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

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

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New post 18 Feb 2019, 06:27
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

19% (01:02) correct 81% (01:19) wrong based on 26 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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How many positive perfect cubes are divisors of 4^6?  [#permalink]

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New post 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

Answer: Option D
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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.
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How many positive perfect cubes are divisors of 4^6?  [#permalink]

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New post 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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New post 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
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Re: How many positive perfect cubes are divisors of 4^6?  [#permalink]

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New post 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

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

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


The correct answer is (D).


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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