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For positive integer n, is the product (n)(n+1)(n+2) divisible by 24?

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For positive integer n, is the product (n)(n+1)(n+2) divisible by 24?  [#permalink]

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New post 31 Dec 2017, 14:58
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A
B
C
D
E

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Re: For positive integer n, is the product (n)(n+1)(n+2) divisible by 24?  [#permalink]

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New post 31 Dec 2017, 14:59
Bunuel wrote:
For positive integer n, is the product (n)(n+1)(n+2) divisible by 24?

(1) n is even
(2) n + 2 = 6


VERITAS PREP OFFICIAL SOLUTION:

Correct Answer: D

Explanation: In this question, it may be easiest to start with statement 2, as it provides a solution for n, which is all you would need to answer the question. Conducting the calculation is actually irrelevant, as having the values for the three terms of the product will guarantee that you get an answer. Statement 1 is a bit more involved, but relatively straightforward when keeping in mind two number properties: In any set of three consecutive integers, one will be divisible by 3, so we can guarantee that this product will be divisible by 3. In any set of consecutive integers, every second number will be even, so we can also guarantee that, if n is even, n+2 will also be even.

Furthermore, every fourth number, and every second even number, will be divisible by 4. Because of that, if n is even, then either it or n+2 will be divisible by 4. In order to be divisible by 24, a number must have the set of prime factors: 2, 2, 2, and 3. Because we can prove that one of the three terms is divisible by 3, and that between n and n+2 one will be divisible by 2 and the other by 4, we can prove that this number is divisible by 24.

Accordingly, statement 1 is sufficient, as well, and the correct answer is D.
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Re: For positive integer n, is the product (n)(n+1)(n+2) divisible by 24?  [#permalink]

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New post 31 Dec 2017, 23:42
Bunuel wrote:
For positive integer n, is the product (n)(n+1)(n+2) divisible by 24?

(1) n is even
(2) n + 2 = 6



24 = 2^3 * 3, so for a number to be divisible by 24, it must be divisible by 8(three times 2) as well as by 3.

(1) If n is even, (n+2) is also even, and since n and n+2 have a difference of 2, one of them will be divisible by 4 also. So basically the product of n(n+2) will give us three 2's, hence divisible by 8. And since n(n+1)(n+2) is product of three consecutive integers, one of them will definitely be divisible by 3. Hence the product will be divisible by 8 as well as by 3, so by 24 also. Sufficient.

(2) n=2 = 6. So the given product is 4*5*6, which is divisible by 24. Sufficient.

Hence D answer
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Re: For positive integer n, is the product (n)(n+1)(n+2) divisible by 24? &nbs [#permalink] 31 Dec 2017, 23:42
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