Chinese Remainder Theorem

How far have you gotten, given what you've learned in class, what you've tried with the Theorem, and what you've been shown in your various other threads?

Please be specific. Thank you.

Eliz.
 
thats the problem, this professor isnt teaching us anything. I try to follow the example in the book but it is only with two congruences. I dont know what Im doing.
 
bigp0ppa1046 said:
x is congruen to 5 (mod 7)
x is congruen to 2 (mod 12)
x is congruen to 3 (mod 13)

Find X
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Use the definition for starters:
x = 5 (mod 7) => 7 | (x-5)
x = 2 (mod 12) => 12 | (x-2)
x = 3 (mod 13) => 13 | (x-3)

Did you know this much?

bigp0ppa1046 said:
thats the problem, this professor isnt teaching us anything.
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LOL, Welocme to college!!
 
bigp0ppa1046 said:
thats the problem, this professor isnt teaching us anything. I try to follow the example in the book but it is only with two congruences. I dont know what Im doing.
Click to expand...
This isn't high-school algebra, when you are provided every formula and extensive worked examples of nearly every possible permutation. By this stage in your studies, you are expected to be able to function at least somewhat on your own.

Eliz.
 
This is a math help board, not a board to diminish people's knowledge or skills. I am a senior applied math major, so i get how it works. Its because of certain "easy A" professors i have had in the past that have not provided me with sufficient teaching that makes some of these problems difficult for me. So either help me, or please refrain from posting a reply to my question.
 
bigp0ppa1046 said:
This is a math help board, not a board to diminish people's knowledge or skills. I am a senior applied math major, so i get how it works. Its because of certain "easy A" professors i have had in the past that have not provided me with sufficient teaching that makes some of these problems difficult for me. So either help me, or please refrain from posting a reply to my question.
Click to expand...

I agree 100%. You pay a lot of money for tuition and get nothing in return :(

The idea is that if you have x Mod Ni = ai

for various Ni, you write down an expression Mod the product of all the Ni which will yield ai when you take Mod Ni. Think about this expression:

a1 [N2*N3*N4*...]{Inverse of [N2*N3*N4*...] mod N1} +

a2 [N1*N3*N4*...]{Inverse of [N1*N3*N4*...] mod N2} +

a3 [N1*N2*N4*...]{Inverse of [N1*N2*N4*...] mod N2} +

In the curly brackets what is meant is that you take the inverse of the number Mod Ni. That is just a number outside the brackets you don't do calculus Mod that number anymore. If you now take Mod Ni of this expression, you see that all the terms except the i-th term is zero, because they are proportional to Ni. Term number i is exactly ai by construction!

This result is valid Mod(N1*N2*N3...). You have to be careful when the Ni are not all relatively prime.
 
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