CAT 2024 Slot 2 QA Q4 Functions & Graphs

Question 4 of the Quantitative Ability (QA) section, CAT 2024 Slot 2, with CAT's official answer and a worked solution.

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

A
1023
B
4095
C
2047
D
8191

Answer: (B) 4095 Functions & Graphs MCQ

Solution

Let \(g(x) = f(x)+1\). Then \(g(xy) = f(x)f(y)+f(x)+f(y)+1 = g(x)g(y)\), so \(g\) is multiplicative over every factorisation.
For a prime \(p\), \(g(p) = 2\). So \(g(n) = 2^{k}\), where \(k\) is the number of prime factors of \(n\) counted with repetition.
\(160000 = 16 \times 10^4 = 2^8 \cdot 5^4\), so \(k = 12\) and \(g = 2^{12} = 4096\).
\(f(160000) = 4096 - 1 = 4095\).

Shortcut. Recognise \(f(xy)+1 = (f(x)+1)(f(y)+1)\); all options are of the form \(2^k-1\).