Problem 25 · 2006 AMC 8
Hard
Number Theory
parity-of-prime
Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? (Visible sides: 44, 59, 38.)
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Answer: B — 14.
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Hint 1 of 3
Look at the three visible numbers: 44 and 38 are even, but 59 is odd. Since all three card-sums are EQUAL, that odd one out is the lever — what kind of prime must hide behind it?
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Hint 2 of 3
Almost every prime is odd; the only even prime is 2. Use parity: even + (odd prime) is odd, while odd + (odd prime) is even — so the sums can only all match if one special card hides the 2.
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Hint 3 of 3
Once parity pins down which card hides 2, you get the common sum, then subtract each visible number to recover the other two hidden primes — and check they're actually prime.
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Approach: parity of the visible numbers forces where 2 hides
- The three sums are equal, so they're all the same parity (all odd or all even). Behind 44 and 38 (even), an odd prime gives an odd sum; behind 59 (odd), an odd prime gives an even sum. Those can't match — so something must break the "odd prime" assumption.
- The only escape is the one even prime, 2. To make 59's sum match the others' parity, the 2 must hide behind 59: 59 + 2 = 61 (odd). Then 44 and 38 need odd primes to reach 61 (odd), which is consistent.
- Common sum = 61. Behind 44: 61 − 44 = 17 (prime ✓). Behind 38: 61 − 38 = 23 (prime ✓). All six numbers (44, 59, 38, 2, 17, 23) are different, as required.
- Average of the hidden primes: (2 + 17 + 23) ÷ 3 = 42 ÷ 3 = 14.
- The transferable move: "2 is the only even prime" is a parity wildcard — whenever primes must satisfy an even/odd condition, the 2 is almost always the special case that makes it work. Spot the odd-one-out, and parity tells you where it goes.
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