What is the sum of all the $2$ digit numbers which leaves a remainder of $3$ when divided by $5$ ?

Note: $2$ digit numbers don't include numbers from $0$ to $9$ .

The answer is 999.

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The numbers described in the problem are $13,18,23,28,...,98$ . There are $18$ terms in all, whose sum is

$\dfrac{18}{2}(13+98)=9\times 111=\boxed {999}$ .