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

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==Solution==
 
==Solution==
−
Add the two equations to get that <math>\log x+\log y+2(\log(gcd(x,y))+\log(lcm(x,y)))=630</math>.
+
Add the two equations to get that <math>\log x+\log y+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630</math>.
−
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.
+
Then, we use the theorem <math>\log a+\log b=\log ab</math> to get the equation, <math>\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630</math>.
−
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.
+
Using the theorem that <math>\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y</math>, along with the previously mentioned theorem, we can get the equation <math>3\log(xy)=630</math>.
−
This can easily be simplified to log xy=210, or xy = 10^210.
+
This can easily be simplified to <math>\log(xy)=210</math>, or <math>xy = 10^{210}</math>.
−
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.
+
<math>10^{210}</math> can be factored into <math>2^{210} \cdot 5^{210}</math>, and <math>m+n</math> equals to the sum of the exponents of 2 and 5, which is <math>210+210 = 420</math>.
−
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.
+
Multiply by two to get <math>2m +2n</math>, which is <math>840</math>.
−
Factor this into 2^20 * 5^20, and add the two 20's, resulting in m, which is 40.
+
Then, use the first equation (<math>\log x + 2\log(\gcd(x,y)) = 60</math>) to show that x has to have lower degrees of 2 and 5 than y. Therefore, making the <math>gcd x</math>. Then, turn the equation into <math>3\log x = 60</math>, which yields <math>\log x = 20</math>, or <math>x = 10^20</math>.
−
Add m to 2m + 2n (which is 840) to get 40+840 = 880.
+
Factor this into <math>2^{20} \cdot 5^{20}</math>, and add the two 20's, resulting in m, which is 40.
 +
Add <math>m</math> to <math>2m + 2n</math> (which is 840) to get <math>40+840 = 880</math>.
  
 
==Solution 2 (Crappier Solution)==
 
==Solution 2 (Crappier Solution)==

Revision as of 01:45, 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 $40+840 = 880$.

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.

See Also

2019 AIME I (Problems • Answer Key • Resources)
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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