Could someone please explain to me how to do this question:

3x=5(mod 7)

The answer is 4 but I'm not sure how to get there.

Help would be appreciated.

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- July 19th 2008, 02:29 AM #1

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- July 19th 2008, 02:56 AM #2

- July 19th 2008, 03:36 AM #3

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Ok, I think I more or less follow the previous explanation, but for this question I'm a little confused still:

Q: 6x=2 (mod 8)

but this cancels to:

(not sure if I can cancel like that)

So

But 3*3=9=1 (mod 8) so

This gives x=3 (mod 8)

I know the answer is suppose to be x=3(mod 8) please can you show me where I went wrong.

- July 19th 2008, 03:54 AM #4
Yes, I forgot to mention it. There is an inverse if and only if they are coprime.

What I mean is that a has an inverse modulo n, if and only if a and n are coprime.

Why ?

Here,

Going back to the definition of the congruence, we have : , where .

That is to say .

So this is the same as .

When there is a common factor in , divide a,b, and n by this factor.

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In a general case, how to find an inverse ?

- observe : for example, what number is divisible by 3 and such that it is 1 added to a multiple of 4 ? Answer is 9. It's easy by trial and error when it's small number.

- use the euclidian algorithm... I can give you links where I've done such things, or you can read that : Extended Euclidean algorithm - Wikipedia, the free encyclopedia