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

Hint: In the given problem, we need to find the remainder when the given value of $x$ is divided by $70$. Therefore, in order to find this, we will use the factorization of ${a^3} + {b^3}$. 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, we can say that $\left( {a + b} \right)$ is one factor of ${a^3} + {b^3}$. That means ${a^3} + {b^3}$ is divisible by $\left( {a + b} \right)$. Also we know that the sum of two even numbers is even and the sum of two odd numbers is also even. We will use this fact to get the required answer.

Complete step by step solution: In this problem, given that $x = {16^3} + {17^3} + {18^3} + {19^3}$. First we will rewrite the given equation. Then, we will make two groups of two numbers.
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.

Note: 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, we can say that $\left( {a - b} \right)$ is one factor of ${a^3} - {b^3}$. That means ${a^3} - {b^3}$ is divisible by $\left( {a - b} \right)$. Sum of two odd numbers is even but the product (multiplication) of two odd numbers is odd.