AMC 10 · 2022 · #6

Grade 5 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

prime-numbersprimality-testpattern-recognitiondigit-decompositiondivisibility-rules pattern-recognitionsystematic-enumerationeasier-related-problem ↑ Prerequisites: prime-numbersdivisibility-rules

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Problem

Look at the sequence 121, 11211, 1112111, …, where the n-th term is a string of n ones, then a 2, then another n ones. Among the first ten of these numbers, how many are 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 Look for a Pattern

Computing a₁0 digit-by-digit and primality-testing a 21-digit number is hopeless. Tool #9 (Easier Problem) says: try the small cases n=1, 2, 3 first and look for shared structure. Tool #5 (Pattern) then notices that each small case splits the same way — a string of ones, shifted, plus a string of ones — which is a clean common factor. Tool #7 (Subproblems) breaks the prime question into two: 'find a factorization' and 'check both factors are ＞ 1'. If we can show every term factors into two pieces both bigger than 1, no term is prime.

1STEP 1

Smallest case: a₁ = 121 = 110 + 11 = 11 · (10 + 1) = 11 · 11 — composite.

a₁ = 121 = 11 · 11

2STEP 2

n=2: a₂ = 11211 = 11100 + 111 = 111 · (100 + 1) = 111 · 101 — same shape.

a₂ = 11211 = 111 · 101

3STEP 3

n=3: a₃ = 1112111 = 1111000 + 1111 = 1111 · (1000 + 1) = 1111 · 1001 — pattern holds.

a₃ = 1112111 = 1111 · 1001

4STEP 4

In general a_n = R_n+1 · (10ⁿ + 1): a run of n+1 ones times a 1, then zeros, then a 1.

a_n = R_n+1 · (10ⁿ + 1)

5STEP 5

For every n from 1 to 10 both factors exceed 1: R_n+1 ≥ 11 and 10ⁿ + 1 ≥ 11.

R_n+1 ≥ 11 ＞ 1 and 10ⁿ + 1 ≥ 11 ＞ 1

6STEP 6

Every a_n (n = 1…10) is a product of two factors ＞ 1 → all composite, none prime. Answer (A).

prime count = 0 → (A)

Answer

Spot-check n=1: 11 · 11 = 121 ✓. Spot-check n=2: 111 · 101 = 11211 ✓. Spot-check n=3: 1111 · 1001 = 1112111 ✓. The factorization is verified on three independent cases, and the inequality R_n+1 ≥ 11, 10ⁿ + 1 ≥ 11 holds for all n ≥ 1, so the conclusion covers n = 1 through n = 10 without exception. The other answer choices 1, 2, 3, 4 would each require pointing to a specific n where the factorization fails — which it never does.

💡Key takeaway

This AMC 10 problem only needs Grade 5 place value and factor pairs you already know — split each number at the 2, peel out the common string of ones, and every single one of the ten terms is a product of two whole numbers bigger than 1, so none are prime.

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