The sum of all two-digit numbers which,when d | vedclass

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(C) The two-digit numbers that leave a remainder of $1$ when divided by $4$ are of the form $4k + 1$.The smallest two-digit number of this form is $13

Vedclass · Accepted Answer

$1210$

Vedclass · Answer

(C) The two-digit numbers that leave a remainder of $1$ when divided by $4$ are of the form $4k + 1$.The smallest two-digit number of this form is $13$ $(4 	imes 3 + 1)$ and the largest is $97$ $(4 	imes 24 + 1)$.This forms an arithmetic progression $(AP)$ with first term $a = 13$,last term $l = 97$,and common difference $d = 4$.Using the formula for the $n$-th term: $l = a + (n - 1)d$$97 = 13 + (n - 1)4$$84 = (n - 1)4$$n - 1 = 21$$n = 22$The sum of the $AP$ is given by $S_n = \frac{n}{2}(a + l)$.$S_{22} = \frac{22}{2}(13 + 97) = 11(110) = 1210$.

The sum of all two-digit numbers which,when divided by $4$,yield unity as a remainder is:

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