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| 1. |
Find the number of natural numbers between 101 and 999 which are devisible by both 2 and 5 |
| Answer» we know that 1st number div by 2 and 5 = 110thus a = 110d = 10 (2 x 5 = 10)last number div by 2 and 5 = an = 990now we know that an = a + (n-1)d990 = 110 + (n-1)10880 = 10n - 10890 = 10nn = 89thus number of numbers between 101 and 999 are 89 | |