Question

Find the sum of all three-digit natural numbers, which on being divided by 5, leave a remainder equal to 4.

• A. 57,270
• B. 96,780
• C. 49,880
• D. 99,270

Answer 1

The correct option is D 99,270

Say 'N' is a three digit number which on being divided by $5$ leaves remainder $=4$.

$\therefore N=5Q+4$.......(i); where Q is an integer.

The first three digit that obeys the equation (i) is $104$

The subsequent three digit numbers exhibiting similar properties will lie at regular intervals of $5$.

Thus, we have an A.P. with the first term as $104$ and a common difference of $5$.

$=>a=104andd=5S_n=\{2a+(n-1)d\}\times \frac{n}{2}$

The last number in this series shall be $999$.

$\therefore 999=104+(n-1)d[T_n=a+(n-1)d]=>\frac{999-104}{5}+1=n=>n=180$

$S_n=\frac{180}{2}\times (2\times 104+179\times 5)=>S_n=90\times (208+895)=>S_n=90\times 1103=>S_n=99270$