How to solve this problem by making a table where vertically you have the multiples of $3$, horizontally, you have the multiples of $4$ and five, then you cross out any sum of the $3m+4n$ or $3m+5k$ such that its under $10$ or above $40$, and then look at the multiples of six only you haven't crossed out?
Club $X$ has more than $10$ but fewer than $40$ members. Sometimes the members sit at tables with $3$ members at one table and $4$ members at each of the other tables, and sometimes they sit at tables with $3$ members at one table and $5$ members at each of the other tables. If they sit at tables with $6$ members at each table except one and fewer than $6$ members at that one table, how many members will be at the table that has fewer than $6$ members?
I cannot really see what you have in mind for the table with multiples of $3$, $4$ and $5$. However, I feel like the riddle has a very simple solution: it's telling you that $(X-3)$ ($X$ being the number of members) is a multiple of both $4$ and $5$. Only their least common multiple is within the bounds, meaning that $X = 23$. Thus, the last table will have $5$ members.