Question

Let $f\left(x\right)=x^{100}$. If $f\left(x\right)$ is divided by $x^2+x,$ then the remainder is $r\left(x\right)$. Find the value of $r\left(5\right)$?

• A. $1$
• B. $5$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$5$

$f\left(x\right)$ when divided by $x^2+x$, the remainder is $r\left(x\right)$.

Given that $f\left(x\right)=x^{100}$

We have to find the value of $r\left(5\right)$.

Now, using $x=5$, we get $f\left(5\right)=5^{100}$

When $x=5$, the divisor becomes $5^2+5=30$

Hence, if we divide $f\left(5\right)$ by $30$, we should get remainder as $r\left(5\right)$.

So, we have to find the remainder when $5^{100}$ is divided by $30$.

$\frac{5^{100}}{30}=\frac{5^{99}}{6}$

Now, if we check for the remainder for different powers of $5$, we get:
$5^1$ when divided by $6$, the remainder is $5$.
$5^2$ when divided by $6$, the remainder is $1$.
$5^3$ when divided by $6$, the remainder is $5$.
$5^4$ when divided by $6$, the remainder is $1$.
.
.
So, if we notice a pattern is followed: odd powers of $5$ when divided by $6$, give a remainder of $5$, and even powers of $5$ when divided by $6$, give a remainder of $1$.
Hence $5^{99}$ when divided by $6$ will give a remainder of $5$.
Hence $r\left(5\right)=5$.