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

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 , this makes the first and second equation equal to ,  respectively. In this step, we switched the log's into exponential form. Now we multiply both equations to get *=()()=. Now we take the cube root of both sides to get .

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

@@ Line 25: / Line 25: @@
 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==
@@ Line 66: / Line 74: @@
 ~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) ==
@@ Line 77: / Line 100: @@
 [[Category:Intermediate Number Theory Problems]]
+[[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

2019 AIME I (Problems • Answer Key • Resources)
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