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If r is a positive integer, n=r^3 and 4,14, and 27 are facto

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If r is a positive integer, n=r^3 and 4,14, and 27 are facto

by Max@Math Revolution » Wed Sep 05, 2018 12:59 am

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[Math Revolution GMAT math practice question]

If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, which of the following must be a factor of n?

A. 16
B. 32
C. 36
D. 48
E. 64
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by GMATGuruNY » Wed Sep 05, 2018 2:12 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If r is a positive integer, n=r^3 and 4, 14, and 27 are factors of n, which of the following must be a factor of n?

A. 16
B. 32
C. 36
D. 48
E. 64
Since 4, 14 and 27 are all factors of n, the prime-factorization of n must include 2*2, 2*7 and 3*3*3.
n = r³ implies that n is a perfect cube.
The prime-factorization of a perfect cube must include at least 3 of each prime factor.
Thus, the prime-factorization of n must include at least three 2's, three 3's, and three 7's, implying that the least possible value of n = 2³3³7³.
Since the least possible value of n is not divisible by 16, 32, 48, or 64, eliminate A, B, D and E.

The correct answer is C.
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by Max@Math Revolution » Fri Sep 07, 2018 12:13 am
=>

Since 4 = 2^2,14 = 2*7, 27 = 3^3 are factors of n and n is a perfect cube, the smallest possible value of n is 2^3*3^3*7^3.

When n = 2^3*3^3*7^3, 16 = 2^4, 32 = 2^5, 48=2^4*3 and 64 = 2^6 can't be factors of n. The only answer choice that is a factor of n is 36 = 2^2*3^2.

Therefore, the answer is C.
Answer: C
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