2019 AIME I Problems/Problem 7: Difference between revisions
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==Solution== | ==Solution== | ||
One immediately sees that <math>x = 10^{20}</math> and <math>y = 10^{190}</math>, so the answer is <math>3 * 40 + 2 * 380 = \boxed{880}</math> because <math>10^{20} = 2^{20} * 5^{20}</math> and similarly for <math>10^{190}</math>. | |||
==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 20:00, 14 March 2019
The 2019 AIME I takes place on March 13, 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
One immediately sees that 
and 
, so the answer is 
because 
and similarly for 
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
