If $f(ab) = f(a)f(b)$ for all positive integers $a$ and $b$ , then the largest possible value of $f(1)$ is

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✅ Correct Answer: 1

We're given that $f(ab) = f(a)f(b)$ for all positive integers $a$ and $b$ . This is a multiplicative function - when you multiply the inputs, you multiply the outputs.

Our goal is to find the largest possible value of $f(1)$ .

Here's the key insight: Let's see what happens when we choose $a = 1$ and $b = 1$ .

Since $1 \times 1 = 1$ , we can create an equation involving only $f(1)$ .

Substituting into our given condition:

$f(1 \times 1) = f(1) \times f(1)$

This simplifies to:

$f(1) = f(1) \times f(1)$

Or more clearly:

$f(1) = [f(1)]^2$

Let's call $f(1) = x$ to make this easier to work with.

Our equation becomes:

$x = x^2$

$x^2 - x = 0$

Factoring out $x$ :

$x(x - 1) = 0$

Using the Zero Product Property, either $x = 0$ or $x - 1 = 0$ , which means $x = 1$ .

We found that $f(1)$ can be either $0$ or $1$ .

Since we want the largest possible value, we choose:

$f(1) = 1$

Therefore, the largest possible value of $f(1)$ is $1$ .

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If f(ab)=f(a)f(b)f(ab) = f(a)f(b)f(ab)=f(a)f(b) for all positive integers aaa and bbb, then the largest possible value of f(1)f(1)f(1) is

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If $f(ab) = f(a)f(b)$ for all positive integers $a$ and $b$ , then the largest possible value of $f(1)$ is