ZETA_in_olympiad wrote: $f$ is a one-to-one function from the set of positive integers to itself such that $$f(xy) = f(x) × f(y)$$Find the minimum possible value of $f(2020)$. $f(2020)=f(2)^2f(5)f(101)$ And so, since injective, min is $\boxed{2^2\times 3\times 5=60}$

pco wrote: ZETA_in_olympiad wrote: $f$ is a one-to-one function from the set of positive integers to itself such that $$f(xy) = f(x) × f(y)$$Find the minimum possible value of $f(2020)$. $f(2020)=f(2)^2f(5)f(101)$ And so, since injective, min is $\boxed{2^2\times 3\times 5=60}$ Don't we need to back this up with an example of such function?

the example is $f(2^{a_1}3^{a_2}5^{a_3}\ldots101^{a_{26}}\ldots)=2^{a_1}3^{a_3}5^{a_{26}}7^{a_4}11^{a_5}\ldots101^{a_2}103^{a_{27}}\ldots$

Why $f(101)$ can't be $4$?

because f(4)=f(2)^2=4

Ohhhhhhhhhhh Thanks!