CAT 2024 Slot 2 · Quant · Question 4

The relation f\left( xy \right)=f\left( x \right)f\left( y \right)+f\left( x \right)+f\left( y \right) is almost multiplicative, and adding 1 to both sides makes it exactly so.

Indeed f\left( xy \right)+1=f\left( x \right)f\left( y \right)+f\left( x \right)+f\left( y \right)+1=\left( f\left( x \right)+1 \right)\left( f\left( y \right)+1 \right).

So define g\left( x \right)=f\left( x \right)+1; then g\left( xy \right)=g\left( x \right)g\left( y \right) for every x and y.

For a prime p, f\left( p \right)=1 gives g\left( p \right)=2.

Any natural number N is a product of primes, and if it has \Omega \left( N \right) prime factors counted with repetition, then g\left( N \right)={2}^{\Omega \left( N \right)}.

Factorising, 160000=16\times 10000={2}^{4}\times \left({2}^{4}\times{5}^{4}\right)={2}^{8}\times{5}^{4}.

Counting with repetition, \Omega \left( 160000 \right)=8+4=12, so g\left( 160000 \right)={2}^{12}=4096.

Therefore f\left( 160000 \right)=4096-1=4095, which is the second option.

Shortcut. All four options are one less than a power of 2, which is the tell-tale sign of the substitution g=f+1. Count the prime factors with repetition and the answer is {2}^{\Omega \left( N \right)}-1 immediately.