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How many two-digit numbes leave the remainder 1 when divided by 5? Find the sum of all these numbers. |
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Answer» Solution :The first two-digit number divisible by 5 is `5xx2=10`. The LAST two digit number divisible by 5 is `5xx9=95` ( `:'5xx20=100` is a three digig number) We want two-digit numbers which would LEAV REMAINDER 1 when dividied by 5. `:.` the first two-digit number is `10+1=11` and the last one is `95+1=96` Here `t_(1)=a=11` and `t_(n)=96, d=5` `t_(n)=a+(n-1)d`....(FORMULA) `:.96=11+(n-1)xx5` ...(Substituting the values) `:.96-11=(n-1)xx5` `:.85=(n-1)xx5` `:.n-1=85/5:.n-1=17:.n=17+1:.n=18` Now `S_(n)=n/2[a+l]` `:.S_(18)=18/2(11+96)` .........( `:' a=11` and `t_(18)=96`) `-9xx107` `=963` Ans. There are 18 two-digit numbers which leave remainder 1 when divided by 5. The sum of all these number is 963. |
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