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?

Note that $x=-4$ modulo $60$ is an obvious solution to the three congruences you list.  Now you just have to add in $x\equiv 6 (\bmod{7})$.  But $-4 \equiv -64 (\bmod{60})$ and $-64 \equiv -1 \equiv 6 (\bmod{7})$.  So one solution is $x \equiv -64 (\bmod{420})$.  Add any multiple of $420$ you want for the final answer.  Say $-64+420=356.$ 
