Square, cube and fifth power

[melguy]

Please help. Thanks

[Brent@GMATPrepNow]

Hi Melguy,

I should point out that, if you post an image as an attachment, only those who are logged in can see it. If you post it as an image (as I've done below), everyone who visits the site can see it.


Now, onto the question.
The correct answer is not among the answer choices.
The correct answer is actually zero. When we multiply N by zero, the product is zero, and zero is a perfect square, a perfect cube and a perfect 5th power.

The next smallest number is [a^(-3)][b^(-4)][c^(-5)]
Notice that (a^3)(b^4)(c^5) x [a^(-3)][b^(-4)][c^(-5)] = 1, and 1 is a perfect square, a perfect cube and a perfect 5th power.

There are actually several other (smaller) numbers that work, but let's find the smallest number among the answer choices.

First we need to know what perfect squares, cubes and 5th powers look like.

For example, 7^10 is a perfect square, since we can rewrite 7^10 as (7^5)^2
7^10 is also a perfect 5th power, since we can rewrite 7^10 as (7^2)^5
However, 7^10 is NOT a perfect cube since we CANNOT rewrite 7^10 as (something)^3

Another example:
7^12 is a perfect square, since we can rewrite 7^12 as (7^6)^2
7^12 is also a perfect cube, since we can rewrite 7^12 as (7^4)^3
However, 7^12 is NOT a perfect 5th power since we CANNOT rewrite 7^12 as (something)^5

So, if 7^k is to be a perfect square, a perfect cube and a perfect 5th power, k must be divisible by 2, 3 and 5
In other words, k must be divisible by 30.

So, among the answer choices, only E will yield a product that's a a perfect square, a perfect cube and a perfect 5th power.
(a^3)(b^4)(c^5) x [(a^27)(b^26)(c^25)] = (a^30)(b^30)(c^30)

(a^30)(b^30)(c^30) is is a perfect square, a perfect cube and a perfect 5th power.

Cheers,
Brent

[sanjoy18]

Actual answer to this question is 0..but unfortunately it is not there.

To get smallest number such that when multiplied with N ..then derived number would be a expressed as perfect square, cube and fifth power.
now N=a^3*b^4*c^5
LCM(2,3,5)=30
thus any number with 30th power can be written as square, cube and fifth power
e.g....M^30=(M^15)^2..square format
M^30=(M^10)^3..cube format
M^30=(M^6)^5..fifth power format

So we need to multiply a numbers with N so that new numbers become..(a*b*c)^30

hence required number is a^27*b^26*c^25.Hence E