AMC 8 · 2023 · #25

Grade 6 arithmetic

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

sequences-arithmeticlinear-equations-two-vardigit-sum bound-inequality-then-enumeratesystematic-enumerationpattern-recognition ↑ Prerequisites: sequences-arithmeticmulti-digit-arithmetic

📏 Medium solution 💡 4 insights

Problem

Fifteen integers a₁, a₂, …, a₁5 sit on a number line in equally spaced order. We know the first term satisfies 1 ≤ a₁ ≤ 10, the second term satisfies 13 ≤ a₂ ≤ 20, and the last term satisfies 241 ≤ a₁5 ≤ 250. Find the sum of the digits of a₁4.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

Equally spaced integers immediately signal Tool #5 (Look for a Pattern) — the arithmetic-sequence pattern a_n = a₁ + (n-1)d lets us express every term using just a₁ and d. From there, Tool #7 (Identify Subproblems) cleanly splits the work into two smaller hunts: first pin down d, then pin down a₁. Differences like a₁5 - a₂ = 13d erase a₁ from the picture and squeeze d between two bounds. Tool #3 (Eliminate Possibilities) finishes the job — only integer values inside both ranges survive, and only one a₁ makes all three given ranges true.

1STEP 1

Equally spaced integers form an arithmetic sequence, so every term follows the rule a_n = a₁ + (n-1)d.

a_n = a₁ + (n-1)d → a₂ = a₁ + d, a₁4 = a₁ + 13d, a₁5 = a₁ + 14d

2STEP 2

Subtract a₂ from a₁5 to cancel a₁, leaving 221 ≤ 13d ≤ 237.

a₁5 - a₂ = 13d, 241 - 20 ≤ 13d ≤ 250 - 13, 221 ≤ 13d ≤ 237

3STEP 3

Dividing by 13 gives d ∈ {17, 18}; d = 18 forces a₁5 ≥ 253, too big, so d = 17 is the only survivor.

\frac{221}{13}

≤ d ≤

\frac{237}{13}

→ d ∈ {17, 18}; d = 18 → a₁5 ≥ 253 ＞ 250 → d = 17

4STEP 4

With d = 17 the three given ranges collapse onto a single integer: a₁ = 3.

1 ≤ a₁ ≤ 10, -4 ≤ a₁ ≤ 3, 3 ≤ a₁ ≤ 12 → a₁ = 3

5STEP 5

So a₁4 = 3 + 13 × 17 = 224, whose digits give 2 + 2 + 4 = 8, choice (A).

a₁4 = 3 + 13 · 17 = 224, 2 + 2 + 4 = 8 → (A)

Answer

Sanity check: with a₁ = 3 and d = 17, a₂ = 20 (inside [13, 20] ✓), and a₁5 = 3 + 14 · 17 = 3 + 238 = 241 (inside [241, 250] ✓). Notice both endpoints are pressed against the boundary — the constraints really do force the unique solution. The arithmetic step d = 17 is steady (no jumps), all 15 terms are integers between 3 and 241, and a₁4 = a₁5 - d = 241 - 17 = 224 confirms the calculation. Digit sum 2+2+4 = 8 is reasonable for a 3-digit number near 200.

💡Key takeaway

This AMC 8 problem only needs Grade 6 inequality reasoning — combining range conditions to pin down one whole-number answer — that you already know!

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