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 $x<y$ 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 $a<b$ . 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 $3(20+20)+2(190+190)=\boxed{880}$ .

Solution 4

We will use the notation $(a, b)$ for $\gcd(a, b)$ and $[a, b]$ as $\text{lcm}(a, b)$ . We can start a similiar way to Solution 1. We have, by logarithm properties, $\log_{10}{x}+\log_{10}{(x, y)^2}=60$ or $x(x, y)^2=10^{60}$ . We can do similiar to the second equation and our two equations become $x(x, y)^2=10^{60}$ $y[x, y]^2=10^{570}$ Adding the two equations give us $xy(x, y)^2[x, y]^2=10^{630}$ . Since we know that $(a, b)\cdot[a, b]=ab$ , $x^3y^3=10^{630}$ , or $xy=10^{210}$ . We can express x as $2^a5^b$ and y as $2^c5^d$ . Another way to express (x, y) is now $2^{min(a, c)}5^{min(b, d)}$ , and [x, y] is now $2^{max(a, c)}5^{max(b, d)}$ . We know that x<y, and thus, a<c, and b<d. Our equations for lcm and gcd now become $2^a5^b(2^a5^a)^2=10^{60}$ or $a=b=20$ . Doing the same for the lcm equation, we have $c=d=190$ , and 190+20=210, satifies xy=210. Thus, $3m+2n=3(20+20)+2(190+190)=\boxed{880}$ . ~awsomek

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