How many functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that:
$f(20160) = 1$
for all $x$ and $y$ , $f(xy) = f(x)f(y)$
for any $n$ whose last digit is 3, $f(n) = 1$ .
Classification:
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Since $f^6(2)f^2(3)f(5)f(7) = f(20160) = 1$ we must have $f(2) = f(3) = f(5) =f(7) = 1$ .
Since $f(2) = f(2\times 1) = f(2)f(1)$ we also have $f(1) = 1$ .
If $n$ is a positive integer that is not a multiple of 2 or 5 then for some $m$ , $nm$ has last digit 3.
Therefore $f(n)f(m) = f(nm) = 1$ and hence $f(n) = 1$ .
If $n = 2^a5^bn_0$ where $n_0$ is not a multiple of 2 or 5 then
$f(n) = f^a(2)f^b(5)f(n_0) = 1$
So $f(n)$ must equal $1$ for all non-zero $n$ .
Finally note that $f(0) = f(0)f(0)$ so $f(0) = 0$ or $f(0) = 1$ .
So, there are $\boxed{2}$ functions such that all of above conditions.