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If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, w
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05 Sep 2018, 01:18
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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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If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, w
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05 Sep 2018, 01:28
MathRevolution 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 \(n\)? \(A. 16\) \(B. 32\) \(C. 36\) \(D. 48\) \(E. 64\) If n has factors \(27 = 3^3, 4 = 2^2,\) and \(14 = 2*7\)  r must have \(2,3,\) and \(7\) as it's prime factors Evaluating answer options(on primefactorizing them) \(A. 16 = 2^4\)(We only know for sure that n will contain \(2^3\)) \(B. 32 = 2^5\)(We only know for sure that n will contain \(2^3\)) \(C. 36 = 2^2 *3^2\) \(D. 48 = 2^4 * 3\)(We only know for sure that n will contain \(2^3\)) \(E. 64 = 2^6\)(We only know for sure that n will contain \(2^3\)) Only Option C(36) can definitely be a factor of \(n\) and is the correct answer!
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Re: If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, w
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05 Sep 2018, 02:25
MathRevolution 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 primefactorization of n must include 2*2, 2*7 and 3*3*3. n = r³ implies that n is a perfect cube. The primefactorization of a perfect cube must include at least 3 of each prime factor. Thus, the primefactorization 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.
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If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, w
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05 Sep 2018, 03:11
MathRevolution 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 \(n\)? \(A. 16\) \(B. 32\) \(C. 36\) \(D. 48\) \(E. 64\) \(n = r^3\) n is a perfect cube.
prime factorization will give us the hint:
4 = 2 * 2
14 = 7* 2
27 = 3*3*3
we have :\(2^3 * 3^3\) but only a 7.
but n is a perfect cube. Thus it must be : \(2^3 *3^3*7^3\)
In this option c becomes the factor of n.
\(36 = 2^2*3^3\)
The best answer is C. Test the other options if u are suspicious.



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Re: If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, w
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05 Sep 2018, 03:44
pushpitkc wrote: MathRevolution 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 \(n\)? \(A. 16\) \(B. 32\) \(C. 36\) \(D. 48\) \(E. 64\) If n has factors \(27 = 3^3, 4 = 2^2,\) and \(14 = 2*7\)  r must have \(2,3,\) and \(7\) as it's prime factors Evaluating answer options(on primefactorizing them) \(A. 16 = 2^4\)(We only know for sure that n will contain \(2^3\)) \(B. 32 = 2^5\)(We only know for sure that n will contain \(2^3\)) \(C. 36 = 2^2 *3^2\) \(D. 48 = 2^4 * 3\)(We only know for sure that n will contain \(2^3\)) \(E. 64 = 2^6\)(We only know for sure that n will contain \(2^3\)) Only Option C(36) can definitely be a factor of \(n\) and is the correct answer! I don't understand how you made the deduction that 36 would work, can you elaborate?



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Re: If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, w
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05 Sep 2018, 11:52
memyselfi wrote: pushpitkc wrote: MathRevolution 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 \(n\)? \(A. 16\) \(B. 32\) \(C. 36\) \(D. 48\) \(E. 64\) If n has factors \(27 = 3^3, 4 = 2^2,\) and \(14 = 2*7\)  r must have \(2,3,\) and \(7\) as it's prime factors Evaluating answer options(on primefactorizing them) \(A. 16 = 2^4\)(We only know for sure that n will contain \(2^3\)) \(B. 32 = 2^5\)(We only know for sure that n will contain \(2^3\)) \(C. 36 = 2^2 *3^2\) \(D. 48 = 2^4 * 3\)(We only know for sure that n will contain \(2^3\)) \(E. 64 = 2^6\)(We only know for sure that n will contain \(2^3\)) Only Option C(36) can definitely be a factor of \(n\) and is the correct answer! I don't understand how you made the deduction that 36 would work, can you elaborate? i am happy to respond here. \(n = 3^3*2^3*7^3\)..................As n is a perfect cube, we need to bring another 7. what we are looking for ? we are looking for factor of n. it means we need a number through which we can divide n. Option C: \(2^2*3^3 = 36.\) n has \(2^3\) and \(3^3\) ................Ultimately u can cancel out 36 . Check other options. u see none of the number is fit to divide n except C. why? prime number doesn't match. Hope it helps .



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If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, w
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05 Sep 2018, 21:41
memyselfi wrote: pushpitkc wrote: MathRevolution 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 \(n\)? \(A. 16\) \(B. 32\) \(C. 36\) \(D. 48\) \(E. 64\) If n has factors \(27 = 3^3, 4 = 2^2,\) and \(14 = 2*7\)  r must have \(2,3,\) and \(7\) as it's prime factors Evaluating answer options(on primefactorizing them) \(A. 16 = 2^4\)(We only know for sure that n will contain \(2^3\)) \(B. 32 = 2^5\)(We only know for sure that n will contain \(2^3\)) \(C. 36 = 2^2 *3^2\) \(D. 48 = 2^4 * 3\)(We only know for sure that n will contain \(2^3\)) \(E. 64 = 2^6\)(We only know for sure that n will contain \(2^3\)) Only Option C(36) can definitely be a factor of \(n\) and is the correct answer! I don't understand how you made the deduction that 36 would work, can you elaborate? Hey memyselfiWelcome to GMATClub  I am merely building upon selim 's excellent answer Since n has factors \(27 = 3^3, 4 = 2^2,\) and \(14 = 2*7\), the minimum value of n must contain \(2^2 * 3^3 * 7\) Now, that we are told \(n = r^3\)  the minimum value of r will contain a 2, a 3, and a 7 and n  r's cube  has \(2^3 * 3^3 * 7^3\) Of the answer options, Only Option C(which has a power less than that of n) can be a possible answer option for a factor. ll of the remaining four options have a power greater than what the minimum value of n will contain. Hope that clears your confusion.
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Re: If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, w
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07 Sep 2018, 00:19
=> 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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Re: If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, w &nbs
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