Question

Use the chinese remainder theorem to solve the systems of congruences:

Answer 1

(a) Suppose we let

$x=2+4=6$

Modulo 7, we're left with $x\equiv2+4\pmod7=6\mod7$, but we want a remainder of 2, so multiply 4 by 7 to assure that that remainder vanishes. So now

$x=2+4\cdot7=30$

and $x\equiv2\pmod7$, but modulo 11, we have $x=\equiv2+28\equiv2+4\equiv6\pmod{11}$. But we want the remainder to be 4, so multiply the first term by 11 to guarantee this. So we write

$x=2\cdot11+4\cdot7=50$

but now, we get a remainder of 1 modulo 7 and 6 modulo 11. To fix the first case, multiply the first term by 2. For the second case, first find the inverse of 6 modulo 11. We have $2\cdot6=12$, and $12\equiv1\pmod{11}$, so the inverse is 2. Multiply the second term by 2, so that the second term's remainder modulo 11 becomes 1. Then multiply by 4, so that now

$x=2\cdot11\cdot2+4\cdot7\cdot2\cdot4=268$

We have

$x=268=3\cdot77+37\implies x\equiv37\pmod{77}$

(b) This is done similarly. Modulo 3, we want a remainder of 2, so we can start with

$x=2+3+3=8$

Then taken modulo 4, we need to multiply the first term by 2 and the third term by 4 to ensure the remainder becomes 3. The second term can be left alone.

$x=2\cdot2+3+3\cdot4=19$

Now taken modulo 5, we can multiply the first two terms by 5 and the third term by the inverse of $3\times2\equiv12\equiv2\pmod5$. We have $3\cdot2\equiv6\equiv1\pmod5$, so we multiply by 3.

$x=2\cdot2\cdot5+3\cdot5+3\cdot4\cdot3=71$

Now

$x=71=1\cdot(3\cdot4\cdot5)+11=1\cdot60+11$

which means the smallest positive solution for the system would be $x=11$.