(x\cdot y\cdot\text{gcd}^2\cdot\text{lcm}^2)=630$$ 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.

## Solution 3 (Easy Solution)

Let $x=10^a$ and $y=10^b$ and $a. Then the given equations become $3a=60$ and $3b=570$. Therefore, $x=10^{20}=2^{20}\cdot5^{20}$ and $y=10^{190}=2^{190}\cdot5^{190}$. Our answer is $3(20+20)+2(190+190)=\boxed{880}$.