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)} $
$= \frac{169-256}{169 - 256} = 1$