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How many divisors of 72^72 are perfect cubes?

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How many divisors of 72^72 are perfect cubes?  [#permalink]

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New post 11 Jul 2017, 13:42
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How many divisors of 72^72 are perfect cubes?
a) 3672
b) 3577
c) 2812
d) 3600
e) 7200

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Re: How many divisors of 72^72 are perfect cubes?  [#permalink]

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New post 11 Jul 2017, 14:44
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IanStewart wrote:
The exponents in the prime factorization of a perfect cube must all be divisible by 3, so a number like 2^15 * 3^6 is a perfect cube (it is the cube of 2^5 * 3^2), while 2^8 * 3^7 is not

If we prime factorize 72^72, we get (2^3 * 3^2 )^72 = 2^216 * 3^144

For a divisor of this to be a perfect cube, it needs to look like this:

2^a * 3^b

where a and b must both be divisible by 3 (and either exponent could be zero). So a must be in this list:

0, 3, 6, 9, ..., 213, 216

and b must be in this list:

0, 3, 6, ...., 141, 144

There are 73 numbers in the first list (if you just divide everything by 3, the list becomes 0, 1, 2, 3, .., 72, which has 73 numbers in it) and similarly there are 49 numbers in the second list, so we have 73*49 choices in total for a and b together, and since the units digit of 73*49 is 7, the only possible answer among the choices is 3577.

Overall it's too inelegant a question to be a realistic GMAT problem, but the ingredients in the solution can all be tested in simpler ways.


Thanks IanStewart for the solution.I have found an alternate solution to solve it fast:

72^72 = (2^3 * 3^2 )^72 = (2^72)^3 * (3^48)^3
as both (2^72)^3 and (3^48)^3 are perfect cubes the total number of factor using factor property would be (72+1)*(48+1)=73 *49 = 3577
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How many divisors of 72^72 are perfect cubes?  [#permalink]

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New post 11 Jul 2017, 14:13
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The exponents in the prime factorization of a perfect cube must all be divisible by 3, so a number like 2^15 * 3^6 is a perfect cube (it is the cube of 2^5 * 3^2), while 2^8 * 3^7 is not

If we prime factorize 72^72, we get (2^3 * 3^2 )^72 = 2^216 * 3^144

For a divisor of this to be a perfect cube, it needs to look like this:

2^a * 3^b

where a and b must both be divisible by 3 (and either exponent could be zero). So a must be in this list:

0, 3, 6, 9, ..., 213, 216

and b must be in this list:

0, 3, 6, ...., 141, 144

There are 73 numbers in the first list (if you just divide everything by 3, the list becomes 0, 1, 2, 3, .., 72, which has 73 numbers in it) and similarly there are 49 numbers in the second list, so we have 73*49 choices in total for a and b together, and since the units digit of 73*49 is 7, the only possible answer among the choices is 3577.

Overall it's too inelegant a question to be a realistic GMAT problem, but the ingredients in the solution can all be tested in simpler ways.
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Re: How many divisors of 72^72 are perfect cubes?  [#permalink]

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New post 11 Jan 2018, 09:14
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Prime factorization of \(72^{72}\) = \((2^3 * 3^2)^{72}\)
= \(((2^3)^{72} * 3^{144})\)
to find factors which are perfect cubes, we represent it as = \(8^{72} * (3^3)^{48}\) = \(8^{72} * 27^{48}\)
since 8, 27 are perfect cubes, any combination powers of 8,27 must be perfect cube
A: { \(8^0\), \(8^1\), \(8^2\) ,...........................\(8^{72}\)} = total = 73
B: { \(27^0\), \(27^1\), ...............................\(27^{48}\)} = total = 49
number of factors which are perfect cubes: \(73 * 49\) = \(3577\) (B)
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How many divisors of 72^72 are perfect cubes?  [#permalink]

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New post 11 Jan 2018, 09:24
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Supermaverick wrote:
How many divisors of 72^72 are perfect cubes?
a) 3672
b) 3577
c) 2812
d) 3600
e) 7200



such Qs when we are looking for ODD factors or EVEN factors or the one asked here, following method is best..

Number of factors of a number
\(x=a^p*b^r*c^s...\) is (\(p+1)(q+1)(s+1)...\)

here we are looking for CUBES..
so \(72^{72}\) can be written as \((2^3*3^2)^{24*3} = (2^{3*24}*3^{2*24})^3 = (2^{72}*3^{48})^3\)
so cubes will be \((72+1)(48+1) = 73*49 = 3577\)
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Re: How many divisors of 72^72 are perfect cubes?  [#permalink]

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New post 24 Jul 2019, 08:52
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Re: How many divisors of 72^72 are perfect cubes?   [#permalink] 24 Jul 2019, 08:52
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