Question

Consider $f\left(x\right)=90x^2+20x+1$ then sum of digits of $f\left(111111\right)$ is

• A. $11$
• B. $12$
• C. $13$
• D. $14$

Answer 1

Answer: (C)

$13$

The given function is $f\left(x\right)=90x^2+20x+1$.

$f\left(1\right)=90+20+1$

$=111$

The sum of digits of $f\left(1\right)$ that is 111 is 3.

$f\left(11\right)=90{\left(11\right)}^2+20\left(11\right)+1$

$=11111$

The sum of digits of $f\left(11\right)$ that is 11111 is 5.

In the same way, the sum of digits of $f\left(111\right)$ is 7.

The sum of digits of $f\left(1111\right)$ is 9.

The sum of digits of $f\left(11111\right)$ is 11.

The sum of digits of $f\left(111111\right)$ is 13.