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

(Solution)
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==Solution==
 
==Solution==
One immediately sees that <math>x = 10^{20}</math> and <math>y = 10^{190}</math>, so the answer is <math>3 * 40 + 2 * 380 = \boxed{880}</math> because <math>10^{20} = 2^{20} * 5^{20}</math> and similarly for <math>10^{190}</math>.
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Add the two equations to get that log x+log y+2(log(gcd(x,y))+log(lcm(x,y)))=630.
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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.
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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.
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This can easily be simplified to log xy=210, or xy = 10^210.
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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.
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Multiply by two to get 2m +2n, which is 840.
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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.
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Factor this into 2^20 * 5^20, and add the two 20's, resulting in m, which is 40.
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Add m to 2m + 2n (which is 840) to get 40+840 = 880.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=I|num-b=6|num-a=8}}
 
{{AIME box|year=2019|n=I|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:59, 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

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.

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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