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

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This can easily be simplified to <math>\log(xy)=210</math>, or <math>xy = 10^{210}</math>.
 
This can easily be simplified to <math>\log(xy)=210</math>, or <math>xy = 10^{210}</math>.
  
<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>.
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<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 <math>2</math> and <math>5</math>, which is <math>210+210 = 420</math>.
 
Multiply by two to get <math>2m +2n</math>, which is <math>840</math>.
 
Multiply by two to get <math>2m +2n</math>, which is <math>840</math>.
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>.
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Then, use the first equation (<math>\log x + 2\log(\gcd(x,y)) = 60</math>) to show that <math>x</math> has to have lower degrees of <math>2</math> and <math>5</math> than <math>y</math>  (you can also test when <math>x>y</math>, which is a contradiction to the restrains you set before). Therefore, <math>\gcd(x,y)=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>.
Factor this into <math>2^{20} \cdot 5^{20}</math>, and add the two 20's, resulting in m, which is 40.
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Factor this into <math>2^{20} \cdot 5^{20}</math>, and add the two 20's, resulting in <math>m</math>, which is <math>40</math>.
Add <math>m</math> to <math>2m + 2n</math> (which is 840) to get <math>40+840 = \boxed{880}</math>.
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Add <math>m</math> to <math>2m + 2n</math> (which is <math>840</math>) to get <math>40+840 = \boxed{880}</math>.
  
==Solution 2 (Crappier Solution)==
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==Solution 2 (Bashier Solution)==
  
 
First simplifying the first and second equations, we get that  
 
First simplifying the first and second equations, we get that  
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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>.
 
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>.
 
==Solution 4 ==
 
==Solution 4 ==
We will use the notation <math>(a, b)</math> for <math>\gcd(a, b)</math> and <math>[a, b]</math> as <math>\text{lcm}(a, b)</math> We can start a similiar way to Solution 1. We have, by logarithm properties <math>\log_{10}{x}+\log_{10}{(x, y)^2}=60</math> or <math>x(x, y)^2=10^{60}</math>. We can do similiar to the second equation and our two equations become <cmath>x(x, y)^2=10^{60}</cmath> <cmath>y[x, y]^2=10^{570}</cmath>Adding the two equations give us <math>xy(x, y)^2[x, y]^2=10^{630}</math>. Since we know that <math>(a, b)\cdot[a, b]=ab</math>, <math>x^3y^3=10^{630}</math>, or <math>xy=10^{210}</math>. We can express x as <math>2^a5^b</math> and y as <math>2^c5^d</math>. Another way to express (x, y) is now <math>2^{min(a, c)}5^{min(b, d)}</math>, and [x, y] is now <math>2^{max(a, c)}5^{max(b, d)}</math>. We know that x<y, and thus, a<c, and b<d. Our equations for lcm and gcd now become <cmath>2^a5^b(2^a5^a)^2=10^60</cmath> or <math>a=b=20</math>. Doing the same for the lcm equation, we have <math>c=d=190</math>, and 190+20=210, satifies xy=210. Thus, <math>3m+2n=3(20+20)+2(190+190)=\boxed{880}</math>.
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We will use the notation <math>(a, b)</math> for <math>\gcd(a, b)</math> and <math>[a, b]</math> as <math>\text{lcm}(a, b)</math>.
 +
We can start with a similar way to Solution 1. We have, by logarithm properties, <math>\log_{10}{x}+\log_{10}{(x, y)^2}=60</math> or <math>x(x, y)^2=10^{60}</math>. We can do something similar to the second equation and our two equations become <cmath>x(x, y)^2=10^{60}</cmath> <cmath>y[x, y]^2=10^{570}</cmath>Adding the two equations gives us <math>xy(x, y)^2[x, y]^2=10^{630}</math>. Since we know that <math>(a, b)\cdot[a, b]=ab</math>, <math>x^3y^3=10^{630}</math>, or <math>xy=10^{210}</math>. We can express <math>x</math> as <math>2^a5^b</math> and <math>y</math> as <math>2^c5^d</math>. Another way to express <math>(x, y)</math> is now <math>2^{min(a, c)}5^{min(b, d)}</math>, and <math>[x, y]</math> is now <math>2^{max(a, c)}5^{max(b, d)}</math>. We know that <math>x<y</math>, and thus, <math>a<c</math>, and <math>b<d</math>. Our equations for <math>lcm</math> and <math>gcd</math> now become <cmath>2^a5^b(2^a5^a)^2=10^{60}</cmath> or <math>a=b=20</math>. Doing the same for the <math>lcm</math> equation, we have <math>c=d=190</math>, and <math>190+20=210</math>, which satisfies <math>xy=210</math>. Thus, <math>3m+2n=3(20+20)+2(190+190)=\boxed{880}</math>.
 
~awsomek
 
~awsomek
  
 
==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}}
 +
 +
[[Category:Intermediate Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:38, 16 October 2020

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 $y$ (you can also test when $x>y$, which is a contradiction to the restrains you set before). Therefore, $\gcd(x,y)=x$. 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 (Bashier 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 with a similar 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 something similar 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 gives 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 $190+20=210$, which satisfies $xy=210$. Thus, $3m+2n=3(20+20)+2(190+190)=\boxed{880}$. ~awsomek

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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