2006 AMC 10A Problem 21

"At least one" is hard to count directly, because a number might have one, two, three or four such digits. It is much easier to count the numbers that have no $2$ and no $3$, then take those away from all four-digit numbers.

1. All four-digit numbers. They run from $1000$ to $9999$: that is $9000$ numbers. (Or: $9$ choices for the first digit, since it cannot be $0$, and $10$ for each of the others: $9 \times 10 \times 10 \times 10 = 9000$.)
2. Numbers with no 2 and no 3.
- First digit: not $0$, not $2$, not $3$. That leaves $1, 4, 5, 6, 7, 8, 9$: $7$ choices.
- Each of the other three digits: anything except $2$ and $3$: $8$ choices.
- Total: $7 \times 8 \times 8 \times 8 = 7 \times 512 = 3584$.
3. Take away. $9000 - 3584 = 5416$.

The answer is $\boxed{\textbf{(E)}\ 5416}$.

"At least one" questions often break into many overlapping cases, but the opposite, "none at all," is a single multiplication. Count all, count none, subtract. The only care needed is the first digit, which can never be $0$.

Estimate. Each digit is a $2$ or a $3$ about one time in five, so a number avoids both about $\left(\frac45\right)^4 \approx 0.41$ of the time. So roughly $59\%$ of $9000$, around $5300$, should work. Only choice (E) is close.