AMC 8 · 2006 · #25

Grade 6 number-theory

Archetype Pair-Sum Invariant └ Sub-archetype Uniform Pair-Sum Recovery

prime-numbersparitymean-median-mode-rangeset-partition identify-subproblemscasework ↑ Prerequisites: prime-numbersparity

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

Three cards lie on a table showing 44, 59, and 38. The hidden side of each card holds a prime number, and the three card totals (visible + hidden) are all the same. Find the average of the three hidden primes.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Find an Invariant

The actual hidden primes are unknown, but their parity (odd or even) is locked down by the visible numbers. Tool #11 (Find an Invariant) catches the parity fact: 44 + h₁ and 38 + h₃ share their parity with h₁ and h₃, while 59 + h₂ flips the parity of h₂. Since all three sums must be equal, they must share one parity — and primes have only one even option (2), which forces exactly one hidden side. Tool #4 (Introduce a Variable) names the common sum S so we can recover the other two primes by subtraction.

1STEP 1

Read each visible number's parity: 44 and 38 are even, but 59 is odd — the odd one out drives everything.

44 (even), 59 (odd), 38 (even)

2STEP 2

If 59's hidden prime were odd, two cards would need even primes — but 2 is the only even prime, so the 2 must sit on the 59 card.

h₂ even, and the only even prime is 2 → h₂ = 2

3STEP 3

With 2 placed on the 59 card, the common sum falls out directly: 59 + 2 = 61.

S = 59 + 2 = 61

4STEP 4

Subtract each visible number from 61: 61 - 44 = 17 and 61 - 38 = 23, both prime as required.

h₁ = 61 - 44 = 17 (prime), h₃ = 61 - 38 = 23 (prime)

5STEP 5

Average the three hidden primes: $\frac{17 + 2 + 23}{3}$ = $\frac{42}{3}$ = 14.

\frac{h₁ + h₂ + h₃}{3}

\frac{17 + 2 + 23}{3}

\frac{42}{3}

= 14 → (B)

Answer

Verify all six numbers are different and primes are primes: visible {44, 59, 38}, hidden {17, 2, 23} — six distinct numbers, and 17, 2, 23 are all prime. Each card sum: 44 + 17 = 61, 59 + 2 = 61, 38 + 23 = 61 — all equal, as required. The average 14 also lands inside the answer choices {13, 14, 15, 16, 17} and is plausible: the three primes {2, 17, 23} have mean near the middle of 2 and 23, which is 12.5, and the heavier weight from 17 and 23 pulls the average up to 14.

💡Key takeaway

Parity is the lock — the three card sums must share one parity, and 2 is the only even prime in town. That single fact pins the 2 to the odd card (59), and the rest of the primes drop out by subtraction. Average them and you get 14.

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