Problem 25 · 1985 AJHSME
Stretch
Logic & Word Problems
contrapositivefind-counterexample
Five cards are lying on a table as shown.
P Q
3 4 6
Each card has a letter on one side and a whole number on the other side. Jane said, “If a vowel is on one side of any card, then an even number is on the other side.” Mary showed Jane was wrong by turning over one card. Which card did Mary turn over?
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Answer: A — 3.
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Hint 1 of 3
The rule is one-directional: 'vowel → even'. The ONLY way to break it is to find a single card that is a vowel on one side and an ODD number on the other. So ask: which visible faces could possibly be hiding that forbidden pairing?
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Hint 2 of 3
Sort the five visible faces. P and Q are consonants — the rule says nothing about consonants, so flipping them proves nothing. 4 and 6 are even — even is exactly what the rule allows, so they're safe too. That leaves only the odd-numbered face.
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Hint 3 of 3
To catch a violation, flip the card whose visible side is ODD: if a vowel is hiding behind it, the rule is busted. Which card is that?
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Approach: test only the cards that could falsify the implication
- Breaking 'vowel → even' requires a card that's a vowel paired with an odd number. Check what flipping each face could reveal: P, Q (consonants) — the rule doesn't restrict consonants, useless to flip; 4, 6 (even) — the rule permits any letter behind an even number, also useless; 3 (odd) — if a vowel hides behind it, that's a vowel with an odd number, a direct contradiction.
- Only the odd-numbered card can expose the forbidden vowel-with-odd combination, so Mary turned over 3.
- Why this transfers: to test an 'if A then B' rule, you only ever check the A-cases (is B true?) and the not-B-cases (is A false behind it?). Cases that are already not-A or already B can never disprove it — a sharp filter that saves you from flipping every card.
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