2023 AMC 10B Problem 23

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

The sum formula makes $n$ a divisor of $2S$, and the constraints $a \ge 1$, $d \ge 2$ give a cheap upper bound $n \le \sqrt{S}$. Factoring the two candidate values of $2S$ then leaves almost nothing to check. Problems of the form "someone made a small error" are solved by listing the possible true values and letting divisibility choose among them.

• Using 222 as the actual sum of the sequence; Carl's total is off by 1, so the real sum is 221 or 223.
• Forgetting $d > 1$, which admits the solution $n = 17$, $d = 1$, $a = 5$ (or $n = 13$, $d = 1$, $a = 11$) and a wrong total.