Chinese Remainder Theorem Puzzle

Question

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?

Answer 1

You have interpreted three of the conditions properly, but you missed the original fight when there were seven men and six coins left, so you should add in $x \equiv 6 \pmod 7$ to your system.

Answer 2

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 $-64+420=356.$

Answer 3

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

Answer 4

Hint $ $ Assumng that $\ a_i - m_i = c\ $ is constant for all $\,i\,$ it follows easily that

$$\begin{align} x&\equiv a_1\!\!\!\pmod{\!m_1}\\ x&\equiv a_2\!\!\!\pmod{\!m_2}\\ x&\equiv a_3\!\!\!\pmod{\!m_3} \end{align}\iff\, x\equiv c\!\!\! \pmod{\! m},\ \ m = {\rm lcm}(m_1,m_2,m_3)$$