TMUA-P2-PRF2-003 — Paper P2 · Proof by cases · Level 2

Options

1. A
   Write $n^2 = 4q$ always, so the remainder is $0$.
2. B
   If $n$ is a multiple of $4$ the remainder is $0$, otherwise it is $1$.
3. C
   Check $n = 0, 1, 2, 3$ only.
4. D
   If $n$ even, $n = 2k$ gives $n^2 = 4k^2$ (remainder $0$); if $n$ odd, $n = 2k+1$ gives $n^2 = 4k(k+1)+1$ (remainder $1$).
5. E
   If $n$ is even, $n^2$ is divisible by $4$; no other case is possible.

Letters only — the answer key is hidden. Work it out, then check the solution below.

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