Eight balls, one heavier, two weighings.
The eight balls puzzle: one ball is slightly heavier than the other seven, and you have a balance scale and two weighings. Splitting in half feels natural and fails. The fix is to notice that a balance has three outcomes, not two.
Eight balls look identical. Seven weigh exactly the same; one is slightly heavier, by too little to feel by hand. You have a two-pan balance scale, the kind that tips toward the heavier side and shows no numbers, and you may use it exactly twice.
Find the heavy ball with certainty in two weighings.
The instinctive plan is to halve: four against four, then two against two, then one against one. That is three weighings, and it is what most people say first. Before opening hints, ask what a single weighing actually tells you. A balance does not answer a yes-or-no question. It can tip left, tip right, or stay level, and each of those outcomes is information you paid for. Halving throws one of the three away every time.
A second useful question: how many balls could you possibly resolve in one weighing? Answer that honestly and the two-weighing plan writes itself.
Stuck? Open hints one at a time
Each hint gives away one more layer. Stop the moment something clicks.
Hint 1 a nudge
Do not put every ball on the scale. Balls left on the table are still being tested: if the pans balance, the heavy one is among them.
Hint 2 the key move
One weighing has three outcomes, so it can single out the heavy ball from at most three candidates: one on each pan, one on the table. Two weighings can therefore handle up to nine. Split eight into groups that respect that.
The solution
Reveal the full solution
Split the balls into groups of 3, 3, and 2. Weigh the two groups of three against each other.
- If one pan drops, the heavy ball is among those three. Weigh two of them against each other: a drop names it, a balance means it is the third one.
- If the pans balance, the heavy ball is one of the two on the table. Weigh them against each other.
Two weighings in every branch. The design principle is to make each weighing's three outcomes carry roughly equal loads: after the first weighing you are left with three, three, or two candidates, and each of those is resolvable in one more weighing.
The halving plan fails because a balance with four against four can only tip, never stay level, so it collapses three possible outcomes into two and wastes a third of the weighing's information. Counting states makes this precise: two weighings produce at most 3 times 3 equals 9 distinguishable outcomes, so two weighings can never resolve more than nine balls, and halving reaches only 2 times 2 equals 4 of them. Eight balls sit inside the ternary bound and outside the binary one, which is the entire trick.
What this puzzle is really testing
Whether you count outcomes before designing a procedure. The bound "n weighings distinguish at most 3 to the power n cases" is the same argument that shows a comparison sort needs on the order of n log n comparisons, since each comparison has two outcomes and n factorial orderings must be told apart. The counting sort deep-dive walks that lower-bound argument, and the three-switches puzzle is the same count applied to a light bulb.
Follow-ups to expect
Nine balls, still two weighings (the bound is tight, so the same plan works with 3, 3, 3). Twelve coins where the odd one may be heavier or lighter, in three weighings: 24 possible answers against 27 outcomes, so it is possible but every weighing must be chosen so that all three outcomes stay live, which takes real care. And the interview twist: what if the scale reports the weight difference numerically? Then one weighing carries far more than three outcomes, and the whole analysis changes.