What is the sum of all two-digit numbers whic | vedclass

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Question

(C) two-digit number $x$ leaves a remainder of $4$ when divided by $6$ if $x = 6n + 4$ for some integer $n$.For two-digit numbers,$10 \le 6n + 4 \le 9

Vedclass · Accepted Answer

$780$

Vedclass · Answer

(C) two-digit number $x$ leaves a remainder of $4$ when divided by $6$ if $x = 6n + 4$ for some integer $n$.For two-digit numbers,$10 \le 6n + 4 \le 99$.Solving for $n$: $6 \le 6n \le 95$,which gives $1 \le n \le 15.83$.Thus,$n$ can take integer values from $1$ to $15$.The sequence of such numbers is $10, 16, 22, \ldots, 94$.This is an Arithmetic Progression $(AP)$ with first term $a = 10$,last term $l = 94$,and number of terms $n = 15$.The sum $S_n$ of an $AP$ is given by $S_n = \frac{n}{2}(a + l)$.$S_{15} = \frac{15}{2}(10 + 94) = \frac{15}{2}(104) = 15 	imes 52 = 780$.

What is the sum of all two-digit numbers which give a remainder of $4$ when divided by $6$?

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