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Given that N=a3b4c5 where a, b and c are distinct prime no

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by GMATGuruNY » Sun Dec 11, 2016 3:36 am
The following reflect the intent of the problem:
Given that N = a³b�c� where a, b and c are distinct prime numbers, what is the smallest POSITIVE INTEGER by which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?

a) a³b�c�
b) a�b�c³
c) a²b³c�
d) a�b�c�
e) a²�b²�c²�
The exponent for a perfect square must be a MULTIPLE OF 2.
The exponent for a perfect cube must be a MULTIPLE OF 3.
The exponent for a perfect fifth must be a MULTIPLE OF 5.
Thus, the exponent for an integer that is a perfect square, cube and fifth must be a multiple of 2*3*5 = 30.

Implication:
For N to become a perfect square, cube and fifth, the LEAST POSSIBLE OPTION FOR THE NEW VALUE OF N = a³�b³�c³�.
Multiplying N= a³b�c� by answer choices A, B, C and D will not yield a³�.
Thus, the correct answer must be E:
a²�b²�c²� * a³b�c� = a³�b³�c³�.

The correct answer is E.
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by [email protected] » Sun Dec 11, 2016 10:31 am
Hi sydneyuni2014,

This question is built around a few Number Properties involving exponents.

For a number to be a perfect square, all of the "exponent terms" must be EVEN.

for example....
25 is a perfect square because 25 = 5^2
16 is a perfect square because 16 = 4^2 = 2^4

For a number to be a perfect cube, all of the "exponent terms" must be A MULTIPLE OF 3.

8 is a perfect cube because 8 = 2^3
64 is a perfect cube because 64 = 4^3 = 2^6

For a number to be a perfect fifth power, all of the "exponent terms" must be A MULTIPLE OF 5.

32 is a perfect fifth power because 32 = 2^5
1024 is a perfect fifth power because 1024 = 4^5 = 2^10

Here, we need each exponent to become a multiple of 2, 3 and 5. The prompt refers to the SMALLEST number, so we need the Least Common Multiple of 2, 3 and 5 ......which is 30. The correct answer will multiply by the given prompt to equal 30.

Since we're starting with (A^3)(B^4)(C^5), we'll need to multiply with something that will end with (A^30)(B^30)(C^30).

Final Answer: E

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by Jay@ManhattanReview » Sun Dec 11, 2016 11:18 pm
sydneyuni2014 wrote:Please help me with the attached question. Thanks

Source: VeritasPrep
Hi sydneyuni2014,

Given that N is a perfect square, a perfect cube and a perfect fifth power number, N must be a number with an exponent of 2*3*5 = 30.

=> N = K^30, where K is an integer

We know that N = a^3b^4c^5, thus a needs minimum 27 as, b needs minimum 26 bs, and c needs minimum 25 cs to make N a perfect 30th power number.

Hope this helps!

-Jay

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