November 21st
Today I learned a nice perspective on completely multiplicative functions on $\ZZ^+\to\ZZ^+,$ from HMMT Guts. Essentially, we're interested in functions satisfying $f(mn)=f(m)f(n),$ which is equivalent to saying that $f$ is a multiplicative homomorphism sending the monoid $\left(\ZZ^+,\times\right)$ to itself. However, this group is\[\left(\ZZ^+,\times\right)\cong\prod_p\NN\]by unique prime factorization. So our homomorphism is really defined by sending primes to other primes in some sane way.
The HMMT problem also asserted that $f(101!)=101!,$ which puts a constraint on which primes we can send where. In particular, if we prime factor $101!=\prod_pp^{\alpha_p},$ then we see\[\prod_pf(p)^{\alpha_p}=\prod_pp^{\alpha_p}.\]So $f$ is allowed to send $p\to q$ provided that $\alpha_p=\alpha_q,$ in order to preserve this prime factorization. Of course, there are infinitely many possible $f$—note that there is no constraint on the primes $p \gt 101$—but we can ask how where, say, $f(2020\cdot2021)=f(20)f(101)f(43)f(47)$ goes.
Note $2$ and $5$ are fixed because $\nu_2(101!)$ and $\nu_5(101!)$ are both unique. Then $f(101)$ can be sent to any of the primes with $\nu_p(101!)=1,$ which is any of the primes above $50.$ Then $f(43)$ and $f(47)$ must be taken to a prime with $\nu_p(101!)=2,$ which are the primes from $37$ to $47.$ Tallying these up, we get $10\cdot4\cdot3=120$ total possibilities.