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Difference between revisions of "2005 AMC 12B Problems/Problem 21"

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Revision as of 09:42, 4 July 2013

Problem

A positive integer $n$ has $60$ divisors and $7n$ has $80$ divisors. What is the greatest integer $k$ such that $7^k$ divides $n$?

$\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}$

Solution

If $n$ has $60$ factors, then $n$ is a product of $2\times2\times3\times5$ powers of (not necessarily distinct) primes. When multiplied by $7$, the amount of factors of $n$ increased by $\frac{80}{60}=\frac{4}{3}$, so there are $4$ possible powers of $7$ in the factorization of $7n$, and $3$ possible powers of $7$ in the factorization of $n$, which would be $7^0$, $7^1$, and $7^2$. Therefore the highest power of $7$ that could divide $n$ is $2\Rightarrow\boxed{C}$.

See also

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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All AMC 12 Problems and Solutions

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