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lana [24]
3 years ago
7

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

Mathematics
1 answer:
kaheart [24]3 years ago
5 0
(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.
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