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

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==Problem 7==
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==Problem==
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There are positive integers <math>x</math> and <math>y</math> that satisfy the system of equations <cmath>
+
There are positive integers <math>x</math> and <math>y</math> that satisfy the system of equations  
 +
<cmath>
 
\begin{align*}
 
\begin{align*}
 
\log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\
 
\log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\
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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>.
 
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>.
  
−
==Solution==
+
==Solution 1==
−
Add the two equations to get that <math>\log x+\log y+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630</math>.
+
Add the two equations to get that <math>\log x+\log y+2(\log(\gcd(x,y))+2(\log(\text{lcm}(x,y)))=630</math>.
 
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>.
 
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>.
 
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>.
 
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 <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>.
+
<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>.
+
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 we 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.
+
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>.
+
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)==
+
~minor mistake fix by virjoy2001
 +
~minor mistake fix by oralayhan
 +
 
 +
Remark: You can obtain the contradiction by using LTE. If <math>\nu_2{(x)}\geq{\nu_2{(y)}}, \nu_2{(y^2x)}=60</math>. However, <math>\nu_2{(xy)}=210</math> a contradiction. Same goes with taking <math>\nu_5{(x,y)}</math>
 +
 
 +
===Easier Approach to Finish===
 +
After noting that <math>xy=10^{210},</math> notice that we can let <math>x=10^a</math> and <math>y=10^b.</math> Thus, we have from the given equations (1) and (2) respectively, that:
 +
<cmath>a+2a=60</cmath>
 +
<cmath>b+2b=570</cmath>
 +
Solving, we get <math>(a,b)=(20,190).</math> This matches with our constraint that <math>xy=10^{210}</math> (this constraint can actually be rederived by adding the two equations) so we finish from here.
 +
 
 +
<math>x=2^{20}\cdot 5^{20}, y=2^{190}\cdot 5^{190}.</math> Thus, the answer desired is <math>3(20+20)+2(190+190)=880.</math> ~mathboy282 (minor addition by Technodoggo)
 +
 
 +
==Solution 2==
  
 
First simplifying the first and second equations, we get that  
 
First simplifying the first and second equations, we get that  
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==Solution 3 (Easy Solution)==
 
