Divisibilty-Perfect Square

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by Brent@GMATPrepNow » Sat Jul 05, 2014 6:11 am
In a perfect SQUARE, all of the exponents are divisible by 2 (i.e., they are even). For example, x²y�z� is a perfect square.
In a perfect CUBE, all of the exponents are divisible by 3. For example, x³y³z� is a perfect cube.
In a perfect 5th POWER, all of the exponents are divisible by 5. For example, x�y�z�� is a perfect cube.

So, to meet ALL 3 conditions, all of the exponents MUST BE divisible by 2, 3 AND 5.
If we take (a^3)(b^4)(c^5) and multiply it by answer choice E, we get the product (a^30)(b^30)(c^30), where all of the exponents are divisible by 2, 3 AND 5.

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by GMATinsight » Sat Jul 05, 2014 6:39 am
The perfect Squares have all the powers of distinct Prime factor a multiple of 2
The perfect Cube have all the powers of distinct Prime factor a multiple of 3
The perfect Fifths have all the powers of distinct Prime factor a multiple of 5

To satisfy all above the powers of all prime numbers must be multiple of LCM of (2,3 and 5)=30

Therefore (a^3)(b^4)(c^5) should be multiplied with something to become (a^30)(b^30)(c^30)

Therefore the number to multiplied = (a^30)(b^30)(c^30) / (a^3)(b^4)(c^5) = [spoiler](a^27)(b^27)(c^25) Option E[/spoiler]
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