Difference between revisions of "2020 AIME I Problems/Problem 12"
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Since <math>3^2</math>, <math>7^5</math> and <math>4\cdot 5^4</math> all divide <math>n</math>, the smallest value of <math>n</math> working is their LCM, also <math>3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. Thus the number of divisors is <math>(2+1)(2+1)(4+1)(5+1) = \boxed{270}</math>. | Since <math>3^2</math>, <math>7^5</math> and <math>4\cdot 5^4</math> all divide <math>n</math>, the smallest value of <math>n</math> working is their LCM, also <math>3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. Thus the number of divisors is <math>(2+1)(2+1)(4+1)(5+1) = \boxed{270}</math>. | ||
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==See Also== | ==See Also== |
Revision as of 17:15, 12 March 2020
Note: Please do not post problems here until after the AIME.
Problem
Let be the least positive integer for which is divisible by Find the number of positive integer divisors of
Solution
Lifting the Exponent shows that so thus, divides . It also shows that so thus, divides .
Now, multiplying by , we see and since and then meaning that we have that by LTE, divides .
Since , and all divide , the smallest value of working is their LCM, also . Thus the number of divisors is .
~kevinmathz
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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