Question

What will be the remainder when $\left( {{67}^{67}}+67 \right)$ is divided by 68.
(A) 52
(B) 60
(C) 66
(D) 70

Answer 1

Hint: First of all, assume $67=x$ . Using this transform $\left( {{67}^{67}}+67 \right)$ as $f(x)=\left( {{x}^{x}}+x \right)$ . Similarly, transform 68 as $x-\left( -1 \right)$ . According to the polynomial remainder theorem we have, “if a polynomial $f(x)$ is divided by another polynomial $\left( x-c \right)$ then, the remainder is always equal to $f\left( c \right)$”. Now, compare $\left( x-c \right)$ and $x-\left( -1 \right)$ , we can say that $c=-1$ . We have $f(x)=\left( {{x}^{x}}+x \right)$ . Now, use the polynomial remainder theorem and put $x=-1$ in $f(x)=\left( {{x}^{x}}+x \right)$ . Since we are dividing $\left( {{67}^{67}}+67 \right)$ by 68 so, we cannot have a negative number as the remainder. Therefore, to convert the negative remainder into positive, we have to add the divisor to the negative remainder. Now, solve further and get the remainder.

Complete step-by-step answer:
According to the question, we have the expression
$\left( {{67}^{67}}+67 \right)$ …………………………(1)
Now, let us assume,
$67=x$ …………………………..(2)
Using equation (2), we can transform equation (1).
On transforming equation (1), we get
$f(x)=\left( {{x}^{x}}+x \right)$ ……………………….(3)
We can write the number 68 as,
$68=67-\left( -1 \right)$ ………………………..(4)
Now, using equation (2), we can transform equation (4).
On transforming equation (4), we get
$68=x-\left( -1 \right)$ ……………………..(5)
We have to find the remainder when the expression $\left( {{67}^{67}}+67 \right)$ is divided by 68. In other words, we can say that we have to find the remainder when $f\left( x \right)$ is divided by $x-\left( -1 \right)$ .
According to the polynomial remainder theorem we have,
“if a polynomial $f(x)$ is divided by another polynomial $\left( x-c \right)$ then, the remainder is always equal to $f\left( c \right)$”.
On comparing $\left( x-c \right)$ and $x-\left( -1 \right)$ , we can say that $c=-1$ .
From equation (3), we have the $f(x)=\left( {{x}^{x}}+x \right)$ .
Now, using the polynomial remainder theorem, we get

& f\left( -1 \right)=\left\\{ {{\left( -1 \right)}^{-1}}-1 \right\\} \\\ & \Rightarrow f\left( -1 \right)=\dfrac{1}{-1}-1 \\\ & \Rightarrow f\left( -1 \right)=-1-1 \\\ \end{aligned}$$ $$\Rightarrow f\left( -1 \right)=-2$$ ………………………..(6) Since we are dividing $$\left( {{67}^{67}}+67 \right)$$ by 68 so, we cannot have a negative number as the remainder. But from equation (6), we have -2 as our remainder and we can not have a negative number as the remainder. So, to convert it into a positive remainder, the divisor is added to the negative remainder. Since we are dividing by 68 now, 68 is our divisor here. Now, adding the divisor to the negative remainder, we get $$-2+68=66$$ . Therefore, the remainder is 66. Hence, the correct option is (C). Note: We can also solve this question by using the property of the polynomial expansion. According to the question, we have the expression $$\left( {{67}^{67}}+67 \right)$$ …………………………(1) Let us assume, $$67=x$$ ……………..(2) Using equation (2), we can transform equation (1) as, $$\left( {{x}^{67}}+1+66 \right)$$ ……………………..(3) We can write 68 as $$68=67+1$$ . Now, using equation (2), we can transform it as $$\left( x+1 \right)$$ ……………………….(4) We have to find the remainder when $$\left( {{67}^{67}}+67 \right)$$ is divided by 68. Now, using equation (3) and equation (4), we can say that we have to find the remainder of $$\dfrac{\left( {{x}^{67}}+1+66 \right)}{x+1}$$ ………………….(5) Simplifying equation (5), we get $$\dfrac{\left( {{x}^{67}}+1 \right)}{\left( x+1 \right)}+\dfrac{66}{\left( x+1 \right)}$$ ……………………(6) According to the property of the polynomial expansion, $$\left( {{x}^{n}}+1 \right)$$ is divisible by $$\left( x+1 \right)$$ if n is odd. Using this property, we can say that $$\left( {{x}^{67}}+1 \right)$$ is divisible by $$\left( x+1 \right)$$ . The number 66 is not divisible by $$\left( x+1 \right)$$ . Therefore, the remainder is 66. Hence, the correct option is (C).