Question

Let $f(n)$ denote the square of the sum of the digits of natural number $n$ & $f^2(n)$ denotes $f(f(n)),f^3(n)$ denotes $f(f(f(n)))$ and so on, then the value of
$\frac{f^{2021}\left(2021\right)-f^{2020}\left(2021\right)}{f^{2023}\left(2021\right)-f^{2022}\left(2021\right)}$ equals

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (C)

Given $f(n)=$ square of the sum of the digits of $n$
$f^2\left(n\right)=f\left(f\left(n\right)\right),f^3\left(n\right)=f\left(f\left(f\left(n\right)\right)\right)$ and so on
Now $f(2021)=(2+0+2+1{)}^2=25$
$\therefore f^2(2021)=f(25)=49$,
$f^3(2021)=f(49)=(4+9{)}^2=169$
$f^4(2021)=f(169)=(1+6+9{)}^2=256$,
$f^5(2021)=f(256)=(2+5+6{)}^2=169$
It is periodic function where
$f^{2n}(2021)=256$ & $f^{2n+1}(2021)=169\forall n\ge 2$
$\frac{f^{2021}\left(2021\right)-f^{2020}\left(2021\right)}{f^{2023}\left(2021\right)-f^{2022}\left(2021\right)}$
$=\frac{169-256}{169-256}=1$