Difference between revisions of "2005 Alabama ARML TST Problems/Problem 6"

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==Problem==
 
==Problem==
How many of the positive divisors of 3,240,000 are perfect cubes?
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How many of the [[positive]] [[divisor]]s of <math>3,240,000</math> are [[perfect cube]]s?
 
==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 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]].
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<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.
  
*[[2005 Alabama ARML TST]]
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==See Also==
*[[2005 Alabama ARML TST/Problem 5 | Previous Problem]]
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{{ARML box|year=2005|state=Alabama|num-b=5|num-a=7}}
*[[2005 Alabama ARML TST/Problem 7 | Next Problem]]
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[[Category:Intermediate Number Theory Problems]]

Latest revision as of 12:46, 11 December 2007

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

2005 Alabama ARML TST (Problems)
Preceded by:
Problem 5
Followed by:
Problem 7
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