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number theory
[#permalink]
30 May 2018, 00:44

Question Stats:

If n is a positive integer and (n + 1)(n + 3) is odd, then (n + 2)(n + 4) must be a multiple of which one of the following?

(A) 3 (B) 5 (C) 6 (D) 8 (E) 16

(A) 3 (B) 5 (C) 6 (D) 8 (E) 16

Retired Moderator

Joined: **07 Jun 2014 **

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Re: number theory
[#permalink]
01 Jun 2018, 15:31

Expert Reply

shahul wrote:

If n is a positive integer and (n + 1)(n + 3) is odd, then (n + 2)(n + 4) must be a multiple of which one of the following?

(A) 3 (B) 5 (C) 6 (D) 8 (E) 16

(A) 3 (B) 5 (C) 6 (D) 8 (E) 16

For \((n + 1)(n + 3)\) to be odd \((n+1)\) is odd and \((n+3)\) is also odd.

Since only \(odd \times odd=odd\). So \(n\) must be even since only \(even + odd = odd\).

\((n + 2)(n + 4)= 2 \times (\frac{n}{2}+1) \times 2 \times (\frac{n}{2}+2)\).

Either \((\frac{n}{2}+1)\) is even or \((\frac{n}{2}+2)\) is even.

So this means there must be a factor of 8 (\(2 \times 2 \times 2\)).

_________________

Sandy

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Re: number theory
[#permalink]
12 Jun 2019, 07:10

sandy wrote:

shahul wrote:

If n is a positive integer and (n + 1)(n + 3) is odd, then (n + 2)(n + 4) must be a multiple of which one of the following?

(A) 3 (B) 5 (C) 6 (D) 8 (E) 16

(A) 3 (B) 5 (C) 6 (D) 8 (E) 16

For \((n + 1)(n + 3)\) to be odd \((n+1)\) is odd and \((n+3)\) is also odd.

Since only \(odd \times odd=odd\). So \(n\) must be even since only \(even + odd = odd\).

\((n + 2)(n + 4)= 2 \times (\frac{n}{2}+1) \times 2 \times (\frac{n}{2}+2)\).

Either \((\frac{n}{2}+1)\) is even or \((\frac{n}{2}+2)\) is even.

So this means there must be a factor of 8 (\(2 \times 2 \times 2\)).

---------------------------------------

From the question stem, n is positive

and further logic shows that n is even.

Therefore, (n+2)(n+4) must be even.

the above entails as n^2+6n+8

as n is positive and even, we can, for sure, assume that n must be equal to 2 [ n may be other values like 4,6,8,10, etc]

if we take n=2 and plug in the above equation we 24 which is a multiple of 3,6,and 8.

Kindly correct my understanding of the above. Regards .

gmatclubot

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