# Chinese Remainder Theorem Calculator

• Nov 21st 2009, 09:46 AM
mathceleb
Chinese Remainder Theorem Calculator
With priceless advice from Grandad here on MHF, I think I finally worked out the Chinese Remainder Theorem Calculator.

Chinese Remainder Theorem Calculator

Located on the discrete math page, this allows you to enter several modulus equations and the calculator will solve them using the Chinese Remainder Theorem. One piece of the calculator will contain a link which opens up to our Euclidean Algorithm Calculator in another window to determine x and y in Bezouts identity.

The lesson was also built in our new format which contains a tab for automated quiz generation.

As always, let me know if you see errors or want enhancements. Have a great weekend.
• Nov 21st 2009, 12:45 PM
Unenlightened
Have you the code you used for the algorithm?

I'd just be interested in messing around with it, checking efficiency etc.
• Nov 21st 2009, 04:51 PM
mathceleb
Quote:

Originally Posted by Unenlightened
Have you the code you used for the algorithm?

I'd just be interested in messing around with it, checking efficiency etc.

Do you mean Euclid's Extended Algorithm? If so, that lesson is here:

Euclid's Algorithm and Euclid's Extended Algorithm Calculator

I list out the math in each column. Nothing else special is going on behind the scenes.
• Nov 21st 2009, 04:52 PM
mathceleb
Enhancement Update:

I've included a section just after we determine the solution to the modulo equations which plugs in your answer and confirms that is the answer to the (n) equations that you enter.

I also updated the random generator to the Teachers Corner quiz generator to have more solutions which are not in the millions.
• Aug 7th 2010, 09:18 AM
mathceleb
Enhancement Update:

Our Chinese remainder theorem calculator now has shorcut commands. Simply enter your set of modulo equations separating each modulus statement by a comma like below into the search engine:
x = 1 mod 2,x = 2 mod 3,x = 3 mod 5,x = 4 mod 11
• Mar 15th 2011, 02:17 PM
mathceleb
Thanks to a college student, we have built one enhancement to this:

If you enter two equations and they are not pairwise coprime with respect to the n<sub>i</sub>, we will attempt to solve the modulo equations using the Method of Successive Substitution.