I know that for an arithmetic function to be multiplicative then $f(nm)=f(n)f(m)$ for $(n,m)=1$
I have just proved that:
$$f(n) = \left\{ \begin{array}{l l} 0 & \quad \text{if 10|n}\\ 1 & \quad \text{if 10 doesn't divide n} \end{array} \right.$$
isn't multiplicative. I just want to firstly check that I did this correctly. I chose $n=2$ and $m=5$ so $f(mn)=f(m)f(n)$?
$f(2 \cdot 5) = f(2)f(5)$
$f(10)=f(2)f(5)$
$0 \neq 1$ $\quad$ $\Box$
I am now trying to prove that
$$g(n) = \left\{ \begin{array}{l l} 0 & \quad \text{if 9|n}\\ 1 & \quad \text{if 9 doesn't divide n} \end{array} \right.$$
is multiplicative.
How do I prove this for every $n$ and $m$?
Any guidance would be great! Thank you
Suppose $\gcd(a, b) = 1$. Then at most one of $a$ and $b$ is divisible by $3$.
If neither is divisible by $3$, then $ab$ isn't divisible by $3$, and $f(a)f(b)=1 \cdot 1 = f(ab)$.
If one of them is divisible by $3$, without loss of generality, let it be $a$ that's divisible by $3$.
If $a$ is not divisible by $9$, then $ab$ isn't divisible by $9$, and $f(a)f(b) = 1 \cdot 1 = f(ab)$.
Otherwise, $a$ and $ab$ are divisible by $9$, and $f(a)f(b) = 0 \cdot 1 = f(ab)$.