Problem
Exactly one statement is true
Three boxes each carry one written sentence. Exactly one box hides a jewel. Exactly one of the three sentences is true. Find the box with the jewel.
Reasoning · Grade 6-1 Truth and Falsehood
Your answer
Watch out Believing the sentences is tempting, but two are false — test each case instead.
How to solve
Strategy Make a Systematic List — The jewel has only three possible homes, so instead of trying to reason about the sentences directly I make a small table: one row for 'the jewel is in A', 'the jewel is in B', 'the jewel is in C', and one column for each sentence. Filling a row is easy, because once I have decided where the jewel is, every sentence becomes plainly true or plainly false with nothing left to argue about. Then I count the trues in each row: the counting condition says the count must be exactly 1, so any row with a different count is impossible and gets crossed off, and any row that hits exactly 1 is a genuine, fully consistent arrangement. Afterwards I change what I look at -- noticing that box A's sentence and box C's sentence are exact opposites -- which gets the same answer in one line and confirms the table was filled in correctly.
1STEP 1
Read the boxes and set out the three possibilities
Split into three cases by box.
3 possible boxes × 3 sentences = 9 true-or-false judgements
Three rows and three columns is a table a fourth grader can draw and fill in on one line of a notebook, and organising it this way means no possibility can be quietly skipped.
1.MD.C.4Make A Systematic List2STEP 2
Learn how to fill in one row
For each, mark the sentences true or false.
Sorting three sentences into a 'true' pile and a 'false' pile and counting the true pile is classify-and-count, the same thing you do sorting buttons by colour -- the only care needed is that 'empty' means the jewel is elsewhere.
K.MD.B.3Make A Systematic List3STEP 3
Row 1: suppose the jewel is in box A -- contradiction
Box A gives two true sentences — contradiction.
A: T, B: T, C: F → 2 ≠ 1
The contradiction is a plain miscount -- two true sentences where the problem promises one -- so a young solver can see the case is dead without any clever argument.
K.MD.B.3Eliminate PossibilitiesSupposing the jewel is in box A makes the wrong number of statements true, so that case is dead.
Why?
The puzzle promises that exactly one statement is true, so a case producing any other count is refuted on the spot.
🧱One counterexample refutes an always-claimA claim about every case dies the moment you exhibit one case where it fails, and exhibiting is easier than arguing.
Why?
The jewel is in exactly one box, so the three cases cover every possibility and never overlap.
🧱Disjoint cases addSplit the possibilities so no case overlaps another and none is left out, and the case counts simply add.
4STEP 4
Row 3: suppose the jewel is in box C -- contradiction
Box C also gives two true.
A: F, B: T, C: T → 2 ≠ 1
Same test, same kind of failure: counting to two when the rules allow only one is enough to throw the case out, and it needs nothing beyond careful reading.
K.MD.B.3Eliminate Possibilities5STEP 5
Row 2: suppose the jewel is in box B -- everything fits
Box B gives exactly one true.
A: F, B: F, C: T → 1 = 1
Checking the survivor against every rule -- one jewel, one true sentence -- is what turns 'the only one left' into 'the one that works', and it takes no more effort than the rows you just rejected.
K.MD.B.3Eliminate Possibilities6STEP 6
A one-line check using the opposite pair
The opposite pair also points to box B.
1 true among {A, C} → 1 - 1 = 0 true left for B
Looking at a pair of opposites instead of one sentence at a time settles the count before you know anything about the jewel, and then the leftover sentence hands you the answer for free.
K.MD.B.3Change Focus Count The ComplementAnswer
box B
3 × 3 = 9
The answer names one of the three boxes, which is what the question asked for, and it is reached with certainty rather than as a best guess -- important, since only one box may be opened. Both counting conditions are met in the surviving arrangement: exactly one jewel (in B) and exactly one true sentence (box C's). The two rejected rows failed for a concrete, checkable reason -- each had two true sentences instead of one -- so nothing was discarded on a hunch. A useful cross-check is that box B's sentence is a lie about box B itself, and a box that says 'I am empty' while holding the jewel is exactly the kind of trap this puzzle is built on. The one-line argument using the opposite pair A and C reaches box B independently of the table, so two different routes agree.
Takeaway
Pretend the jewel is in each box in turn and count how many sentences come out true -- and if two sentences are exact opposites, one of them is already the true one, so everything else must be false.
- Read the boxes and set out the three possibilities
- Learn how to fill in one row
- Row 1: suppose the jewel is in box A -- contradiction
- Row 3: suppose the jewel is in box C -- contradiction
- Row 2: suppose the jewel is in box B -- everything fits
- A one-line check using the opposite pair