Answer :

The numbers between 101 and 999 that are divisible by both 2 and 5 are 110, 120, 130,…, 990.

This forms an AP with first term a = 110


and common difference = d = 10


Last term is 990.


Now, number of terms in this AP are given as:


990 = a + (n - 1)d


990 = 110 + (n - 1)10


990 - 110 = 10n - 10


880 + 10 = 10n


890 = 10n


n = 89


There are 89 numbers between 101 and 999 that are divisible by both 2 and 5.


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