Question

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

• A. $86889$
  
• B. $78442$
• C. $85339$
  
• D. $98910$

Answer 1

Answer: (D)

$98910$

Three digit number which leaves remainder $2$ when divided by $5$ is
$102,107,112,\mathrm{117......997}$
This form an AP whose first term is $=102$, $d=5$
Let $997$ is the n th term of AP i.e. $a_n=997$
$a_n=a+(n-1)d$
$997=102+(n-1)5$
$5n=900$
$n=180$
Sum of all three digits which leaves remainder $2$ when divided by $5$ is
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$=\frac{180}{2}[2\times 102(180-1\left)5\right]$
$=90[204+179\times 5]$
$=90[204+895]$
$=90\times 1099$

$=98910$