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

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.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Easier Related Problem

Tool #9 (Easier Problem) first: p and q differ by only 2, so p ≈ q and the cube gap is close to 3q² · 2 = 6q². Setting 6q² ≈ 31106 gives q² ≈ 5184, so q is near √(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 73³ - 71³ directly. Tool #5 (Pattern) uses a³ - b³ = (a-b)(a²+ab+b²) 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.

1STEP 1

Since p−q=2 is tiny, p³−q³ ≈ 6q²; setting 6q² ≈ 31106 gives q² ≈ 5184, so q ≈ 72.

6q² ≈ 31106 → q² ≈ 5184 → q ≈ 72

2STEP 2

The twin primes nearest 72 are (71, 73) — both prime — so q = 71, p = 73.

q = 71, p = 73 (twin primes)

3STEP 3

Check via a³−b³=(a−b)(a²+ab+b²): 73³−71³ = 2·(5329+5183+5041) = 2·15553 = 31106, an exact match.

73³ - 71³ = 2 · (5329 + 5183 + 5041) = 2 · 15553 = 31106 ✓

4STEP 4

Walk 74–78 (all composite: even, ends in 5, or 7×11) and reach 79, the next prime.

74, 75, 76, 77, 78 composite; 79 prime

5STEP 5

Digit sum of 79: 7 + 9 = 16.

7 + 9 = 16

6STEP 6

Among the five choices, 16 is (E).

16 → (E)

Answer

Sanity-check the estimate. 6 · 72² = 6 · 5184 = 31104, very close to 31106 — so q near 72 was the right ballpark. Direct verification: 71³ = 357911 and 73³ = 389017, giving 73³ - 71³ = 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).

💡Key takeaway

This AMC 10 problem only needs Grade 8 square-root estimation you already know — the cube gap forces q ≈ √(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.

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