Question

The sum of all two digit natural numbers which leave a remainder $5$ when they are divided by $7$ is equal to

• A. 715
• B. 702
• C. 615
• D. 602

Answer 1

Answer: (B)

702

Answer 2

The required natural numbers are $12,19,26,33,40,47,54,61,68,75,82,89,96$ $\therefore$ $Sum=12+19+26+33+40+47+54$ $+61+68+75+82+89+96$ $=702$ Alternate The required natural numbers are $12,\text{ }19,\text{ }26,\text{ }...,96$ Let there are n terms in this series. Then last term $=96=12+(n-1)7$ $\Rightarrow$ $n-1=\frac{84}{7}=12$ $\Rightarrow$ $n=13$ Now, sum of these terms ${{S}_{13}}=\frac{13}{2}[24+12\times 7]=\frac{13}{2}\times 108$ $=13\times 54=702$