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2005 Alabama ARML TST Problems/Problem 6

Revision as of 12:55, 20 November 2006 by MCrawford (talk | contribs) (Exponent change -- it was my error in the solutions packet, though it doesn't change the answer)

Problem

How many of the positive divisors of 3,240,000 are perfect cubes?

Solution

$3240000=2^7\cdot 3^4\cdot 5^4$. We want to know how many numbers are in the form $2^{3a}3^{3b}5^{3c}$ which divide $3,240,000$. This imposes the restrictions $0\leq a\leq 2$,$0 \leq b\leq 1$ and $0 \leq c\leq 1$, which lead to 12 solutions and thus 12 such divisors.

See Also

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