AMC 10 · 2022 · #13

Grade 8 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

prime-numberstwin-primesdifference-of-cubespolynomial-factoringdigit-sum easier-related-problemguess-and-checksystematic-enumeration ↑ Prerequisites: prime-numberspolynomial-factoring

📏 Medium solution 💡 2 insights

Problem

The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 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.

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Toolkit + CCSS Solution

Understand

Restated: Two prime numbers differ by 2 (twin primes). Their cubes differ by 31106. Find the smallest prime that is larger than both of them, then add the digits of that prime.

Givens: $p > q$, both prime, with $p - q = 2$ (twin primes); $p^3 - q^3 = 31106$; Need the smallest prime greater than $p$, then its digit sum; Answer choices: (A) 8, (B) 10, (C) 11, (D) 13, (E) 16

Unknowns: The two primes $p, q$; The next prime after $p$; The digit sum of that next prime

Understand

Restated: Two prime numbers differ by 2 (twin primes). Their cubes differ by 31106. Find the smallest prime that is larger than both of them, then add the digits of that prime.

Plan

Primary tool: #9 Easier Related Problem

Secondary: #6 Guess and Check, #5 Look for a Pattern, #2 Make a Systematic List, #3 Eliminate Possibilities

Tool #9 (Easier Problem) first: $p$ and $q$ differ by only 2, so $p \approx q$ and the cube gap is close to $3q^2 \cdot 2 = 6q^2$. Setting $6q^2 \approx 31106$ gives $q^2 \approx 5184$, so $q$ is near $\sqrt{5184} = 72$. That estimate turns a giant cube-difference equation into the easy question "which twin prime sits near 72?" Tool #6 (Guess and Check) then verifies the natural candidate $q = 71, p = 73$ by computing $73^3 - 71^3$ directly. Tool #5 (Pattern) uses $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ to make that arithmetic clean. Tool #2 (Systematic List) finishes the job: walk 74, 75, 76, 77, 78 and rule each out, leaving 79 as the next prime.

Execute — Answer: E

#9 Easier Related Problem 8.EE.A.2 Step 1

Estimate the size of $p$ and $q$.
Since $p - q = 2$ is tiny compared to the size of $p$ and $q$, we have $p^3 - q^3 \approx 6q^2$ (this is roughly $2 \cdot 3q^2$, the derivative-style estimate).
Setting $6q^2 \approx 31106$: $q^2 \approx 5184$, so $q \approx \sqrt{5184} = 72$.

$$6q^2 \approx 31106 \;\Rightarrow\; q^2 \approx 5184 \;\Rightarrow\; q \approx 72$$

💡 Grade 8 square roots: turning a giant cube-difference into a small square-root estimate shrinks the search to "primes near 72."

#6 Guess and Check 4.OA.B.4 Step 2

Pick the twin-prime pair near 72.
The closest twin primes to 72 are $(71, 73)$ — both are prime, and they differ by 2.
(Check: 71 is prime — not divisible by 2, 3, 5, 7.
73 is prime — same checks.) So our guess is $q = 71, p = 73$.

$$q = 71, \; p = 73 \quad (\text{twin primes})$$

💡 Grade 4 primality: with primes near 72, check 71 and 73 for divisors; both pass.

#5 Look for a Pattern 6.EE.A.3 Step 3

Verify with the cube-difference identity.
Use $p^3 - q^3 = (p - q)(p^2 + pq + q^2)$.
With $p - q = 2$, $p^2 = 5329$, $pq = 71 \cdot 73 = 5183$, $q^2 = 5041$: $p^2 + pq + q^2 = 5329 + 5183 + 5041 = 15553$.
Multiply: $2 \cdot 15553 = 31106$.
Matches exactly.

$$73^3 - 71^3 = 2 \cdot (5329 + 5183 + 5041) = 2 \cdot 15553 = 31106 \;\checkmark$$

💡 Grade 6 algebraic identity: the difference-of-cubes pattern $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ avoids computing two giant cubes.

#2 Make a Systematic List 4.OA.B.4 Step 4

Find the next prime after 73.
Walk forward and rule each candidate out.
74 even; 75 ends in 5 (div by 5); 76 even; 77 = 7 $\times$ 11; 78 even.
79 — check 2, 3, 5, 7 (since $\sqrt{79} < 9$): not divisible by any.
So 79 is the next prime.

$$74, 75, 76, 77, 78 \text{ composite}; \; 79 \text{ prime}$$

💡 Grade 4 primes: list the candidates and knock them out with divisibility shortcuts (even, ends in 5, sums to a multiple of 3).

#2 Make a Systematic List 2.NBT.B.5 Step 5

Compute the digit sum of 79: $7 + 9 = 16$.

$$7 + 9 = 16$$

💡 Grade 2: digit sum is just one addition once the digits are pulled out.

#3 Eliminate Possibilities 6.EE.B.5 Step 6

Match 16 to the answer choices: that is (E).

$$16 \;\Rightarrow\; \textbf{(E)}$$

💡 Final compare against the five options — only (E) lands on 16.

[1] #9 8.EE.A.2 Estimate the size of $p$ and $q$. Since $p - q = 2$ is tiny compared to the size

[2] #6 4.OA.B.4 Pick the twin-prime pair near 72. The closest twin primes to 72 are $(71, 73)$ —

[3] #5 6.EE.A.3 Verify with the cube-difference identity. Use $p^3 - q^3 = (p - q)(p^2 + pq + q^

[4] #2 4.OA.B.4 Find the next prime after 73. Walk forward and rule each candidate out. 74 even;

[5] #2 2.NBT.B.5 Compute the digit sum of 79: $7 + 9 = 16$.

[6] #3 6.EE.B.5 Match 16 to the answer choices: that is (E).

Review

Reasonableness: Sanity-check the estimate. $6 \cdot 72^2 = 6 \cdot 5184 = 31104$, very close to $31106$ — so $q$ near 72 was the right ballpark. Direct verification: $71^3 = 357911$ and $73^3 = 389017$, giving $73^3 - 71^3 = 31106$ exactly. Both 71 and 73 are well-known twin primes. The next prime 79 is famously the first prime past the "prime gap" at 73 (since 74-78 are all composite). Digit sum $16$ matches choice (E).

Alternative: Tool #13 (Algebra): set $p = q + 2$ and substitute into $p^3 - q^3 = 31106$. Expand to $6q^2 + 12q + 8 = 31106$, then $q^2 + 2q = 5183$, complete the square to $(q+1)^2 = 5184 = 72^2$, giving $q = 71$. Same twin-prime pair, but it requires expanding a cube and completing the square; the estimate-and-check path stays in plain arithmetic.

CCSS standards used (min grade 8)

2.NBT.B.5 Fluently add and subtract within 100 (Adding the digits $7 + 9 = 16$ to finish the problem.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Confirming 71 and 73 are prime, then walking 74-79 to find the next prime.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Using the difference-of-cubes identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ to verify $73^3 - 71^3 = 31106$ without computing two giant cubes.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Matching the digit sum 16 to the answer choices and selecting (E).)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Estimating $q \approx \sqrt{5184} = 72$ from the approximation $6q^2 \approx 31106$.)

⭐ This AMC 10 problem only needs Grade 8 square-root estimation you already know — the cube gap forces $q \approx \sqrt{5184} = 72$, the twin-prime pair nearby is 71 and 73, and the next prime after 73 is 79 with digit sum $7+9=16$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 8)

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