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If n is an integer, is (n-1)(n+1) a multiple of 24?

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If n is an integer, is (n-1)(n+1) a multiple of 24?  [#permalink]

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New post 05 Oct 2017, 03:15
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A
B
C
D
E

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67% (02:01) correct 33% (01:31) wrong based on 49 sessions

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[GMAT math practice question]

If n is an integer, is (n-1)(n+1) a multiple of 24?

1) n is odd.
2) n is not divisible by 3.

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Re: If n is an integer, is (n-1)(n+1) a multiple of 24?  [#permalink]

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New post 05 Oct 2017, 04:56
According to the question we need to prove (n^2-1)/24 = an integer

1. If n is odd its not sufficient since it can take value as 3
2. If is n is not divisible by 3 then n can be even also hence, in sufficient

taking both 1 & 2 we get n^2-1 is a multiple of 24

hence C answer
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Re: If n is an integer, is (n-1)(n+1) a multiple of 24?  [#permalink]

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New post 08 Oct 2017, 18:00
=>
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember that equal number of variables and independent equations ensures a solution.
We have 1 variable and 0 equation from the original condition. Therefore, D is most likely to be the answer.

Condition 1)
Since n is odd, n-1 and n+1 are consecutive even integers.
One of two consecutive even integers must be a multiple of 4. For example, (2,4), (6,8), (8,10) and (10,12).
Thus (n-1)(n+1) is a multiple of 4, but its divisibility by 3 is not identified.
Counterexamples are n = 3 and n = 5.
(3-1)(3+1) = 2*4 = 8, which is not a multiple of 24.
(5-1)(5+1) = 4*6 = 24, which is a multiple of 24.
This is not sufficient.

Condition 2)
Since n is not a multiple of 3, n = 3k +1 or n = 3k + 2 for some integer k.
For the case, n = 3k +1, (n-1)(n+1) = (3k+1-1)(3k+1+1) = 3k(3k+2) is a multiple of 3. However, we can’t identify if n is a multiple of 24. Counterexamples are n = 4 for which (n-1)(n+1) = 3*5 = 15 is not a multiple of 24 and n = 7 for which (n-1)(n+1) = 6*8 = 48 is a multiple of 24.
For the case, n = 3k +2, (n-1)(n+1) = (3k+2-1)(3k+2+1) = (3k+1)(3k+3) = 3(3k+1)(k+1) is a multiple of 3. However, we can’t identify if n is a multiple of 24. Counterexamples are n = 5 for which (n-1)(n+1) = 4*6 = 15 is a multiple of 24 and n = 6 for which (n-1)(n+1) = 5*7 = 35 is a not multiple of 24.

Condition 1) & 2)
From the condition 1), (n-1)(n+1) is a multiple of 8 and from the condition 2), (n-1)(n+1) is a multiple of 3.
Thus, (n-1)(n+1) is a multiple of 24.
Therefore, unlike our expectation, C is the answer.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both con 1) and con 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using con 1) and con 2) separately according to DS definition. Obviously, there may be cases where the answer is A, B, C or E.

Answer: C
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Re: If n is an integer, is (n-1)(n+1) a multiple of 24? &nbs [#permalink] 08 Oct 2017, 18:00
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