Now that you have completed the first two Research Questions, you know the exact nature of the solutions to the pair of congruences

xa_{1}(modm_{1}) andxa_{2}(modm_{2}).Recall that the cases where there is a

uniquesolution modulom_{1}m_{2}is covered by the Chinese Remainder Theorem for two congruences. Suppose instead you had three or more congruences. Is there a version of the Chinese Remainder Theorem in this case? There is indeed, and (surprise, surprise) it's your job to find it.To help get you headed in the right direction, let's look at an example involving three specific congruences, such as

x1 (mod 5) x2 (mod 6) x3 (mod 7). One way of proceeding is to begin by solving the first pair of congruences,

x1 (mod 5) andx2 (mod 6).By applying the algebraic method of the Prelab, we can show that this pair of congruences is equivalent to the single congruence

x26 (mod 30). Let's check this with our applet:

OK, now we have reduced our problem to just two congruences:

x26 (mod 30) andx3 (mod 7).We can employ our method for pairs again to reach the single congruence

x206 (mod 210). Again, let's check this with the applet:

So, we did it! We solved three congruences by twice applying what we know about solving a pair of congruences. Here's your chance to generalize this process.

## Research Question 3

Give a statement of the Chinese Remainder Theorem (as in Research Question 1) for

ncongruences.## 7.3.1 Using Java to Solve Lots of Congruences

We now introduce a new and improved Java applet that can be used to solve more than two congruences simultaneously. For example, to solve the system of congruences

x1 (mod 5) x2 (mod 6) x3 (mod 7), we enter

Notice that the output agrees with what we found above. Try it with different numbers.

Section 7.1 | Section 7.2 | Section 7.3 | Section 7.4

Copyright © 2001 by W. H. Freeman and Company