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 00:23, 15 March 2019
Problem 7
There are positive integers and that satisfy the system of equations 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
Thus, when the two equations are added, we have that
When simplified, this equals
so this means that
so
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 and Therefore, and and if we plug these values back, it works! has total factors and has so
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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