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

Question

GMATH  practice exercise (Quant Class 16)

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

GMATinsight · Answer

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

fskilnik · Answer

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.

Archit3110 · Answer

4^6 = 2 ^ 12
2^12 = 1^3, 2^3, 2^6,2^9,2^12) 
total 5 cubes
IMO D

fskilnik · Answer

?\,\,\,:\,\,\,\# \,\,{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.

CrackverbalGMAT · Answer

W know the method of finding the number of divisors of any number N. i.e if we break up N into its prime factors and powers. 
If N = a^p * b^q * c^r and so on, the number of divisors of N = (p+1) * (q + 1) (r + 1)

For eg if N = 60 =  2^2  *  3^1  *  5^1 , then the number of factors = (2 + 1) * (1 + 1) * (1 + 1) = 9 (the factors being 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60)

This Rule of counting can be taken forward for finding the number of divisors of N who
(a) Are perfect squares 
(b) Perfect cubes
(c) end in 1 zero
(d) end in 2 zeros

The only process to be done is to ensure, we convert N into prime numbers and their powers.

For eg: how many factors of 360 are perfect squares: 360 =  2^3  *  3^2  *  5^1
 
Since we want perfect squares, we convert the primes into squares and find the possible powers =  (2^2)^1  * 2 * ( 3^2) 1 * 5

so  2^2  has 1 power and  3^2  has 1 power. Therefore number of perfect squares = (1 + 1) * (1 + 1) = 4

1, 4, 9 and 36 are the perfect squares which are also divisors of 360.
 
_________________________
 
In the example above,  (4)^6  =  (2^2)^6  =  (2)^{12}

Since we want the number of perfect cubes, we convert the base of 2 into a cube and find the powers =  (2^3)^4
 
The power = 4, and by the rule of counting above, the number of perfect cubes = 4 + 1 = 5

Option D

Arun Kumar
____________________

EgmatQuantExpert · Answer

Solution

Given

• The number  4^6

To Find

• The number of divisors that are cubes.

Approach and Working Out

• We have,  4^6   
  =  (2^2)^6 
=  (2)^{12} 
=  (2^4)^3

From here, we need to find the number of factors of  2^4 .
 • number of factors of  2^4  is 5.

Correct Answer: Option D

bumpbot · Answer

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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?

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