==Solution 3 (Easy Solution)==
 
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 ==
 +
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
 +
 +
==Solution 5==
 +
Let <math>x=d\alpha, y=d\beta, (\alpha, \beta)=1</math>. Simplifying, <math>d^3\alpha=10^{60}, d^3\alpha^2\beta^3=10^{510} \implies \alpha\beta^3 = 10^{510}=2^{510} \cdot 5^{510}</math>. Notice that since <math>\alpha, \beta</math> are coprime, and <math>\alpha < 5^{90}</math>(Prove it yourself !) , <math>\alpha=1, \beta = 10^{170}</math>. Hence, <math>x=10^{20}, y=10^{190}</math> giving the answer <math>\boxed{880}</math>.
 +
 +
(Solution by Prabh1512)
 +
==Solution 6 (Official MAA)==
 +
The two equations are equivalent to <math>x(\gcd(x,y))^2=10^{60}</math> and <math>y(\operatorname{lcm}(x,y))^2=10^{570}</math> respectively. Multiplying corresponding sides of the equations leads to <math>xy(\gcd(x,y)\operatorname{lcm}(x,y))^2=(xy)^3=10^{630}</math>, so <math>xy=10^{210}</math>. It follows that there are nonnegative integers <math>a,\,b,\,c,</math> and <math>d</math> such that <math>(x,y)=(2^a5^b,2^c5^d)</math> with <math>a+c=b+d=210</math>. Furthermore, <cmath>\frac{(\operatorname{lcm}(x,y))^2}{x}=\frac{y(\operatorname{lcm}(x,y))^2}{xy}=\frac{10^{570}}{10^{210}}=10^{360}.</cmath> Thus <math>\max(2a,2c)-a=\max(2b,2d)-b=360.</math> Because neither <math>2a-a</math> nor <math>2b-b</math> can equal <math>360</math> when <math>a+c=b+d=210,</math> it follows that <math>2c-a=2d-b=360</math>. Hence <math>(a,b,c,d)=(20,20,190,190</math>, so the prime factorization of <math>x</math> has <math>20+20=40</math> prime factors, and the prime factorization of <math>y</math> has <math>190+190=380</math> prime factors. The requested sum is <math>3\cdot40+2\cdot380=880.</math>
 +
 +
==Solution 7==
 +
Add the two equations and use the fact that <math>\gcd\left(x,y\right)\cdot\mathrm{lcm}\left(x,y\right)=xy</math> to find that <math>xy=10^{210}</math>. So let <math>x=2^a5^b</math> and <math>y=2^{210-a}5^{210-b}</math> for <math>0\leq a,b\leq210</math>. If <math>a\geq105</math> then the exponent of <math>2</math> in <math>x\cdot\gcd\left(x,y\right)^2=10^{60}</math> is <math>a+2\left(210-a\right)=420-a</math>, so <math>a=360</math>, contradiction. So <math>a<105</math>. Then the exponent of <math>2</math> in <math>x\cdot\gcd\left(x,y\right)^2</math> is <math>a+2a=3a</math>, so <math>a=20</math>. Similarly, <math>b=20</math>. Then <math>3m+2n=3\left(a+b\right)+2\left(420-a-b\right)=\boxed{880}</math> as desired.
 +
 +
~from trumpeter in the AoPS Forums Contest Discussion
 +
 +
==Solution 8==
 +
We can simplify the equations step by step. The first equation simplifies to <math>log(</math>(x)(<math>(gcd(x,y))^2</math>)<math>)=60</math>. The second equation simplifies to log(<math>(y)</math>(<math>(lcm(x,y)^2</math>)<math>)=570</math>. Up to here, we used the exponent and addition log identities.
 +
 +
Now before we move on to the next few simplification steps, we must remember that <math>gcd(a,b)</math>*<math>lcm(a,b)</math>=<math>a*b</math>.
 +
 +
Due to the fact that these log's are in base <math>10</math>, this makes the first and second equation equal to <math>10^{60}</math>, <math>10^{570}</math> respectively. In this step, we switched the log's into exponential form. Now we multiply both equations to get <math>x*y</math>*<math>(xy)^2</math>=(<math>x^3</math>)(<math>y^3</math>)=<math>10^{630}</math>. Now we take the cube root of both sides to get <math>xy=10^{210}</math>.
 +
 +
We've now gotten to the crucial part of this equation. Though this wouldn't pass for full points in a proof-based contest, this is AIME. So, we assume that <math>x<y</math>. We also let <math>x</math>=<math>10^a</math> and <math>y</math>=<math>10^b</math> That means that <math>gcd(x,y)</math> is <math>x</math> and the <math>lcm(x,y)</math> is <math>y</math> due to the fact that we are also assuming that both <math>x,y</math> are <math>10^a</math>, <math>10^b</math> respectively.
 +
 +
If we put our last few insights together into the first and second equation, we see that <math>x</math>=<math>10^{20}</math>. We also see that <math>y</math>=<math>10^{190}</math>. We could check these if wanted (don't worry they work), but if you were very limited on time for this question, just assume these values work and move on.
 +
 +
Now <math>10^{20}</math> factors as <math>2^{20}</math>*<math>5^{20}</math>. This has <math>40</math> prime factors. <math>2</math>, <math>20</math> times and <math>5</math>, <math>20</math> times. <math>10^{190}</math> factors as <math>2^{190}</math>*<math>5^{190}</math>. This has <math>380</math> prime factors. <math>2</math>, <math>190</math> times and <math>5</math>, <math>190</math> times. Now it's just <math>40*3+380*2=880</math> as our final answer.
 +
 +
-Schintalpati
 +
 +
==Video Solution(Pretty Straightforward) ==
 +
https://www.youtube.com/watch?v=NOLk9-A4eDo
 +
Remember to subscribe!
 +
 +
~North America math Contest Go Go Go
  
 
==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]]
 +
[[Category:Intermediate Algebra Problems]]
 +
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:44, 23 October 2024

Problem

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 $\log x+\log y+2(\log(\gcd(x,y))+2(\log(\text{lcm}(x,y)))=630$. 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 we 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 $40+840 = \boxed{880}$.

~minor mistake fix by virjoy2001 ~minor mistake fix by oralayhan

Remark: You can obtain the contradiction by using LTE. If $\nu_2{(x)}\geq{\nu_2{(y)}}, \nu_2{(y^2x)}=60$. However, $\nu_2{(xy)}=210$ a contradiction. Same goes with taking $\nu_5{(x,y)}$

Easier Approach to Finish

After noting that $xy=10^{210},$ notice that we can let $x=10^a$ and $y=10^b.$ Thus, we have from the given equations (1) and (2) respectively, that: \[a+2a=60\] \[b+2b=570\] Solving, we get $(a,b)=(20,190).$ This matches with our constraint that $xy=10^{210}$ (this constraint can actually be rederived by adding the two equations) so we finish from here.

