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# How many positive perfect cubes are divisors of 4^6?

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GMATH Teacher
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Joined: 12 Oct 2010
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How many positive perfect cubes are divisors of 4^6?  [#permalink]

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18 Feb 2019, 06:27
00:00

Difficulty:

75% (hard)

Question Stats:

18% (00:54) correct 82% (01:20) wrong based on 33 sessions

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GMATH practice exercise (Quant Class 16)

How many positive perfect cubes are divisors of 4^6?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
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How many positive perfect cubes are divisors of 4^6?  [#permalink]

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Updated on: 18 Feb 2019, 07:51
fskilnik wrote:
GMATH practice exercise (Quant Class 16)

How many positive perfect cubes are divisors of 4^6?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

CONCEPT : Perfect cube $$= PrimeNumber^{3a}$$

i.e. Perfect Cube is defined as a number which when prime factorised has all exponents (powers of prime numbers) multiples of 3

$$4^6 = (2^2)^6 = 2^{12}$$

the factors of $$2^{12}$$ which are perfect cubes = {$$1^3$$, $$2^3$$ , $$2^6$$ , $$2^9$$ , $$2^{12}$$}

Hence, 5 such numbers

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Originally posted by GMATinsight on 18 Feb 2019, 06:36.
Last edited by GMATinsight on 18 Feb 2019, 07:51, edited 1 time in total.
GMATH Teacher
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How many positive perfect cubes are divisors of 4^6?  [#permalink]

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18 Feb 2019, 07:42
Definition: N is a perfect cube if (and only if) N can be expressed as the cube of an integer, that is, N = M^3 , where M is an integer.

Important: 0 and 1 are examples of perfect cubes that cannot be factorized into primes. I explain: the Fundamental Theorem of Arithmetic tells us that the prime decomposition of an integer is possible for every integer greater than 1. (*)

(*) The theorem also tells us that this decomposition is unique if we do not take into account the order of the primes present in the decomposition.
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How many positive perfect cubes are divisors of 4^6?  [#permalink]

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18 Feb 2019, 10:03
fskilnik wrote:
GMATH practice exercise (Quant Class 16)

How many positive perfect cubes are divisors of 4^6?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

4^6 = 2 ^ 12
2^12 = 1^3, 2^3, 2^6,2^9,2^12)
total 5 cubes
IMO D
GMATH Teacher
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Re: How many positive perfect cubes are divisors of 4^6?  [#permalink]

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18 Feb 2019, 18:36
fskilnik wrote:
GMATH practice exercise (Quant Class 16)

How many positive perfect cubes are divisors of 4^6?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

$$?\,\,\,:\,\,\,\# \,\,{M^3}\,\,\,\left( {M \ge 1\,\,{\mathop{\rm int}} } \right)\,\,\,{\rm{such}}\,\,{\rm{that}}\,\,{{{4^6}} \over {{M^3}}}\,\,\mathop = \limits^{\left( * \right)} \,\,{\mathop{\rm int}}$$

$${4^6} = {2^{12}} = {\left( {{2^4}} \right)^3}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{\left( {{{{2^4}} \over M}} \right)^3} = {\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,M = {2^0},{2^1},{2^2},{2^3},{2^4}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {\rm{D}} \right)$$

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: How many positive perfect cubes are divisors of 4^6?   [#permalink] 18 Feb 2019, 18:36
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