2023 AMC 10B Problem 23

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 $\boxed{\textbf{(B)}\ 20}$.

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.

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 $\boxed{\textbf{(B)}\ 20}$.