Éveriste listed all of the positive integers from 1 to 90 . He then crossed out all of the multiples of 3 from the list. Of the remaining numbers, he then crossed out all of the multiples of 5 . How many numbers were not crossed out?
Since 90 is a multiple of 3 , there are $\frac{90}{3}=30$ multiples of 3 in the list.

Since 90 is a multiple of 5 , there are $\frac{90}{5}=18$ multiples of 5 in the list.

A multiple of 5 is a multiple of 3 exactly when it is a multiple of 15 , which means the multiples of 15 were already crossed out when the multiples of 3 were crossed out.

There are $\frac{90}{15}=6$ multiples of 15 in the list, so this means there are $18-6=12$ multiples of 5 remaining in the list once the multiples of 3 have been crossed out.

Therefore, there are $90-30-12=48$ numbers in the list once all of the multiples of 3 and 5 have been crossed out.

ANSWER: 48