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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?

by fskilnik@GMATH » Mon Feb 18, 2019 5:12 am

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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

Answer: [spoiler]_____(D)__[/spoiler]
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by Brent@GMATPrepNow » Mon Feb 18, 2019 6:08 am
fskilnik@GMATH 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� = (2²)� = 2¹²
Since 2¹² = (2³)(2�), we know 2³ is a divisor of 2¹²
Since 2¹² = (2�)(2�), we know 2� is a divisor of 2¹²
Since 2¹² = (2�)(2³), we know 2� is a divisor of 2¹²
Since 2¹² = (2¹²)(2�), we know 2¹² is a divisor of 2¹²
Since 2¹² = (2�)(2¹²), we know 2� is a divisor of 2¹²

2³ is a perfect cube.
2� = (2²)³. So, 2� is a perfect cube.
2� = (2³)³. So, 2� is a perfect cube.
2¹² = (2�)³. So, 2¹² is a perfect cube.
2� = 1 = 1³. So, 2� is a perfect cube.
There are 5 perfect cubes.

Answer: D

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Brent
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by fskilnik@GMATH » Mon Feb 18, 2019 5:26 pm
fskilnik@GMATH 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
Thank you for your nice contribution, Brent!

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


The correct answer is (D).

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

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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