TMUA-P2-PRF5-012 — Paper P2 · Following proofs · Level 3

Options

1. A
   If $n$ is odd then $n^3$ is odd (write $n = 2k+1$ and expand); this one direction suffices for the equivalence.
2. B
   Chain: $n$ odd $\Rightarrow$ $n^2$ odd (direct: $(2k+1)^2 = 2m+1$) $\Rightarrow$ $n^3 = n^2 \cdot n$ odd (product of two odds); and $n^3$ odd $\Rightarrow$ $n$ odd (contrapositive: $n$ even $\Rightarrow$ $n^3 = 2(4k^3)$ even). Both directions hold, so the equivalence follows.
3. C
   Check $n = 3$: $27$ is odd, so the equivalence holds for all $n$.
4. D
   If $n^3$ is odd then $n^2$ is odd, and examples show odd cubes come from odd $n$.

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

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