Difference between revisions of "2005 AMC 12B Problems/Problem 21"
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− | 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 | + | 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 there are <math>4</math> possible powers of <math>7</math> in the factorization of <math>7n</math>, and <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 15:05, 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 there are
possible powers of
in the factorization of
, and
possible powers of
in the factorization of
, which would be
,
, and
. Therefore the highest power of
that could divide
is
.