Let $b_m$ be numbers of factors $2$ of the number $m!$ (that is, $2^{b_m}|m!$ and $2^{b_m+1}\nmid m!$). Find the least $m$ such that $m-b_m = 1990$.
Source: Turkey TST 1990 - P5
Tags: number theory proposed, number theory
Let $b_m$ be numbers of factors $2$ of the number $m!$ (that is, $2^{b_m}|m!$ and $2^{b_m+1}\nmid m!$). Find the least $m$ such that $m-b_m = 1990$.