Find the numbers of natural number between 101 & 999 which are divisible by both 2 & 5
Answers
Answered by
1
the least common multiple of 2 and 5 are 10 so all the numbers ending with 0 will be divisible by 2 and 5 both so the number will start from 110,120,130......990. It forms an AP so we will find the nth term.
A_n = a + (n-1) × d
990= 110 + (n-1) × 10
990= 110 + 10n -10
990 - 100 = 10n
n = 890 ÷ 10
n = 89
hence there are 89 numbers divisible by 2 and 5 between 101 and 999
A_n = a + (n-1) × d
990= 110 + (n-1) × 10
990= 110 + 10n -10
990 - 100 = 10n
n = 890 ÷ 10
n = 89
hence there are 89 numbers divisible by 2 and 5 between 101 and 999
Similar questions