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 .

If n is a positive integer and (n + 1)(n + 3) is odd , that means n must be even= 2,4,6,8 .........etc.

When n=2 then (n + 1)(n + 3) = 15 and (n + 2)(n + 4) = 24 that value are divisible by : 3 , 6 & 8

When n=4 then (n + 1)(n + 3) = 35 and (n + 2)(n + 4) = 48 that value are divisible by : 3 , 6 & 8

When n=6 then (n + 1)(n + 3) = 63 and (n + 2)(n + 4) = 80 that value are divisible by : 8 & 16

When n=8 then (n + 1)(n + 3) = 99 and (n + 2)(n + 4) = 120 that value are divisible by only 6 & 8

Question mentioned " must be a multiple" not could be, so can see in above calculation 8 is a must be multiple of (n + 2)(n + 4)

So, Answer must be only " 8 "