Difference between revisions of "2005 AMC 12B Problems/Problem 21"
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== Solution == | == Solution == | ||
+ | If <math>n</math> has <math>60</math> factors, then <math>n</math> is a product of <math>2\times2\times3\times5</math> powers of (not necessarily distinct) primes. When multiplied by <math>7</math>, the amount of factors of <math>n</math> increased by <math>\frac{80}{60}=\frac{4}{3}</math>, so before there were <math>3</math> possible powers of <math>7</math> in the factorization of <math>n</math>, which would be <math>7^0</math>, <math>7^1</math>, and <math>7^2</math>. Therefore the highest power of <math>7</math> that could divide <math>n</math> is <math>2\Rightarrow\boxed{C}</math>. | ||
== See also == | == See also == | ||
* [[2005 AMC 12B Problems]] | * [[2005 AMC 12B Problems]] |
Revision as of 14:03, 4 February 2011
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Problem
A positive integer has divisors and has divisors. What is the greatest integer such that divides ?
Solution
If has factors, then is a product of powers of (not necessarily distinct) primes. When multiplied by , the amount of factors of increased by , so before there were possible powers of in the factorization of , which would be , , and . Therefore the highest power of that could divide is .