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

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The 2019 AIME I takes place on March 13, 2019.
 
 
 
==Problem 7==
 
==Problem 7==
 
There are positive integers <math>x</math> and <math>y</math> that satisfy the system of equations <cmath>\log_{10} x + 2 \log_{10} (\gcd(x,y)) = 60</cmath><cmath>\log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) = 570.</cmath> Let <math>m</math> be the number of (not necessarily distinct) prime factors in the prime factorization of <math>x</math>, and let <math>n</math> be the number of (not necessarily distinct) prime factors in the prime factorization of <math>y</math>. Find <math>3m+2n</math>.
 
There are positive integers <math>x</math> and <math>y</math> that satisfy the system of equations <cmath>\log_{10} x + 2 \log_{10} (\gcd(x,y)) = 60</cmath><cmath>\log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) = 570.</cmath> Let <math>m</math> be the number of (not necessarily distinct) prime factors in the prime factorization of <math>x</math>, and let <math>n</math> be the number of (not necessarily distinct) prime factors in the prime factorization of <math>y</math>. Find <math>3m+2n</math>.

Revision as of 22:41, 14 March 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 $10^{190}$.

See Also

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 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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