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×odd=odd. So
n must be even since only
even+odd=odd.
(n+2)(n+4)=2×(n2+1)×2×(n2+2).
Either
(n2+1) is even or
(n2+2) is even.
So this means there must be a factor of 8 (
2×2×2).
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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 .