Question 10:
Pigeonhole principle

Step-by-step answer

1. To minimise the winner's tally while ensuring a unique maximum, give the other candidates as many votes as possible without reaching .
2. Each of the other can receive at most votes, so .
3. Thus .
4. The inequality forces only if equality of the bound is achievable; check: if , total used at most . The two leftover votes can be given without creating a second candidate on (e.g. pad losers still at ). Wait — actually , and spare: if any loser reaches there is a tie. So spare votes must not lift a loser to . Giving both spare to already-filled losers is impossible without exceeding . One may leave votes unassigned? No — all members voted. So we need as well for feasibility with all votes cast...
5. Standard pigeonhole approach for a unique winner: minimise such that the remaining votes can be distributed to candidates with each getting at most . Need .
6. , so .
7. Check : remaining and , feasible (e.g. thirteen get and one gets , or similar).
8. Check : remaining and , impossible without some other candidate getting at least (a tie or worse).
9. Answer: C (). (NESA multiple-choice key: C.)

Common mistake: choosing a familiar formula before checking that its conditions, signs, interval and units match this question.

The paper records 1 total mark for this group. Where the official marking guide provides useful partial-credit criteria, those criteria are stated in the steps above.