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If n is a positive integer, is 91 a factor of n?

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If n is a positive integer, is 91 a factor of n? [#permalink]

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[GMAT math practice question]

If \(n\) is a positive integer, is \(91\) a factor of \(n\)?

1) \(91\) is a factor of \(n^2\)
2) \(91\) is a factor of \(2n\)
[Reveal] Spoiler: OA

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Re: If n is a positive integer, is 91 a factor of n? [#permalink]

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(1) 91 is a factor of \(n^2\)
It means \(n^2\) is divisible by 91. It means if \(n^2\) is divisible by 91, then n must be divisible by 91. In other words, 91 is a factor of n [Suff]

(2) 91 is a factor of 2n
It means 2n is divisible by 91. But 91 is not divisible by 2. So n must be divisible by 91. In other words, 91 is a factor of n [Suff]

Answer: D.
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Re: If n is a positive integer, is 91 a factor of n? [#permalink]

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New post 06 Feb 2018, 22:04
If n is a positive integer, is 91 a factor of n?

1) 91 is a factor of n^2
If 91 is a factor of n * n, then 91 is a factor of n as well - Sufficient

2) 91 is a factor of 2n
If 91 is a factor of 2 * n, , then 91 is a factor of n as well - Sufficient

Both option sufficient.
Ans - Option D
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Re: If n is a positive integer, is 91 a factor of n? [#permalink]

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New post 08 Feb 2018, 01:09
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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (n) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
Since \(91 = 7*13\), \(n^2\) is a multiple of both \(7\) and \(13\). Since \(7\) and \(13\) are prime numbers, \(n\) must be a multiple of both \(7\) and \(13\).
Condition 1) is sufficient.


Condition 2)
If \(91\) is a factor of \(2n\), then \(91k = 2n\) for some integer \(k\). Since \(91\) is odd, \(k\) must be an even integer. Write \(k = 2a\), for some integer \(a\). Then
\(2n = 91*2a\). It follows that \(n = 91*a\).
Thus, \(n\) is a multiple of \(91\).
Condition 2) is sufficient.

This is a CMT(Common Mistake Type) 4(B) question. Condition 2) is easy to understand and condition 1) is difficult to figure out. If you are unable to figure out condition 2), you should choose D as the answer.

Therefore, the answer is D.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A, B, C or E.

Answer: D
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Re: If n is a positive integer, is 91 a factor of n? [#permalink]

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New post 12 Feb 2018, 17:14
MathRevolution wrote:
[GMAT math practice question]

If \(n\) is a positive integer, is \(91\) a factor of \(n\)?

1) \(91\) is a factor of \(n^2\)
2) \(91\) is a factor of \(2n\)



Notice that 91 = 7 x 13. So if 91 is a factor of n, n must be divisible by both 7 and 13.

Statement One Alone:

91 is a factor of n^2

Since n^2/91 = integer, we see that n must have factors of 7 and 13 and thus n/91 or n/(7*13) = integer.

Statement one alone is sufficient to answer the question.

Statement Two Alone:

91 is a factor of 2n

Since neither 7 nor 13 divides into 2, it must be true that 7 and 13 both divide into n. In other words, their LCM, which is 91, divides into n also. So 91 is a factor of n.

Answer: D
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Re: If n is a positive integer, is 91 a factor of n? [#permalink]

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New post 25 Mar 2018, 14:26
Hey guys,

I got stuck on this question...I chose E because of the following reasoning:

1. e.g. 25 is a factor of 5^2 but 25 is not a factor of 5, but 5 is a factor of 5^2 and of 5, hence insuff

2. e.g. 42 is a factor of 21*2 but 41 is not a factor of 21, but 21 is a factor of 21*2 and of 21, hence insuff

Where is the flaw?
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Re: If n is a positive integer, is 91 a factor of n? [#permalink]

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New post 25 Mar 2018, 21:05
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truongvu31 wrote:
Hey guys,

I got stuck on this question...I chose E because of the following reasoning:

1. e.g. 25 is a factor of 5^2 but 25 is not a factor of 5, but 5 is a factor of 5^2 and of 5, hence insuff

2. e.g. 42 is a factor of 21*2 but 41 is not a factor of 21, but 21 is a factor of 21*2 and of 21, hence insuff

Where is the flaw?


Why are you choosing 25 and 42 when the question asks about 91? You cannot arbitrarily change numbers in a question and expect that they will behave the same way original numbers would.

1. 25 is itself a perfect square. So, if 25 = 5^2 is a factor of n^2, it's not necessary 25 to be a factor of n. But 91 is NOT a perfect square: 91 = 7*13. For 91 = 7*13 to be a factor of n^2, it should be a factor of n because how else 7 and 13 would appear in n^2? Exponentiation does not produce primes, meaning that x^2, where x is a positive integer, has as many different prime factors as x itself.

2. Here again, you chose an even number, 42 and the factor that it's a factor of 2n does not necessarily mean that 42 is a factor of n. But 91 is NOT even. For 91 to be a factor of 2n, it should be a factor of n because how else 7 and 13 would appear in 2n?
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Re: If n is a positive integer, is 91 a factor of n? [#permalink]

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New post 26 Mar 2018, 04:31
Bunuel wrote:
truongvu31 wrote:
Hey guys,

I got stuck on this question...I chose E because of the following reasoning:

1. e.g. 25 is a factor of 5^2 but 25 is not a factor of 5, but 5 is a factor of 5^2 and of 5, hence insuff

2. e.g. 42 is a factor of 21*2 but 41 is not a factor of 21, but 21 is a factor of 21*2 and of 21, hence insuff

Where is the flaw?


Why are you choosing 25 and 42 when the question asks about 91? You cannot arbitrarily change numbers in a question and expect that they will behave the same way original numbers would.

1. 25 is itself a perfect square. So, if 25 = 5^2 is a factor of n^2, it's not necessary 25 to be a factor of n. But 91 is NOT a perfect square: 91 = 7*13. For 91 = 7*13 to be a factor of n^2, it should be a factor of n because how else 7 and 13 would appear in n^2? Exponentiation does not produce primes, meaning that x^2, where x is a positive integer, has as many different prime factors as x itself.

2. Here again, you chose an even number, 42 and the factor that it's a factor of 2n does not necessarily mean that 42 is a factor of n. But 91 is NOT even. For 91 to be a factor of 2n, it should be a factor of n because how else 7 and 13 would appear in 2n?


I got it, thanks for the explanation!
Re: If n is a positive integer, is 91 a factor of n?   [#permalink] 26 Mar 2018, 04:31
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