Math Is Fun Forum

Au101

Member

Registered: 2010-12-01

Posts: 353

Arithmetic modulo

Hey ho, I've come up against a question on something that was never actually on my syllabus, so I don't really know where to begin. I googled around looking for things on modular arithmetic and the modulo operation, but I can't apply it to this question, because I can't really parse the question, I don't understand what it wants.

Here it is:

21. Write down all the solutions of 3x(3x + 4) = 0
(i) in arithmetic modulo six; (ii) in arithmetic modulo five.

Obviously the "basic" solutions are x = 0, x = -4/3, but I don't really know what to do with those numbers

Bob

Administrator

Registered: 2010-06-20

Posts: 10,143

Re: Arithmetic modulo

hi Au101,

Modulo arithmetic is sometimes called 'clock arithmetic'.  eg.  23 hours plus 2 hours = 1 hour.  This is modulo twenty four.

So I wrote down the equivalent numbers for mod 6:

0,  6,  12, 18, 24, .....
1,  7,  13, 19, 25, .....
2,  8,  14, 20, 26, .....
3,  9,  15, 21, 27, .....
4,  10, 16, 22, 28, .....
5,  11, 17, 23, 29, .....

As there are only 6 values to test, I tried them all*.

3x(3x+4)  From this point on the 'x's are times signs.
3x0(3x0+4)=0x4=0
3x1(3x1+4)=3x7=3x1=3
3x2(3x2+4)=6x10=0x4=0
3x3(3x3+4)=9x13=3x1=3
3x4(3x4+4)=12x16=0x4=0
3x5(3x5+4)=15x19=3x1=3

Thus the solutions in mod 6 are x=0, 2 and 4

*At the moment I cannot think of a way that avoids 'trying them all'.  If I come up with something, I'll post again.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Arithmetic modulo

Well, for the first one, you can say it's equivalent to

This is equivalent to

So, if 2 must divide either x or 3x+4. For 3x+4 to be divisible by 2, 3x must be too, and then so must be x, so in either case, x or 3x+4 being divisible by 2, it holds that x must be divisible by 2. So the answer is

.

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Au101

Member

Registered: 2010-12-01

Posts: 353

Re: Arithmetic modulo

Okay, I see it now, thank you!