25. A function, defined on the set of positive integers, is such that $f(x y)=f(x)+f(y)$ for all $x$ and $y$. It is known that $f(10)=14$ and $f(40)=20$.
What is the value of $f(500)$ ?
A 29
B 30
C 39
D 48
E 50
SOLUTION
C
ComMENTARY
The factorization of 500 into primes is $500=2^{2} \times 5^{3}$. Since the given function satisfies the formula $f(x y)=f(x)+f(y)$, it follows that $f(500)=f(2 \times 2 \times 5 \times 5 \times 5)=$ $f(2)+f(2)+f(5)+f(5)+f(5)=2 f(2)+3 f(5)$.
Therefore we can find the value of $f(500)$ if we can find the values of $f(2)$ and $f(5)$. In the first method given below we go about this directly. The second method is more systematic, but more complicated.

Method 1
Using the formula $f(x y)=f(x)+f(y)$ with $x=4$ and $y=10$, gives $f(40)=f(4)+f(10)$. Hence $f(4)=f(40)-f(10)=20-14=6$. Using the formula with $x=y=2$, gives $f(4)=f(2)+f(2)$. Therefore $f(2)=\frac{1}{2} f(4)=\frac{1}{2} \times 6=3$. Using the formula with $x=2$ and $y=5$, gives $f(10)=f(2)+f(5)$ and therefore $f(5)=f(10)-f(2)=14-3=11$.
Since $500=2^{2} \times 5^{3}$, we can now deduce that $f(500)=2 f(2)+3 f(5)=2 \times 3+3 \times 11=$ $6+33=39$.

Method 2
Let $f(2)=a$ and $f(5)=b$. Since $10=2 \times 5$, we have $f(10)=f(2)+f(5)$. Since $40=2^{3} \times 5$, we have $f(40)=3 f(2)+f(5)$.
Therefore, as $f(10)=14$ and $f(40)=20$, we obtain the two linear equations
$$
\begin{array}{r}
a+b=14 \\
3 a+b=20
\end{array}
$$

From equations (1) and (2)
$$
(3 a+b)-(a+b)=20-14 .
$$

We deduce that $2 a=6$ and therefore $a=3$. Hence, by equation (1), we have $b=11$.
Since $500=2^{2} \times 5^{3}$, we deduce that $f(500)=2 f(2)+3 f(5)=2 \times 3+3 \times 11=39$.
For INVESTIGATION
25.1 Assume that $f(x y)=f(x)+f(y)$ for all positive integers $x$ and $y$. What is the value of $f(1)$ ?
25.2 Again, assume that $f(x y)=f(x)+f(y)$ for all positive integers $x$ and $y$. Show that if the positive integer $n$ has the factorization $n=p^{a} \times q^{b} \times r^{c}$, then $f(n)=a f(p)+b f(q)+c f(r)$.