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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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If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, w  [#permalink]

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New post 05 Sep 2018, 02:18
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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  [#permalink]

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New post 05 Sep 2018, 02:28
1
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 prime-factorizing 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  [#permalink]

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New post 05 Sep 2018, 03:25
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 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.


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If r is a positive integer, n=r^3 and 4,14, and 27 are factors of n, w  [#permalink]

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New post 05 Sep 2018, 04:11
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  [#permalink]

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New post 05 Sep 2018, 04:44
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 prime-factorizing 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  [#permalink]

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New post 05 Sep 2018, 12:52
1
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 prime-factorizing 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  [#permalink]

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New post 05 Sep 2018, 22:41
1
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 prime-factorizing 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 memyselfi

Welcome 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  [#permalink]

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New post 07 Sep 2018, 01:19
=>

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 [#permalink] 07 Sep 2018, 01:19
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