AMC 8 · 2006 · #25
Grade 6 number-theoryProblem
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?
Pick an answer.
AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Visible numbers: $44$, $59$, $38$ on three separate cards; Hidden numbers are prime, and all six numbers (visible and hidden) are different; The two-number sum on each card is the same constant $S$; Answer choices: (A) $13$, (B) $14$, (C) $15$, (D) $16$, (E) $17$
Unknowns: The three hidden primes $h_1$, $h_2$, $h_3$ and their average $\dfrac{h_1 + h_2 + h_3}{3}$
Understand
Restated: 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.
Givens: Visible numbers: $44$, $59$, $38$ on three separate cards; Hidden numbers are prime, and all six numbers (visible and hidden) are different; The two-number sum on each card is the same constant $S$; Answer choices: (A) $13$, (B) $14$, (C) $15$, (D) $16$, (E) $17$
Plan
Primary tool: #11 Find an Invariant
Secondary: #4 Introduce a Variable
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: $44 + h_1$ and $38 + h_3$ share their parity with $h_1$ and $h_3$, while $59 + h_2$ flips the parity of $h_2$. 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.
Execute — Answer: B
4.OA.B.4 Step 1 - Read off the parity of each visible number.
- The constant card sum $S$ has the same parity on every card.
💡 Parity is the unchanging feature here. Even $+$ even is even and even $+$ odd is odd, so each hidden prime's parity is forced by the visible side and by $S$.
4.OA.B.4 Step 2 - Use the invariant to identify the middle card.
- If $h_2$ were odd, then $59 + h_2$ would be even, so $S$ would be even.
- Then $44 + h_1$ even forces $h_1$ even, and $38 + h_3$ even forces $h_3$ even — meaning two hidden primes are even.
- But $2$ is the only even prime, and the six numbers must all differ, so we can't have two even hidden primes.
- So $h_2$ must be the even one.
💡 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.
6.EE.A.2 Step 3 - Name the common sum $S$ and read it off the middle card.
- With $h_2 = 2$, the card sum is $59 + 2$.
💡 Once one card's hidden side is known, the constant sum drops out for free — and that constant unlocks the other two.
6.EE.B.7 Step 4 - Recover the other two hidden primes by subtracting each visible number from $S$.
- Check that both results are prime, as required.
💡 $h_1 = S - 44$ and $h_3 = S - 38$ are one-step equations; the prime check confirms the parity argument really did land on a valid configuration.
6.SP.B.5 Step 5 Average the three hidden primes.
💡 With all three primes pinned down, the average is just their sum divided by $3$.
4.OA.B.4 Read off the parity of each visible number. The constant card sum $S$ has the sa 4.OA.B.4 Use the invariant to identify the middle card. If $h_2$ were odd, then $59 + h_2 6.EE.A.2 Name the common sum $S$ and read it off the middle card. With $h_2 = 2$, the car 6.EE.B.7 Recover the other two hidden primes by subtracting each visible number from $S$. 6.SP.B.5 Average the three hidden primes. Review
Reasonableness: 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 $14$.
Alternative: Tool #6 (Guess and Check): try the only candidate for which card hides $2$. If $h_1 = 2$, then $S = 46$, so $h_2 = 46 - 59 = -13$, not a positive prime — reject. If $h_3 = 2$, then $S = 40$, so $h_2 = 40 - 59 = -19$ — reject. Only $h_2 = 2$ survives, giving $S = 61$, $h_1 = 17$, $h_3 = 23$, and average $14$. Same answer (B), reached by ruling out the other two slots one at a time.
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs, recognize multiples, and determine whether a whole number is prime or composite (Using the parity of each visible number and the fact that $2$ is the only even prime to force which card hides the even prime.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the common card sum $S$ so each hidden prime can be written as $S$ minus a visible number.)6.EE.B.7Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ (Solving $44 + h_1 = 61$ and $38 + h_3 = 61$ to recover $h_1 = 17$ and $h_3 = 23$.)6.SP.B.5Summarize numerical data sets, including reporting the number of observations and measures of center (Averaging the three recovered primes: $(2 + 17 + 23) \div 3 = 14$.)
⭐ 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 $14$.
⭐ 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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