Art of Problem Solving

2005 Alabama ARML TST Problems/Problem 6: Difference between revisions

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==Solution==
==Solution==


<math>3240000=2^63^45^4</math>. We want to know how many numbers are in the form <math>2^{3a}3^{3b}5^{3c}</math>, <math>a\leq 2A</math>,<math>b\leq 1</math>, <math>c\leq 1</math>; there are 12 such numbers, if we count 1 as a [[perfect cube]].
<math>3240000=2^63^45^4</math>. We want to know how many numbers are in the form <math>2^{3a}3^{3b}5^{3c}</math>, <math>a\leq 2</math>,<math>b\leq 1</math>, <math>c\leq 1</math>; there are 12 such numbers, if we count 1 as a [[perfect cube]].


*[[2005 Alabama ARML TST]]
*[[2005 Alabama ARML TST]]
*[[2005 Alabama ARML TST/Problem 5 | Previous Problem]]
*[[2005 Alabama ARML TST/Problem 5 | Previous Problem]]
*[[2005 Alabama ARML TST/Problem 7 | Next Problem]]
*[[2005 Alabama ARML TST/Problem 7 | Next Problem]]

Revision as of 13:50, 17 November 2006

Problem

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

Solution

$3240000=2^63^45^4$. We want to know how many numbers are in the form $2^{3a}3^{3b}5^{3c}$, $a\leq 2$,$b\leq 1$, $c\leq 1$; there are 12 such numbers, if we count 1 as a perfect cube.