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

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

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]

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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Source: Beat The GMAT — Problem Solving | Mon Feb 18, 2019 5:12 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

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

4â�¶ = (2Â²)â�¶ = 2Â¹Â²
Since 2Â¹Â² = (2Â³)(2â�¹), we know 2Â³ is a divisor of 2Â¹Â²
Since 2Â¹Â² = (2â�¶)(2â�¶), we know 2â�¶ is a divisor of 2Â¹Â²
Since 2Â¹Â² = (2â�¹)(2Â³), we know 2â�¹ is a divisor of 2Â¹Â²
Since 2Â¹Â² = (2Â¹Â²)(2â�°), we know 2Â¹Â² is a divisor of 2Â¹Â²
Since 2Â¹Â² = (2â�°)(2Â¹Â²), we know 2â�° is a divisor of 2Â¹Â²

2Â³ is a perfect cube.
2â�¶ = (2Â²)Â³. So, 2â�¶ is a perfect cube.
2â�¹ = (2Â³)Â³. So, 2â�¹ is a perfect cube.
2Â¹Â² = (2â�´)Â³. So, 2Â¹Â² is a perfect cube.
2â�° = 1 = 1Â³. So, 2â�° is a perfect cube.
There are 5 perfect cubes.

Answer: D

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

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)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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