find the number of natural numbers in 101 and 999 which are divisiblw by both 2 and 5?
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Given
a = 110 {first number to be divided by 2 and 5}
difference d =10
an=990 {last no to be divided by 2 and 5}
an = a + {n - 1}d
990 = 110 + 10n -10
880 + 10 = 10n
890 = 10n
n = 89
∴ number of natural numbers between 101 and 999 is 89.
a = 110 {first number to be divided by 2 and 5}
difference d =10
an=990 {last no to be divided by 2 and 5}
an = a + {n - 1}d
990 = 110 + 10n -10
880 + 10 = 10n
890 = 10n
n = 89
∴ number of natural numbers between 101 and 999 is 89.
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