2003 AMC 10B Problem 25

A four-digit number ending in $23$ is $100k + 23$ where $k$ is its first two digits, $10 \le k \le 99$. That is $90$ candidates: $1023, 1123, 1223, \dots, 9923$.

Consecutive candidates differ by $100$, and $100 \equiv 1 \pmod 3$, so as $k$ increases by $1$ the residue of $100k + 23$ mod $3$ increases by $1$. The residues therefore cycle $r, r+1, r+2, r, \dots$, and in any three consecutive candidates exactly one is a multiple of $3$. Since $90$ is a multiple of $3$, exactly $\frac{90}{3} = 30$ candidates are divisible by $3$.

Concretely, $1023 = 3 \cdot 341$ is the first, and the multiples of $3$ are $1023, 1323, 1623, \dots, 9923$ (step $300$): $\frac{9923 - 1023}{300} + 1 = 30$.

The answer is $\boxed{\textbf{(B)}\ 30}$.

An arithmetic progression whose common difference is coprime to $m$ runs through all residues mod $m$ in turn, so among any $m$ consecutive terms exactly one is divisible by $m$. This makes "how many terms are multiples of $3$" a division problem rather than a digit-sum casework problem, and the same idea handles any modulus and any fixed ending.

Digit sums: with thousands digit $a$ ($1$–$9$) and hundreds digit $b$ ($0$–$9$), divisibility by $3$ requires $a + b + 5 \equiv 0$, i.e. $a + b \equiv 1 \pmod 3$. For each $a$, the valid $b$ form one residue class mod $3$ among $0$–$9$: $4$ values when $b \equiv 0$ (that is $a \equiv 1$: $a = 1, 4, 7$) and $3$ values otherwise. Total $3 \cdot 4 + 6 \cdot 3 = 30$.

• Counting $100$ candidates by allowing a leading zero, or arguing about digit sums and missing that a tens digit of $0$ is allowed while a thousands digit of $0$ is not.
• Using step $100$ instead of $300$ in the final count, or miscounting the residue class sizes among $0$–$9$ (four $0$s, three $1$s, three $2$s).