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We prove L64. Fix $B,C$, write $L=L(K)$, and set $Y=\lceil K L^C\rceil$. For each tuple of digits, first record which digits are zero. There are at most $2^u$ choices. The product of the positive digits is less than or equal to $2^K$, since their bit widths sum to at most $K$. Every prime factor is at most $F\le Y$.

For primes $p\le\sqrt Y$, each exponent in a positive digit is at most $K$. There are at most $\lfloor\sqrt Y\rfloor$ such primes, without using a prime-counting theorem. All their exponents across all slots have at most

$$(K+1)^{u\lfloor\sqrt Y\rfloor}$$

possibilities. This overcounts harmlessly, including digits set to zero.

Each prime factor exceeding $\sqrt Y$ contributes more than $\tfrac12\log_2Y$ bits. The total number of such factors, with multiplicity and across all slots, is at most

$$J=\left\lfloor\frac{2K}{\log_2Y}\right\rfloor.$$

A factor is specified by a prime and its slot, giving at most $uY$ types. The number of multisets of at most $J$ factors is at most $\binom{uY+J}{J}$. Finally $x_0$ has at most $2^{b_0}$ choices. Unique factorization therefore bounds the number of digit tuples, and hence $|S|$, by

$$|S|\le 2^{b_0+u}(K+1)^{u\lfloor\sqrt Y\rfloor} \binom{uY+J}{J}. \tag{1}$$

For large $K$, $J\asymp K/\log K$. The elementary estimate $\binom{N+J}{J}\le[e(1+N/J)]^J$ gives, uniformly over the allowed data,

$$\begin{aligned} \log_2|S| &\le b_0+u+u\sqrt Y\log_2(K+1) +J\log_2\bigl(e(1+uY/J)\bigr)\\ &=O_{B,C}\left( \log K+(\log K)^B +\sqrt K(\log K)^{B+C/2+1} +\frac{K\log\log K}{\log K}\right) =o(K). \end{aligned}$$

This also covers $u=0$, with the binomial factor equal to one. The harmless rounding of $Y$ changes only constants.

Any probability measure on at most $|S|$ residues has an atom of mass at least $1/|S|=2^{-o(K)}$. For each fixed $\gamma>0$, this exceeds $2^{-\gamma K}$ for all sufficiently large $K$, proving the last clause. The argument concerns these separated digits and their total width; it places no such bound on arbitrary sums of factorial divisors.