Three Ages, a Sum She Cannot Use, and 2, 2 and 9

Three children. Their ages, in whole years, multiply to 36. Someone is told the sum of the three ages, thinks about it, and says she cannot name them. She is then told that there is an eldest child, meaning one child older than both of the others, and she answers at once. The ages are 2, 2 and 9.

Most people who meet this puzzle decide it is broken, on the grounds that the sum is never revealed to the reader. The complaint is reasonable and it is also the trap. The sum is not the information you were given. Her failure to answer is.

What the two sentences actually say

Both clues need reading slowly. The first says that a person who knows the sum still cannot identify the ages. For that to be possible, the sum has to be compatible with more than one triple of whole numbers whose product is 36. Any sum belonging to a single triple would have given her the answer immediately and she would not have hesitated. So her hesitation is a statement about the sum: whatever it is, somebody else has it too.

The second clue says nothing about any number. It describes the shape of the answer. To speak of the eldest child presupposes that exactly one child holds that position. If two children were the same age and both older than the third, there would be no single eldest and the sentence would be false.

That second reading is not a convenience adopted to make the sums come out. Allow the eldest to mean one of the joint eldest and the puzzle acquires two answers instead of one, which is the usual sign that the strict reading was intended.

The eight triples

Everything after that is bookkeeping. Write 36 as a product of three positive integers in every possible way, counting rearrangements as the same solution, and there are eight ways. Their sums are

Six of those values appear once each. One value appears twice.

Why the sum has to be 13

Now put the first clue to work. Had she been told 38, or 21, or 16, or 14, or 11, or 10, she would have held exactly one triple and answered on the spot. She did not answer. So the sum she was told is 13, and 13 is the only value her silence permits.

Two triples, one sum, one product. This is the pair the second clue has to separate, and arithmetic has nothing left to separate them with. Every number you can compute from one of them you can compute from the other.

What the eldest rules out

Look at the two survivors as multisets rather than as sums. In $(1,6,6)$ the largest age is 6 and two children have it, so there is no eldest, only two joint eldest and a youngest. In $(2,2,9)$ the largest age is 9 and exactly one child has it. The sentence about an eldest is therefore true of one candidate and false of the other.

Notice how little arithmetic the last step used. Nothing was added and nothing was multiplied. The clue counted the multiplicity of a maximum, which is a fact about the multiset and not about 36 at all.

Neither clue does anything alone

The order of the clues carries weight, and the cheapest way to see it is to apply each one by itself. Ask which triples have a strict eldest and seven of the eight qualify, since only $(1,6,6)$ ties at the top. That clue on its own eliminates a single candidate, which is close to nothing. Ask instead which triples carry a shared sum and only $(1,6,6)$ and $(2,2,9)$ qualify, because 13 is the sole repeated value. Two candidates. Then the weak clue, worthless in isolation, cuts those two down to one.

Reverse the clues and the puzzle still resolves, because a conjunction does not care about order. What changes is the storytelling. The version that reads well puts the weak clue second, since a condition that kills one candidate out of eight feels like padding, while the same condition applied to a shortlist of two feels like the answer.

Why “not enough information” is false

The answer heard most often in a room is that the problem is underdetermined. That answer is wrong rather than merely cautious. A sentence of the form I know the sum and I still cannot tell is hard data. It eliminates six of the eight triples in one stroke, which is more than any explicit number in the problem manages.

Reasoning from somebody else’s ignorance takes practice, because the useful sentence never mentions the quantity it constrains. She talks about herself and the constraint lands on the sum. A whole family of puzzles is built on that move, the sum and product puzzle being the one that has been studied hardest.

Where the construction stops working

Replace 36 by another product and the puzzle usually collapses, because it needs two separate coincidences at the same time. The sum multiset must have exactly one repeated value, so that her silence pins the sum. And among the triples sharing that value, exactly one must have a strict maximum, so that the eldest clue finishes the job.

Below 400 only three products manage both. With 36 the answer is 2, 2 and 9. With 72 the shared sum is 14, held by $(2,6,6)$ and $(3,3,8)$, and the answer is 3, 3 and 8. With 225 the shared sum is 31, held by $(1,15,15)$ and $(3,3,25)$, giving 3, 3 and 25. All three winning pairs have the same shape: one candidate ties at the top, the other ties at the bottom.

The two ways of failing are worth seeing once. Take 96, whose only repeated sum is 21, held by $(1,8,12)$ and $(2,3,16)$. Both have a strict eldest, so the second clue separates nothing and the question has no answer. Take 144, whose sums repeat twice over, at 19 and again at 17. Her silence no longer pins the sum, because two different sums are compatible with it, and the first clue loses its edge before the second one gets a turn.

So 36 is not an arbitrary friendly number picked because it factors tidily. It sits on a short list, and asked about almost any other product the puzzle would either have no answer or several.

Sources and further reading

• Writing an integer as an unordered product of factors — Multiplicative partition
• The best studied puzzle built on someone else’s admitted ignorance — Sum and Product Puzzle
• The logic of reasoning about what other people know — Common knowledge (logic)
• Divisors and the factorisations they generate — Divisor

Every count here was produced by walking all eight triples rather than by hand, which is how the seven with a strict eldest and the six with a private sum were separated. Hand counts of those two quantities disagree with each other more often than you would expect.