Competition · AMC preparation · step 4 of 4

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

Check each number's parity

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

Find the middle card

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

The "only one even prime" fact is what makes parity decisive — it forces the even slot onto a single card, and the odd visible card (59) is the only one that can take it.

The hidden prime on the card showing 59 must be even, which forces it to be the number 2.

▸ Why?

The three card sums are all equal, so they are one number S and must share a single parity — all even, or all odd.

▸ Why?

The problem sets 44 + h1 = 59 + h2 = 38 + h3, and chaining these equalities makes every card sum the same value S.

Transitivity of equalityIf the first equals the second and the second equals the third, the first equals the third.

▸ Why?

That common parity S must be odd, because an even S would demand two different even primes at once, and those do not exist.

▸ Why?

If S were even, the even visible numbers 44 and 38 would each need an even hidden prime, since even plus even is even.

▸ Why?

An even number is one that has 2 as a factor, and adding two numbers that each carry a factor of 2 keeps that factor, so even plus even stays even.

Even and odd (parity)A whole number is even when it splits into two equal whole groups (a factor of 2), else odd; odd×odd stays odd and even×anything is even.

▸ Why?

There is only one even prime, so 44 and 38 cannot each pair with a different even prime, and the six numbers being all different rules that out.

▸ Why?

Being even means having 2 as a factor, so any even number past 2 has 2 as a proper divisor and is composite — leaving 2 as the sole even prime.

Even and odd (parity)A whole number is even when it splits into two equal whole groups (a factor of 2), else odd; odd×odd stays odd and even×anything is even.

▸ Why?

With S odd, the card showing the odd number 59 is the only one whose hidden prime comes out even; the even cards 44 and 38 keep odd hidden primes.

▸ Why?

Odd plus even is odd and odd plus odd is even, so reaching an odd S needs an even partner for 59 but odd partners for 44 and 38.

▸ Why?

An odd number leaves a remainder of 1 past a factor of 2, so a sum turns out odd exactly when one addend is even and the other odd.

Even and odd (parity)A whole number is even when it splits into two equal whole groups (a factor of 2), else odd; odd×odd stays odd and even×anything is even.

▸ Why?

That even hidden prime on the 59 card can only be the number 2.

▸ Why?

Being even means 2 is a factor, so every even number greater than 2 is divisible by 2 and therefore composite, making 2 the only even prime.

Even and odd (parity)A whole number is even when it splits into two equal whole groups (a factor of 2), else odd; odd×odd stays odd and even×anything is even.

4.OA.B.4Work Backwards

3STEP 3

Find the common sum

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

S = 59 + 2 = 61

4STEP 4

Recover the other two primes

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 hidden primes

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

(h₁ + h₂ + h₃)/3 = (17 + 2 + 23)/3 = 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.

Check each number's parity
Find the middle card
Find the common sum
Recover the other two primes
Average the hidden primes

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Check each number's parity

Find the middle card

Find the common sum

Recover the other two primes

Average the hidden primes

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