If n is a positive integer what is the remainder when n(2n - 2)(n + 1)

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Given that n is a positive integer and we need to find what is the remainder when \(n*(2n-2)*(n+1)^2\) is divided by 32

If we simplify \(n*(2n-2)*(n+1)^2\) then we get
\(2 * (n-1) * n * (n+1)^2\) and it has a 2
=> we need to prove that \((n-1) * n * (n+1)^2\) is divisible by \(\frac{32}{2}\) = 16 = \(2^4\)

Let's take n = odd
Ex: n = 3
=> \((n-1) * n * (n+1)^2\) = \(2 * 3 * 4^2\) => divisible by \(2^4\)
Ex: n = 5
=> \((n-1) * n * (n+1)^2\) = \(4 * 3 * 6^2\) => divisible by \(2^4\)
Even if we take other cases then also we will have same conclusion
=> \((n-1) * n * (n+1)^2\) is divisible by \(2^4\) in all the cases when n = odd
=> We CAN answer the problem for sure when n = odd

Let's take n = even
Ex: n = 4
=> \((n-1) * n * (n+1)^2\) = \(3 * 4 * 5^2\) => NOT divisible by \(2^4\)
Ex: n = 16
=> \((n-1) * n * (n+1)^2\) = \(15 * 16 * 17^2\) => divisible by \(2^4\)
=> In some cases it \((n-1) * n * (n+1)^2\) is divisible by \(2^4\) and in some cases it is not
=> We CANNOT answer the problem for sure when n = even

STAT 1: n leaves a remainder 2 when divided by 3
=> n = 3k + 2 (where k is the quotient (integer))
=> n = 2, 5, 8,...
=> in some cases n is odd and in some cases it is even
=> \((n-1) * n * (n+1)^2\) will be divisible by \(2^4\) in some cases and will not be divisible in some cases.
=> NOT SUFFICIENT

STAT 2: n leaves a remainder 1 when divided by 2
=> n = 2t + 1 (where t is the quotient (integer))
=> n = 1, 3, 5,...
=> n = ALWAYS odd
=> \((n-1) * n * (n+1)^2\) will ALWAYS be divisible by \(2^4\)
=> SUFFICIENT

So, Answer will be B
Hope it helps!

Watch the following video to learn the Basics of Remainders