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

Revision as of 15:03, 4 February 2011

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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 before there were $3$ 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 $2\Rightarrow\boxed{C}$ .

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 == 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 ==
 * [[2005 AMC 12B Problems]]

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

Revision as of 15:03, 4 February 2011

Problem

Solution

See also