# Modular Arithmetic

• Sep 19th 2010, 01:12 AM
demode
Modular Arithmetic
http://img718.imageshack.us/img718/9097/97494921.gif

Here is the solution they've given us for (a):

Quote:

We have to solve:
7 × 2 + 3 × 1 + 9 × 1 + 7 × 8 + 3 × 7 + 9 × 2 + 7 × 9 + 3 × 4 + 9 × x = 0 mod 10

196 + 9x = 0 mod 10

6 − x = 0 mod 10

x = 6

So, how did they get from "196+9x" to "6-x"? I know that 196 mod 10 = 6, but why did they change "9x" to "-x"? What did they do to it?

Also this is the given solution to part (b):

Quote:

228 + 3x = 0 mod 10
8 + 3x = 0 mod 10
3x = -8 = 2 = 12 mod 10
x = 4

I know that 228 mod 10=8, but this time they didn't change the other term (3x)...

• Sep 19th 2010, 01:55 AM
Traveller
9x = -x mod 10

The terms are altered so that the coefficient of x divides the other number.
• Sep 19th 2010, 03:32 AM
HallsofIvy
Quote:

Originally Posted by demode
http://img718.imageshack.us/img718/9097/97494921.gif

Here is the solution they've given us for (a):

So, how did they get from "196+9x" to "6-x"? I know that 196 mod 10 = 6, but why did they change "9x" to "-x"? What did they do to it?

Same thing, really, 9= 10- 1 so 9 mod 10= -1.

Quote:

Also this is the given solution to part (b):

I know that 228 mod 10=8, but this time they didn't change the other term (3x)...

They didn't need to. 12 is evenly divisible by 3. Had it reduced to, say 3x= 7, then they would have set 3 mod 10= -7 and had -7x= 7 so x= -1= 9 (mod 10) or 3x= -3 (mod 10).
• Sep 19th 2010, 03:47 AM
demode
So in these type of problems we should always look for a way to make the coefficient of x divide the other number, or make the other number divisible by the coefficient of x?