2003 AMC 10B Problem 25

1. List the four-digit numbers ending in $23$: $1023, 1123, 1223, \ldots, 9923$. The first two digits run from $10$ to $99$, so there are $99 - 10 + 1 = 90$ of them.
2. Find the first one divisible by $3$. A number is divisible by $3$ when its digit sum is. $1023$ has digit sum $1 + 0 + 2 + 3 = 6$, so $1023$ works.
3. See what happens as you go up the list. Each step adds $100$, and $100 = 99 + 1$ where $99$ is a multiple of $3$. So each step raises the remainder (when you divide by $3$) by $1$:

1. The pattern yes, no, no repeats all the way up the list. So exactly one out of every three numbers works.
2. The list has $90$ numbers, which is $30$ groups of three, and each group has one number that works. That gives $30$.

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

If you keep adding the same amount, and that amount leaves remainder $1$ when divided by $3$, the remainders cycle through all three possibilities in turn. So in any three numbers in a row of the list, exactly one is a multiple of $3$. This turns the question into a single division, and the same idea works for any fixed ending and any divisor.

Use digit sums. Call the first two digits $a$ and $b$. The number is divisible by $3$ when $a + b + 2 + 3$ is, that is, when $a + b + 5$ is a multiple of $3$. The possible values of $a + b$ go from $1$ (for $10$) to $18$ (for $99$), and the ones that work are $1, 4, 7, 10, 13, 16$. Count two-digit numbers with each digit sum: $1$ (just $10$), $4$ ($13, 22, 31, 40$), $7$ ($16$ to $70$: $7$ numbers), $10$ ($19$ to $91$: $9$ numbers), $13$ ($49$ to $94$: $6$ numbers), $16$ ($79, 88, 97$: $3$ numbers). Total: $1 + 4 + 7 + 9 + 6 + 3 = 30$.