How many two digit numbers divisible by neither 3 nor 5?

Answer 1

48.

The first two digit number is 10, the last two digit number is 99, so there are 99 - 10 + 1 = 90 two digit numbers

→ 10 ÷ 3 = 31/3 → first two digit number divisible by 3 is 4 x 3 = 12

→ 99 ÷ 3 = 33 → last two digit number divisible by 3 is 33 x 3 = 99

→ 33 - 4 + 1 = 30 two digit numbers divisible by 3

→ 10 ÷ 5 = 2 → first two digit number divisible by 5 is 2 x 5 = 10

→ 99 ÷ 5 = 194/5 → last two digit number divisible by 5 is 19 x 5 = 95

→ 19 - 2 + 1 = 18 two digit numbers divisible by 5

→ 30 + 18 = 48 two digit numbers divisible by 3 or 5 OR BOTH.

The numbers divisible by both are multiples of their lowest common multiple: lcm(3, 5) = 15, and have been counted twice, so need to be subtracted from the total

→ 10 ÷ 15 = 010/15 → first two digit number divisible by 15 is 1 x 15 = 15

→ 99 ÷ 15 = 69/15 → last two digit number divisible by 15 is 6 x 15 = 90

→ 6 - 1 + 1 = 6 two digit numbers divisible by 15 (the lcm of 3 and 5)

→ 48 - 6 = 42 two digit numbers divisible by 3 or 5.

→ 90 - 42 = 48 two digit numbers divisible by neither 3 nor 5.