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

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(Solution 2 (Crappier Solution))
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Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.
 
Please remember that you should only assume on these math contests because they are timed; this would technically not be a valid solution.
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==Solution 3 (Easy Solution)==
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Let <math>x=10^a</math> and <math>y=10^b</math> and <math>a<b</math>. Then the given equations become <math>3a=60</math> and <math>3b=570</math>. Therefore, <math>x=10^{20}=2^{20}\cdot5^{20}</math> and <math>y=10^{190}=2^{190}\cdot5^{190}</math>. Our answer is <math>3(20+20)+2(190+190)=\boxed{880}</math>.
  
 
==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 10:35, 15 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 $\log x+\log y+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630$. 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 \[\log_{10}(x\cdot y\cdot\text{gcd}\cdot\text{lcm}^2)=630\] 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 $3(20+20)+2(190+190)=\boxed{880}$.

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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