Given that N=a3b4c5 where a, b and c are distinct prime

[ideree]

Given that N=a^3b^4c^5
where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?

(A) a^3b^4c^5
(B) a^5b^4c^3
(C) a^2b^3c^5
(D) a^7b^6c^5
(E) a^27b^26c^25

[GMATGuruNY]

The problem should read as follows:

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, a perfect cube and a perfect fifth must be a multiple of 2*3*5 = 30.

Implication:
For N to become a perfect square, a perfect cube and a perfect 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, only E is viable:
a²�b²�c²� * a³b�c� = a³�b³�c³�.

The correct answer is E.

[Jeff@TargetTestPrep]

Given that N=a^3b^4c^5
where a, b and c are distinct prime numbers, what is the smallest number with which N should be multiplied such that it becomes a perfect square, a perfect cube as well as a perfect fifth power?

(A) a^3b^4c^5
(B) a^5b^4c^3
(C) a^2b^3c^5
(D) a^7b^6c^5
(E) a^27b^26c^25

The exponents of a perfect square are all multiples of 2. The exponents of a perfect cube must be all multiples of 3. And the exponents of a perfect fifth power must be all multiples of 5. Since we need N to be a perfect square, perfect cube, and perfect fifth power, we need each exponent to be a multiple of the LCM of 2, 3, and 5, which is 30. Hence, we know that the smallest number of each exponent must be 30.

The original exponent of a is 3; so we need 27 more a's to get to 30: a^3 x a^27 = a^30

The original exponent of b is 4; so we need 26 more b's to get to 30: b^4 x b^26 = b^30

The original exponent of c is 5; so we need 25 more c's to get to 30: c^5 x c^25 = c^30

Therefore, we need to multiply N = a^3b^4c^5 by a^27 x b^26 x c^25 to make it become a perfect square, a perfect cube and a perfect fifth power.

Answer: E

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Hi ideree,

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