Let $f$ be a function such that $f(mn) = f(m)f(n)$ for every positive integers m and n. If $f(1)$, $f(2)$ and $f(3)$ are positive integers, $f(1) < f(2)$, and $f(24) = 54$, then $f(18)$ equals

We're given that $f(mn) = f(m)f(n)$ for all positive integers $m$ and $n$.

This means that the function value of a product equals the product of the function values. This is called a multiplicative function.

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Let's use the given property with $m = 1$ and $n = 2$:

$f(1 \times 2) = f(1) \times f(2)$

$f(2) = f(1) \times f(2)$

Since $f(2)$ is a positive integer, we can divide both sides by $f(2)$ to get:

$1 = f(1)$

Therefore, $f(1) = 1$.

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Since we know $f(1) = 1$ and $f(1) < f(2)$, we have $f(2) > 1$.

Let's call:

$f(2) = a$ (where $a$ is a positive integer greater than 1)

$f(3) = b$ (where $b$ is a positive integer)

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Using $f(mn) = f(m)f(n)$, we can find:

For $f(4)$:

$f(4) = f(2 \times 2) = f(2) \times f(2) = a^2$

For $f(6)$:

$f(6) = f(2 \times 3) = f(2) \times f(3) = ab$

For $f(24)$:

Since $24 = 2^3 \times 3 = 8 \times 3$:

$f(24) = f(8) \times f(3)$

But $f(8) = f(2^3) = f(2 \times 2 \times 2) = f(2)^3 = a^3$

So: $f(24) = a^3 \times b = a^3b$

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We know that $f(24) = 54$, so:

$a^3b = 54$

We need to find positive integer values of $a$ and $b$ such that $a^3b = 54$.

Let's factor 54: $54 = 2 \times 3^3 = 2 \times 27$

Testing values:

If $a = 3$: then $3^3 \times b = 27b = 54$, so $b = 2$

If $a = 2$: then $2^3 \times b = 8b = 54$, so $b = 6.75$ (not an integer)

Therefore: $a = 3$ and $b = 2$

This means $f(2) = 3$ and $f(3) = 2$.

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Since $18 = 2 \times 3^2$:

$f(18) = f(2 \times 3^2) = f(2) \times f(3^2)$

And $f(3^2) = f(3 \times 3) = f(3) \times f(3) = b^2 = 2^2 = 4$

Therefore:

$f(18) = f(2) \times f(3^2) = 3 \times 4 = 12$

Therefore, $f(18) = 12$