# Applying Chinese Remainder Theorem to polynomials

• stgermaine
In summary, in order to find all integers x that satisfy the given congruences, we can use the Chinese Remainder theorem and Bezout's theorem. The first step is to get rid of the coefficients by computing their respective modular multiplicative inverses. This allows us to rewrite the system of congruences in terms of x alone. Then, using the extended Euclidean algorithm, we can solve for x and obtain all possible solutions.

## Homework Statement

Find all integers x such that
$7x \equiv 11 mod 30$ and
$9x \equiv 17 mod 25$

## Homework Equations

I guess the Chinese Remainder theorem and Bezout's theorem would be used here.

## The Attempt at a Solution

I can do this if the x-terms didn't have a coefficient. I'd just rewrite the congruences so that x - 11 = 30k etc and use Euclid's algorithm to solve it, which is not too difficult.
I'm just confused as to what to do since there are the coefficients and I'm not too sure what to do.

Thanks

stgermaine said:

## Homework Statement

Find all integers x such that
$7x \equiv 11 mod 30$ and
$9x \equiv 17 mod 25$

## Homework Equations

I guess the Chinese Remainder theorem and Bezout's theorem would be used here.

## The Attempt at a Solution

I can do this if the x-terms didn't have a coefficient. I'd just rewrite the congruences so that x - 11 = 30k etc and use Euclid's algorithm to solve it, which is not too difficult.
I'm just confused as to what to do since there are the coefficients and I'm not too sure what to do.

Thanks

The first step is to get rid of those coefficients by computing their respective modular multiplicative inverses.

So the system becomes ##x \equiv 7^{-1}.11 \pmod {30}## and ##x \equiv 9^{-1}.17 \pmod {25}##.

You can do this because the coefficients are relatively prime to those moduli.

Do you know how to compute those inverses? If not, you can read: http://en.wikipedia.org/wiki/Modular_multiplicative_inverse and http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm (for the actual algorithm). A quick calculator is available online at: http://www.cs.princeton.edu/~dsri/modular-inversion.html [Broken]

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