$x=2^{20}\cdot 5^{20}, y=2^{190}\cdot 5^{190}.$ Thus, the answer desired is $3(20+20)+2(190+190)=880.$ ~mathboy282 (minor addition by Technodoggo)

Solution 2

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

Solution 5

Let $x=d\alpha, y=d\beta, (\alpha, \beta)=1$. Simplifying, $d^3\alpha=10^{60}, d^3\alpha^2\beta^3=10^{510} \implies \alpha\beta^3 = 10^{510}=2^{510} \cdot 5^{510}$. Notice that since $\alpha, \beta$ are coprime, and $\alpha < 5^{90}$(Prove it yourself !) , $\alpha=1, \beta = 10^{170}$. Hence, $x=10^{20}, y=10^{190}$ giving the answer $\boxed{880}$.

(Solution by Prabh1512)

Solution 6 (Official MAA)

The two equations are equivalent to $x(\gcd(x,y))^2=10^{60}$ and $y(\operatorname{lcm}(x,y))^2=10^{570}$ respectively. Multiplying corresponding sides of the equations leads to $xy(\gcd(x,y)\operatorname{lcm}(x,y))^2=(xy)^3=10^{630}$, so $xy=10^{210}$. It follows that there are nonnegative integers $a,\,b,\,c,$ and $d$ such that $(x,y)=(2^a5^b,2^c5^d)$ with $a+c=b+d=210$. Furthermore, \[\frac{(\operatorname{lcm}(x,y))^2}{x}=\frac{y(\operatorname{lcm}(x,y))^2}{xy}=\frac{10^{570}}{10^{210}}=10^{360}.\] Thus $\max(2a,2c)-a=\max(2b,2d)-b=360.$ Because neither $2a-a$ nor $2b-b$ can equal $360$ when $a+c=b+d=210,$ it follows that $2c-a=2d-b=360$. Hence $(a,b,c,d)=(20,20,190,190$, so the prime factorization of $x$ has $20+20=40$ prime factors, and the prime factorization of $y$ has $190+190=380$ prime factors. The requested sum is $3\cdot40+2\cdot380=880.$

Solution 7

Add the two equations and use the fact that $\gcd\left(x,y\right)\cdot\mathrm{lcm}\left(x,y\right)=xy$ to find that $xy=10^{210}$. So let $x=2^a5^b$ and $y=2^{210-a}5^{210-b}$ for $0\leq a,b\leq210$. If $a\geq105$ then the exponent of $2$ in $x\cdot\gcd\left(x,y\right)^2=10^{60}$ is $a+2\left(210-a\right)=420-a$, so $a=360$, contradiction. So $a<105$. Then the exponent of $2$ in $x\cdot\gcd\left(x,y\right)^2$ is $a+2a=3a$, so $a=20$. Similarly, $b=20$. Then $3m+2n=3\left(a+b\right)+2\left(420-a-b\right)=\boxed{880}$ as desired.

~from trumpeter in the AoPS Forums Contest Discussion

Solution 8

We can simplify the equations step by step. The first equation simplifies to $log($(x)($(gcd(x,y))^2$)$)=60$. The second equation simplifies to log($(y)$($(lcm(x,y)^2$)$)=570$. Up to here, we used the exponent and addition log identities.

Now before we move on to the next few simplification steps, we must remember that $gcd(a,b)$*$lcm(a,b)$=$a*b$. 

Due to the fact that these log's are in base $10$, this makes the first and second equation equal to $10^{60}$, $10^{570}$ respectively. In this step, we switched the log's into exponential form. Now we multiply both equations to get $x*y$*$(xy)^2$=($x^3$)($y^3$)=$10^{630}$. Now we take the cube root of both sides to get $xy=10^{210}$.

We've now gotten to the crucial part of this equation. Though this wouldn't pass for full points in a proof-based contest, this is AIME. So, we assume that $x<y$. We also let $x$=$10^a$ and $y$=$10^b$ That means that $gcd(x,y)$ is $x$ and the $lcm(x,y)$ is $y$ due to the fact that we are also assuming that both $x,y$ are $10^a$, $10^b$ respectively.

If we put our last few insights together into the first and second equation, we see that $x$=$10^{20}$. We also see that $y$=$10^{190}$. We could check these if wanted (don't worry they work), but if you were very limited on time for this question, just assume these values work and move on.

Now $10^{20}$ factors as $2^{20}$*$5^{20}$. This has $40$ prime factors. $2$, $20$ times and $5$, $20$ times. $10^{190}$ factors as $2^{190}$*$5^{190}$. This has $380$ prime factors. $2$, $190$ times and $5$, $190$ times. Now it's just $40*3+380*2=880$ as our final answer.

-Schintalpati

Video Solution(Pretty Straightforward)

https://www.youtube.com/watch?v=NOLk9-A4eDo Remember to subscribe!

~North America math Contest Go Go Go

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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