# What does the whole modulo thing actually mean?Watch

#1
I have to solve for x so two linear congruences are equal, but what does A = B mod C actually mean?

Also, how would I solve it? The question is:

x = A mod B
x = C mod D

and I have to find the smallest integer for x which satisfies these to linear congruences simultaneously.

How do I do this?

Thank you
0
7 years ago
#2
If you write it means that when you divide by you get the same remainder as when you divide by . That is, and . Equivalently, is divisible by .

So you need to find such that is a multiple of and is a multiple of .
7 years ago
#3
a=b (mod c) means a=kc+b, where k is a constant. Basically, it is the remainder when a is divided by c.

You can solve systems of two congruences using the Chinese remainder theorem. If you only need one solution, which you do, then you can write down the solutions for the first congruence until you find one which satisfies the second one (as long as the numbers involved are fairly small).
0
#4
(Original post by j.alexanderh)
a=b (mod c) means a=kc+b, where k is a constant. Basically, it is the remainder when a is divided by c.

You can solve systems of two congruences using the Chinese remainder theorem. If you only need one solution, which you do, then you can write down the solutions for the first congruence until you find one which satisfies the second one (as long as the numbers involved are fairly small).
So does that mean something like: 8 = 2 mod 3, because, using the a=kc +b formula, 8 = 3k + 2, where k=2, right?

What do you mean the numbers are fairly small? Less than 10,20?
0
7 years ago
#5
(Original post by claret_n_blue)
So does that mean something like: 8 = 2 mod 3, because, using the a=kc +b formula, 8 = 3k + 2, where k=2, right?

What do you mean the numbers are fairly small? Less than 10,20?
Yeah, the first part is right. I should have mentioned that k is an integer.

Fairly small in this case means you can write the first ten or fifteen multiples of the number without any effort, and quickly check the remainders of these numbers. I t depends on how good your arithmetic is, but this method should be sufficient for numbers larger than twenty. Just don't try it if the numbers are 1024=175 mod whatever.
0
#6
(Original post by j.alexanderh)
Yeah, the first part is right. I should have mentioned that k is an integer.

Fairly small in this case means you can write the first ten or fifteen multiples of the number without any effort, and quickly check the remainders of these numbers. I t depends on how good your arithmetic is, but this method should be sufficient for numbers larger than twenty. Just don't try it if the numbers are 1024=175 mod whatever.
Cool, thanks

I'm going through stuff on the Chinese remainder theorem now, so if I get stuck, hopefully you won't mind me posting more questions

Thanks once more
0
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