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

We have a special function with two key properties that we need to use strategically.

We have:

• $f(xy) = f(x)f(y) + f(x) + f(y)$ for all $x, y$
• $f(p) = 1$ for every prime number $p$

Since we know the value of $f$ for all primes, we can build up to find $f(160000)$ by breaking it into prime factors.

$160000 = 16 \times 10000$

$= 2^4 \times (10^2)^2$

$= 2^4 \times (2 \times 5)^4$

$= 2^4 \times 2^4 \times 5^4$

$= 2^8 \times 5^4$

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Since $f(2) = 1$ and $f(5) = 1$, let's find values for their powers.

For powers of 2:

Using the functional equation with $x = y = 2$:

$f(4) = f(2 \times 2) = f(2)f(2) + f(2) + f(2) = 1 \times 1 + 1 + 1 = 3$

The functional equation tells us how to combine two numbers. When we multiply the same number by itself, we substitute both $x$ and $y$ with that number.

$f(16) = f(4 \times 4) = f(4)f(4) + f(4) + f(4) = 3 \times 3 + 3 + 3 = 15$

$f(256) = f(16 \times 16) = f(16)f(16) + f(16) + f(16) = 15 \times 15 + 15 + 15 = 255$

For powers of 5:

$f(25) = f(5 \times 5) = f(5)f(5) + f(5) + f(5) = 1 \times 1 + 1 + 1 = 3$

$f(625) = f(25 \times 25) = f(25)f(25) + f(25) + f(25) = 3 \times 3 + 3 + 3 = 15$

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Notice that:

$f(p^1) = 1 = 2^1 - 1$

$f(p^2) = 3 = 2^2 - 1$

$f(p^4) = 15 = 2^4 - 1$

$f(p^8) = 255 = 2^8 - 1$

Pattern: $f(p^n) = 2^n - 1$ for any prime $p$

This pattern emerges because our functional equation has the structure that builds up exponentially.

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Since $160000 = 2^8 \times 5^4 = 256 \times 625$:

$f(160000) = f(256 \times 625) = f(256)f(625) + f(256) + f(625)$

$f(160000) = 255 \times 15 + 255 + 15$

$f(160000) = 3825 + 270 = 4095$

Therefore, $f(160000) = 4095$.