Difference between revisions of "2019 AIME I Problems/Problem 7"
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Factor this into 2^20 * 5^20, and add the two 20's, resulting in m, which is 40. | Factor this into 2^20 * 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. | Add m to 2m + 2n (which is 840) to get 40+840 = 880. | ||
| + | |||
| + | ==Solution 2 (Almost same as above but cleaner)== | ||
| + | |||
| + | First simplifying the first and second equations, we get that | ||
| + | |||
| + | <cmath>\log_{10}(x\cdot\text{gcd}(x,y)^2)=60</cmath> | ||
| + | <cmath>\log_{10}(y\cdot\text{lcm}(x,y)^2)=570</cmath> | ||
| + | |||
| + | |||
| + | Thus, when the two equations are added, we have that | ||
| + | <cmath>\log_{10}(x\cdot y\cdot\text{gcd}\cdot\text{lcm}^2)=630</cmath> | ||
| + | When simplified, this equals | ||
| + | <cmath>\log_{10}(x^3y^3)=630</cmath> | ||
| + | so this means that | ||
| + | <cmath>x^3y^3=10^{630}</cmath> | ||
| + | so | ||
| + | <cmath>xy=10^{210}.</cmath> | ||
| + | |||
| + | Now, the following cannot be done on a proof contest but let's (intuitively) assume that <math>x<y</math> and <math>x</math> and <math>y</math> are both powers of <math>10</math>. This means the first equation would simplify to <cmath>x^3=10^{60}</cmath> and <cmath>y^3=10^{570}.</cmath> Therefore, <math>x=10^{20}</math> and <math>y=10^{190}</math> and if we plug these values back, it works! <math>10^{20}</math> has <math>20\cdot2=40</math> total factors and <math>10^{190}</math> has <math>190\cdot2=380</math> so <cmath>3\cdot 40 + 2\cdot 380 = \boxed{880}.</cmath> | ||
| + | |||
| + | 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== | ==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 23:23, 14 March 2019
Problem 7
There are positive integers 
and 
that satisfy the system of equations ![\[\log_{10} x + 2 \log_{10} (\gcd(x,y)) = 60\]](http://latex.artofproblemsolving.com/1/6/f/16f085cc382249acf20a8b88a9a35f00cbac2fb3.png)
![\[\log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) = 570.\]](http://latex.artofproblemsolving.com/e/1/e/e1eff7ffc812984b5b58c17079adcc6677c3497b.png)
Let 
be the number of (not necessarily distinct) prime factors in the prime factorization of , and let 
be the number of (not necessarily distinct) prime factors in the prime factorization of . Find 
.
Solution
Add the two equations to get that log x+log y+2(log(gcd(x,y))+log(lcm(x,y)))=630. 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. 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. This can easily be simplified to log xy=210, or xy = 10^210. 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. 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. Factor this into 2^20 * 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 (Almost same as above but cleaner)
First simplifying the first and second equations, we get that
![\[\log_{10}(x\cdot\text{gcd}(x,y)^2)=60\]](http://latex.artofproblemsolving.com/e/3/0/e30b5562ae1f9b20a800b049ac435119bbf0b6c3.png)
![\[\log_{10}(y\cdot\text{lcm}(x,y)^2)=570\]](http://latex.artofproblemsolving.com/d/c/b/dcbadc2fbcb278efef05cb4fa95b8710b7b7cb82.png)
Thus, when the two equations are added, we have that
![\[\log_{10}(x\cdot y\cdot\text{gcd}\cdot\text{lcm}^2)=630\]](http://latex.artofproblemsolving.com/a/8/d/a8d70e9e14ee862d70f5de3abc62af0b1cab9253.png)
When simplified, this equals
![\[\log_{10}(x^3y^3)=630\]](http://latex.artofproblemsolving.com/d/e/c/dec609122c3613e7a272badf9b0b9cbb50233317.png)
so this means that
![\[x^3y^3=10^{630}\]](http://latex.artofproblemsolving.com/0/e/5/0e5ba1cf469ca8c156a39808dd58d35f0beda593.png)
so
![\[xy=10^{210}.\]](http://latex.artofproblemsolving.com/8/6/f/86f837490d381ffc8f5d7e7921c363950304f403.png)
Now, the following cannot be done on a proof contest but let's (intuitively) assume that 
and and are both powers of 
. This means the first equation would simplify to ![\[x^3=10^{60}\]](http://latex.artofproblemsolving.com/f/7/8/f78cd2bd12ea58dfb32afc193bd6e44310b252f7.png)
and ![\[y^3=10^{570}.\]](http://latex.artofproblemsolving.com/8/a/4/8a4f4b64335d3d96c3264edc26bfbe906477124a.png)
Therefore, 
and 
and if we plug these values back, it works! 
has 
total factors and 
has 
so ![\[3\cdot 40 + 2\cdot 380 = \boxed{880}.\]](http://latex.artofproblemsolving.com/9/c/b/9cb1628698b47262272100fcfe54d327084f5e42.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
