AMC 10 · 2023 · #23

Grade 8 arithmetic

sequences-arithmeticdivisibility-rulesbound-inequality-then-enumeratesystematic-enumerationlinear-diophantine easier-related-problemsystematic-enumerationcaseworkbound-inequality-then-enumerate ↑ Prerequisites: sequences-arithmeticdivisibility-rules

📏 Long solution 💡 3 insights

Problem

An arithmetic sequence of positive integers has first term a, common difference d ＞ 1, and n ≥ 3 terms. Carl wrote the sum but made one term off by 1, giving total 222. Find a + d + n.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem): the unknown number n of terms is the toughest part, so doubling the sum makes 2S = n(2a + (n-1)d) a clean product of two integers. This turns the problem into 'factor 442 or 446, then check'. Tool #2 (Systematic List) sweeps each factor of 442 = 2 · 13 · 17 and 446 = 2 · 223 as a candidate for n. Tool #6 (Guess and Check) plugs each candidate n into the formula and solves for (a, d). Tool #3 (Eliminate) drops candidates with a ＜ 1 or d ＜ 2 or non-integer.

1STEP 1

Off by ± 1 means the true sum is S = 221 or 223; doubling gives 2S = n · K with K = 2a + (n-1)d, so n divides 2S.

2S = n · (2a + (n-1)d), 2S ∈ {442, 446}

2STEP 2

S = 223 fails: 446 = 2 · 223 (prime), so n ≥ 3 forces n = 223 or 446, both making 2a + (n-1)d far too small.

S=223: 446 = 2 · 223, n ∈ {223, 446} both fail ✗

3STEP 3

So S = 221 = 13 · 17 and 2S = 442 = 2 · 13 · 17; divisors with n ≥ 3 give candidates n ∈ {13, 14, 17, 26, 34, 221, 442}.

2S = 442 = 2 · 13 · 17, n ∈ {13, 14, 17, 26, 34, 221, 442}

4STEP 4

With a ≥ 1, d ≥ 2 the smallest K is 2n, so 2n² ≤ 442 gives n² ≤ 221, i.e. n ≤ 14; only n = 13 and 14 remain.

2n² ≤ 2S = 442 → n ≤ 14

5STEP 5

Test n = 14: K = $\frac{442}{14}$ is not an integer, so reject.

n=14: K =

\frac{442}{14}

∉ Z ✗

6STEP 6

Test n = 13: K = $\frac{442}{13}$ = 34 gives a + 6d = 17, and d ≥ 2 forces the unique positive solution a = 5, d = 2.

n=13: a + 6d = 17 → (a, d) = (5, 2) ✓

7STEP 7

Check: 5, 7, …, 29 (13 terms, d = 2) sums to 221 = 222 − 1, all constraints hold; so a + d + n = 5 + 2 + 13 = 20, choice (B).

a + d + n = 5 + 2 + 13 = 20 → (B)

Answer

Sequence check passes: 5 + 7 + 9 + … + 29 has 13 terms with first 5 and last 29, sum (13(5 + 29))/2 = 221 ✓. All terms are positive integers, d = 2 ＞ 1, n = 13 ≥ 3 — every constraint satisfied. Carl's 222 = 221 + 1 matches the 'off by 1' clue (he could have written 14 instead of 13 for the term that should be 7? Actually any term t written as t + 1 gives +1 — many valid mistakes). The competing case S = 223 was ruled out cleanly. Final answer 20 matches choice (B).

💡Key takeaway

This AMC 10 problem only needs Grade 8 algebra you already know — the off-by-1 clue means the true sum is 221 or 223, so doubling gives n · K = 442 or 446. Factor pairs plus the positivity bound n ≤ 14 leave only n = 13, which forces a = 5, d = 2. Add up: (B) 20.