CAT 2024 Slot 2 QA Question 8: A function $f$ maps the set of natural numbers to whole numbers, such that $f(xy) = f(x)f(y) + f(x) + f(y)$ for all $x$, $y$ and $f(p) = 1$ for every prime number $p$. Then, the value of $f(160000)$ i

Question

A function $f$ maps the set of natural numbers to whole numbers, such that $f(xy) = f(x)f(y) + f(x) + f(y)$ for all $x$, $y$ and $f(p) = 1$ for every prime number $p$. Then, the value of $f(160000)$ i

Accepted Answer

The expression becomes multiplicative if we add $1$ to $f$.
Let $g(n) = f(n) + 1$.
$g(xy) = f(xy) + 1$
$\Rightarrow g(xy) = f(x)f(y) + f(x) + f(y) + 1$
$\Rightarrow g(xy) = (f(x)+1)(f(y)+1) = g(x)g(y)$
For every prime $p$, $f(p) = 1$, so $g(p) = 2$.
$160000 = 2^8 	imes 5^4$
$g(160000) = g(2)^8 	im

A function fff maps the set of natural numbers to whole numbers, such that f(xy)=f(x)f(y)+f(x)+f(y)f(xy) = f(x)f(y) + f(x) + f(y)f(xy)=f(x)f(y)+f(x)+f(y) for all xxx, yyy and f(p)=1f(p) = 1f(p)=1 for every prime number ppp. Then, the value of f(160000)f(160000)f(160000) is

A function $f$ maps the set of natural numbers to whole numbers, such that $f(xy) = f(x)f(y) + f(x) + f(y)$ for all $x$ , $y$ and $f(p) = 1$ for every prime number $p$ . Then, the value of $f(160000)$ is