Difference between revisions of "2005 Alabama ARML TST Problems/Problem 6"
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==Problem== | ==Problem== | ||
| − | How many of the [[positive]] [[divisor]]s of 3,240,000 are [[perfect cube]]s? | + | How many of the [[positive]] [[divisor]]s of <math>3,240,000</math> are [[perfect cube]]s? |
==Solution== | ==Solution== | ||
| − | <math>3240000=2^ | + | <math>3240000=2^7\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== | ==See Also== | ||
| − | + | {{ARML box|year=2005|state=Alabama|num-b=5|num-a=7}} | |
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| − | [[Category: | + | |
| + | [[Category:Intermediate Number Theory Problems]] | ||
Latest revision as of 11:46, 11 December 2007
Problem
How many of the positive divisors of 
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.
See Also
| 2005 Alabama ARML TST (Problems) | ||
| Preceded by: Problem 5 |
Followed by: Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
