2023 AMC 10B Problem 23

The true sum is $S = na + \frac{n(n-1)}{2}d.$ Since one term is off by $1,$ the written total satisfies $222 = S \pm 1,$ so $S = 221$ or $223.$ Also $2S = n\bigl(2a + (n-1)d\bigr),$ so $n$ divides $2S.$ Since $a \ge 1$ and $d \ge 2,$ we have $2S \ge 2n^2,$ hence $n^2 \le S.$ For $S = 223,$ no divisor of $446$ lies between $3$ and $\sqrt{223}.$ For $S = 221 = 13 \cdot 17,$ the only divisor of $442$ in this range is $n = 13.$ Thus $2a + 12d = 34,$ or $a + 6d = 17.$ Since $a$ and $d$ are positive integers with $d \gt 1,$ we get $a = 5,$ $d = 2.$ Then $a + d + n = 5 + 2 + 13 = 20.$ Thus, B is the correct answer.