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number theory

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number theory [#permalink]  30 May 2018, 00:44
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66% (00:48) correct 33% (00:47) wrong based on 6 sessions
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
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Re: number theory [#permalink]  01 Jun 2018, 15:31
Expert's post
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

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 ($$2 \times 2 \times 2$$).
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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

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