jmc

A 5-digit number can have for its leftmost digit anything from 1 to 9 inclusive, and for each of its next four digits anything from 0 through 9 inclusive. Thus there are $9\times 10\times 10\times 10\times 10=90,000$ 5-digit numbers.

A 5-digit number with no 2 or 3 as a digit can have for its first digit 1 or anything from 4 through 9, and can have for each other digit any of those numbers or 0.

There are $7\times 8\times 8\times 8\times 8=28,672$ such 5-digit numbers. Therefore the number of 5-digit numbers with at least one 2 or one 3 as a digit is $90,000-28,672=\boxed{61,328}.$