Facts nobody disputes:
1. The four workers together emptied exactly 120 buckets.
2. Every number of buckets is a whole number.
3. Exactly one of the four statements below is false. The liar is the one who has been selling sap on the side.
The statements:
Walt: “I emptied exactly twice as many buckets as Ira.”
Hank: “Ira and I emptied 41 buckets between us.”
Ira: “I emptied exactly two thirds as many buckets as Dale.”
Dale: “I emptied 10 more buckets than Hank.”
Who is the liar, and how many buckets did each worker empty?
💡 Hint
Try each worker in turn: remove that statement, solve with the rest and the total, and see whether the counts come out whole. One removal behaves differently from the others.
Reveal answer
Ira is the liar. Walt emptied 56 buckets, Hank 13, Ira 28 and Dale 23.
Step-by-step solution
- Use W, H, I and D for the counts. The statements say W = 2I, H + I = 41, 3I = 2D and D = H + 10, and W + H + I + D = 120.
- If Walt lies: H = 41 – I and D = H + 10 = 51 – I. Then 3I = 2D gives 3I = 102 – 2I, so I = 20.4. Not a whole number, so Walt is truthful.
- If Hank lies: W = 2I, and 3I = 2D means D = 1.5I. With D = H + 10, we get H = 1.5I – 10. The total is 2I + (1.5I – 10) + I + 1.5I = 6I – 10 = 120, so I = 21.67. Not whole. Hank is truthful.
- If Dale lies: W = 2I, H = 41 – I and D = 1.5I. The total is 2I + (41 – I) + I + 1.5I = 3.5I + 41 = 120, so I = 22.57. Not whole. Dale is truthful.
- If Ira lies: W = 2I, H = 41 – I, D = 51 – I. The total is 2I + (41 – I) + I + (51 – I) = 92 + I = 120. So I = 28, H = 13, D = 23 and W = 56. All whole, all positive, and they add to 120.
- Check Ira’s own statement: two thirds of Dale’s 23 is about 15.3, not 28, so it is false, as it must be. Ira is the liar. A brute-force search over every way to split 120 buckets among four workers confirms this is the only world.
Why it fools people
The numbers invite you to trust the neat statements and go hunting for the strange one. But a lie does not have to look strange. What exposes Ira is not his claim but the damage it does to the rest: only when his statement is thrown away do the other three agree with the total and with whole numbers. The method is the lesson. When one of several statements is false, test each candidate by removal and let arithmetic, not instinct, do the accusing.
Did you get it?
About This Puzzle
Every March, when the days warm up and the nights stay below freezing, the sugarbush on a Vermont hillside comes alive. Sap drips from the maples into metal buckets, and workers walk the lines each morning to empty them into a gathering tank. At the Hollis Ridge farm, the foreman found that someone had been sneaking sap into a private barrel for a roadside stand. He cannot prove it, so he asks the four workers to state their share. This case is solved with arithmetic alone.
How to Think About It
When exactly one statement is false, you do not have to decide who sounds guilty. You can test each worker in turn: assume that worker is the liar, throw their statement away, and see whether the other three statements plus the total of 120 give whole, positive numbers. Turn each statement into a short equation using a letter for each worker. Substitute until only one letter remains. A suspect is cleared only if the numbers come out as sensible whole counts, and condemned if they do not. Set up all four tests the same way so you can compare them quickly.
Common Traps
The first trap is to solve the first three statements together and stop, as if the order of speaking mattered. The liar can be anyone. The second is to accept any answer that makes the total 120 without checking that every count is a whole number. Fractional answers are a quiet way a wrong suspect gives himself away. The third is to be swayed by Hank’s statement because it uses a nice round figure of 41. Round numbers feel trustworthy, but a lie can contain them too.
Where This Puzzle Comes From
Puzzles in which one person in a group lies belong to a long tradition of logic and liar puzzles, and the arithmetic version asks you to expose the liar through numbers instead of locked rooms. It sits comfortably within the “fair play” habit of Golden Age detective fiction, where every fact the detective uses is shown to the reader. This particular case is an original construction.
Try This Variation
Keep the same four statements, but the foreman recounts the empty buckets and the total is only 100, not 120. Who is the liar now, and how many buckets did each worker empty?
Show the answer to the variation
Ira is still the liar. Without his statement, the other three give a total of 92 plus Ira’s count, so Ira emptied 8 buckets. That makes Hank 33, Dale 43 and Walt 16, and those add to 100. Ira’s claim of two thirds of Dale’s 43 does not hold, and the other three removal tests produce fractions.