AJMO Equal Sums and Invariants: Reason from What Stays the Same

By Ace Achievers · Published 16 October 2026 · Official information checked 14 October 2026

What should we look for before trying every possibility?

The official samples include equal totals from different selections and information about sums of pairs. The common teaching idea is to notice a whole or a repeated contribution. Our models below use new numbers and wording. The fictional points in an example are part of that puzzle, not a claim about AJMO’s competition scoring.

An invariant is a quantity that stays the same under the permitted changes. Do not simply attach that name to a pattern: state what remains unchanged and why. Then use it to reduce the search, while keeping the other conditions visible.

How do complements simplify an equal-score problem?

Now explain why a larger common score is impossible. Completing all six tasks would give the same set, which is excluded. Completing five means omitting one task; different single omitted values cannot give equal totals. Completing three or fewer cannot exceed 4+5+6 = 15. With four tasks completed, two are omitted. The smallest total attainable by two different pairs is 5: the pairs 1+4 and 2+3. Therefore 16 is both achievable and maximal.

The equal-number-of-tasks condition is essential. Equal totals from sets of different sizes would be a different problem. Changing the available values to 1 through 8 gives an unchanged whole of 36; the same two omitted pairs produce a maximum common score of 31 under the same equal-size rules.

How can pair sums reveal a missing total?

Why can we identify the common total? The six pair sums form three opposite pairs, each summing to S. Only one entry is missing, so two complete opposite pairs are still readable. Checking the readable pairs gives the repeated total 42. This is a structural argument, not a guess based on the largest visible entry.

As a check, the six pair sums total 13+17+20+22+25+29 = 126. Each hidden number occurs in exactly three pair sums, so that total is 3S = 3×42. The missing sum is determined even though the individual hidden numbers need not be unique: both 5, 8, 12, 17 and 4, 9, 13, 16 give the same six sums.

How should we check an invariant argument?

State the unchanged quantity, explain every repeated contribution, and verify the restrictions on the objects. Use a complete small-case list if a step relies on a claimed minimum or uniqueness. Ask what would change if a repeated value were permitted, a selection size differed, or more than one pair sum were missing.

These questions build a bridge from getting a number to producing a defensible argument. They are useful before attempting longer AIMO problems, where an unexplained pattern is not enough.