Question

Find the sum of all $2$-digit numbers which when divided by $5$ leave remainder $1$.

• A. $963$
• B. $863$
• C. $983$
• D. $943$

Answer 1

The correct option is A $963$

The smallest and the largest numbers of two digits, which when divided by $5$ leave remainder $1$ are $11$ and $96$ respectively.

So, the sequence of two digit numbers which when divide by $5$ leave remainder $1$ are $11,16,21,...,96$.

Clearly, it is an AP with first term $a=11$ and common difference $d=5$.
Let there be $n$ terms in this sequence.

Then,
$a_n=96\Rightarrow a+(n-1)d=96\Rightarrow 11+(n-1)\times 5=96\Rightarrow n=18$

Now, Required sum$=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{18}{2}[2\times 11+(18-1)\times 5]$

$=9\times 107=963$.