I've been trying to solve a Chinese remainder theorem problem using modulo arithmetic. The example on wikipedia is a case where there is only 3 equations (and therefore 3 unknowns). In the wikipedia example, it seems as though I am expressing the 1st unknown in terms of the 2nd unknown, and then plug this into the 1st equation to get 2 equations (old and new) for the 2nd unknown. Equate equations 2 and 3 to get 3rd unknown in terms of equation 2, then substituting this into the 2nd new equation. I'm sorry if that seems quite confusing, but that's how I feel currently (confused).
In my case, there are 6 unknowns,
(i) $x = 1$ (mod $2$) = $1 + 2.p_{1}$ (mod $2$)
(ii) $x = 2$ (mod $3$) = $2+3.p_{2}$ (mod $3$)
(iii) $x = 3$ (mod $4$) = $3 + 4.p_{3}$ (mod $4$)
(iv) $x = 4$ (mod $5$) = $4+5.p_{4}$ (mod $5$)
(v) $x = 5$ (mod $6$) = $5 + 6.p_{5}$ (mod $6$)
(vi) $x = 0$ (mod $7$) = $7.p_{6}$ (mod $7$)
I have been able to get $x = 5 + 6_p{2}$ (mod $2$), but have been unable progress further. Any pointers in the right direction would be much appreciated! THanks