Difference between revisions of "2005 Alabama ARML TST Problems/Problem 6"
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==Problem== | ==Problem== | ||
| − | How many of the positive | + | How many of the [[positive]] [[divisor]]s of 3,240,000 are [[perfect cube]]s? |
==Solution== | ==Solution== | ||
| − | <math>3240000=2^ | + | <math>3240000=2^6\cdot 3^4\cdot 5^4</math>. We want to know how many numbers are in the form <math>2^{3a}3^{3b}5^{3c}</math> which divide <math>3,240,000</math>. This imposes the restrictions <math>0\leq a\leq 2</math>,<math>0 \leq b\leq 1</math> and <math>0 \leq c\leq 1</math>, which lead to 12 solutions and thus 12 such divisors. |
| + | ==See Also== | ||
*[[2005 Alabama ARML TST]] | *[[2005 Alabama ARML TST]] | ||
| − | *[[2005 Alabama ARML TST/Problem 5 | Previous Problem]] | + | *[[2005 Alabama ARML TST Problems/Problem 5 | Previous Problem]] |
| − | *[[2005 Alabama ARML TST/Problem 7 | Next Problem]] | + | *[[2005 Alabama ARML TST Problems/Problem 7 | Next Problem]] |
| + | |||
| + | [[Category:Introductory Number Theory]] | ||
Revision as of 17:37, 17 November 2006
Problem
How many of the positive divisors of 3,240,000 are perfect cubes?
Solution
. We want to know how many numbers are in the form 
which divide 
. This imposes the restrictions 
,
and 
, which lead to 12 solutions and thus 12 such divisors.
