Chinese Remainder Theorem Puzzle Im working on another CRT problem and I'm having a bit of trouble understanding the question at hand.
a group of seven men have found a stash of silver coins and are trying to share the coins equally among each other. Finally, there are six coins left over, and in a ensuing fight, one man is slain.  The remaining six men, still unable  to  share  equally  because  two  silver coins  are  left  over,  fight  again  —  and  another man lies dead.  The remaining men attempt to share, except that one coin is left over.  One more fight, one more dead man, and now that an equal sharing is possible.
So I'm assuming this means
$$x\equiv 2 \pmod {6}$$
$$x\equiv 1 \pmod {5}$$
$$x\equiv 0 \pmod {4}$$
right?

No need to apply CRT for this problem for

$x\equiv 2 \pmod {6}$
$x\equiv 1 \pmod {5} $
$x\equiv 0 \pmod {4}$

The number is a multiple of $4$, so last digit must be one of $(4,8,2,6,0)$
The number gives remainder $1$ when divided by $5$. So last digit of the number must be one of $(6,1)$
From these two, we know last digit of the number $=6$
Therefore we are looking for number of the form $6k+2$ whose last digit is $6$
For $k=9$ we get $56$ which is the requried number
