Questions & Answers

Question

Answers

A. $0$

B. $1$

C. $69$

D. $35$

Answer
Verified

Let us rewrite the given equation as $x = {16^3} + {19^3} + {17^3} + {18^3}$. Now we will make one group of the first two numbers and one group of the last two numbers. That is, $x = \left( {{{16}^3} + {{19}^3}} \right) + \left( {{{17}^3} + {{18}^3}} \right) \cdots \cdots \left( 1 \right)$.

As we know that the factorization of ${a^3} + {b^3}$is given by ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$. Therefore, $\left( {a + b} \right)$ is one factor of ${a^3} + {b^3}$. That means ${a^3} + {b^3}$ is divisible by $\left( {a + b} \right)$. Let us use this fact in the equation $\left( 1 \right)$. Therefore, we can say that $\left( {{{16}^3} + {{19}^3}} \right)$ is divisible by $16 + 19$ and $\left( {{{17}^3} + {{18}^3}} \right)$ is divisible by $17 + 18$. That is, $\left( {{{16}^3} + {{19}^3}} \right)$ is divisible by $35$ and $\left( {{{17}^3} + {{18}^3}} \right)$ is also divisible by $35$. Therefore, we can say that $35$ is a factor of $x = \left( {{{16}^3} + {{19}^3}} \right) + \left( {{{17}^3} + {{18}^3}} \right)$.

Note that here the unit digit of ${16^3}$ and ${18^3}$ is even number $\left( E \right)$ and the unit digit of ${17^3}$ and ${19^3}$ is odd number $\left( O \right)$. Therefore, we can say that unit digit of $x = \left( {{{16}^3} + {{19}^3}} \right) + \left( {{{17}^3} + {{18}^3}} \right)$ will be an even number $\left( E \right)$ because the sum of two even numbers is even and the sum of two odd numbers is also even. That is, $\left( {E + E} \right) + \left( {O + O} \right) = E + E = E$.

Now $x = \left( {{{16}^3} + {{19}^3}} \right) + \left( {{{17}^3} + {{18}^3}} \right)$ is an even number. Therefore, we can say that $2$ is also a factor of $x = \left( {{{16}^3} + {{19}^3}} \right) + \left( {{{17}^3} + {{18}^3}} \right)$. So, now we have two factors $2$ and $35$ of $x$. Therefore, we can say that $x = {16^3} + {17^3} + {18^3} + {19^3}$ is exactly divisible by $35 \times 2 = 70$. Therefore, $x = {16^3} + {17^3} + {18^3} + {19^3}$ leaves a remainder of $0$ when it is divided by $70$.

Therefore, option A is correct.