2023 AMC 10B Problem 23

P21-25~7 minPrint

Sequences & Series Diophantine Equations Divisibility & Factors Bound the quantity above/below or estimate to pin it down Split into exhaustive cases and handle each

An arithmetic sequence of positive integers has $n\ge3$ terms, initial term $a$ , and common difference $d>1$ . Carl wrote down all the terms in this sequence correctly except for one term, which was off by $1$ . The sum of the terms he wrote down was $222$ . What is $a+d+n$ ?

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Key insight

The real sum is 221 or 223. The number of terms divides twice the sum and is at most 14, since steps are at least 2. Only 13 fits.

Solution

Carl's total is off by exactly $1$ , so the real sum $S$ is either $221$ or $223$ . Use the sum formula to see which one can work.

1. The sum formula. The terms are $a, a + d, \ldots, a + (n-1)d$ . Adding the first and last gives $2a + (n-1)d$ , and there are $n$ terms, so
$2S = n\,\big(2a + (n-1)d\big).$
In particular, $n$ must divide $2S$ .
2. The number of terms cannot be large. The smallest start is $a = 1$ and the smallest step is $d = 2$ , so the smallest possible sum with $n$ terms is $1 + 3 + 5 + \cdots$ , which is $n^2$ . So $n^2 \le S \le 223$ , which means $n \le 14$ (since $15^2 = 225$ is too big).
3. Try $S = 223$ . Then $2S = 446 = 2 \cdot 223$ , and $223$ is prime. The divisors of $446$ are $1, 2, 223, 446$ . None of them is between $3$ and $14$ . So $S = 223$ is impossible.
4. Try $S = 221$ . Then $2S = 442 = 2 \cdot 13 \cdot 17$ . Its divisors are $1, 2, 13, 17, 26, 34, 221, 442$ . The only one between $3$ and $14$ is $13$ , so $n = 13$ .
5. Find $a$ and $d$ . With $n = 13$ : $442 = 13\,(2a + 12d)$ , so $2a + 12d = 34$ and
$a + 6d = 17.$
With $d \ge 2$ : $d = 2$ gives $a = 5$ . If $d = 3$ , then $a = -1$ , which is not positive. So $a = 5$ and $d = 2$ .
6. Check. The sequence is $5, 7, 9, \ldots, 29$ , with $13$ terms. Its sum is $13 \times 17 = 221$ (there are $13$ terms and the middle one is $17$ ). Writing one term $1$ too big gives $222$ .

So $a + d + n = 5 + 2 + 13 = 20$ .

The answer is $\boxed{\textbf{(B)}\ 20}$ .

Why this works

The sum formula says the number of terms divides twice the sum, and the rules "positive terms" and "step at least $2$ " put a ceiling on how many terms there can be. With only two possible sums to try, factoring them leaves just one choice. When a problem says someone made a small error, list the possible true values and test each.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

One term was off by $1$ , so the true sum $S$ is $221$ or $223$ . For an arithmetic sequence,
$2S = n\big(2a + (n-1)d\big).$
Because $d \ge 2$ and $a \ge 1$ , the second factor satisfies $2a + (n-1)d \ge 2 + 2(n-1) = 2n$ , so $2S \ge 2n^2$ and $n \le \sqrt{S} < 15$ .

Case $S = 223$ : $2S = 446 = 2 \cdot 223$ , and $223$ is prime. The only divisors of $446$ below $15$ are $1$ and $2$ , but $n \ge 3$ . No solution.

Case $S = 221$ : $2S = 442 = 2 \cdot 13 \cdot 17$ . Its divisors below $15$ are $1, 2, 13$ , so $n = 13$ . Then $2a + 12d = 34$ , i.e. $a + 6d = 17$ . With $d \ge 2$ : $d = 2$ gives $a = 5$ ; $d \ge 3$ makes $a$ negative. So $a = 5$ , $d = 2$ , $n = 13$ .

Check: the sequence $5, 7, 9, \dots, 29$ has sum $13 \cdot 17 = 221$ , and one wrong term makes $222$ .

Therefore $a + d + n = 5 + 2 + 13 = 20$ .

The answer is $\boxed{\textbf{(B)}\ 20}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)