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

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

GMAT assassins aren't born, they're made,
Rich

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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