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If n is a positive integer and (n + 1)(n + 3)
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Updated on: 11 Dec 2022, 23:29
Updated on: 11 Dec 2022, 23:29
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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
Re: If n is a positive integer and (n + 1)(n + 3)
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01 Jun 2018, 15:31
01 Jun 2018, 15:31
1
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 (\(2 \times 2 \times 2\)).
Re: If n is a positive integer and (n + 1)(n + 3)
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12 Jun 2019, 07:10
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 (\(2 \times 2 \times 2\)).
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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 .
