Question

If $x=({16}^3+{17}^3+{18}^3+{19}^3)$, then x divided by $70$ leaves a remainder of.

• A. $0$
• B. $1$
• C. $69$
• D. $35$

Answer 1

Answer: (A)

$0$

$x={16}^3+{17}^3+{18}^3+{19}^3$

$(a+b)$ is a factor of $(a^3+b^3)$

$({16}^3+{19}^3)$ is divisible by $16+19=35$

$({17}^3+{18}^3)$ is divisible by $17+18=35$

So, $x={16}^3+{17}^3+{18}^3+{19}^3$ has a factor $35$

Also, $x={16}^3+{17}^3+{18}^3+{19}^3$ is even number.

Since, unit digit $=6+3+2+9=0$ so $x$ has a factor $0.$

We know, $(E+O+E+O=E)$

So, $x$ is completely divisible by $35\times 2=70$ leaves a remainder $0.$