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

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- Jul 19th 2008, 02:56 AM #2
Hello,

You have to perform a division :

$\displaystyle x=5 \cdot 3^{-1} \bmod 7$

But you have to calculate $\displaystyle a=3^{-1}$, which is a such that $\displaystyle 3a=1 \bmod 7$

One can see that $\displaystyle 15=3*5=1 \bmod 7$

So $\displaystyle a=5 \bmod 7$

Therefore :

$\displaystyle 3x=5 \bmod 7 \Longleftrightarrow \overbrace{3a}^{1}x=5a \bmod 7 $

$\displaystyle \implies x=5a \bmod 7 \implies x=25 \bmod 7 \implies x=4 \bmod 7 $

- Jul 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)

$\displaystyle x=2 \cdot 6^{-1} \bmod 8 $

but this cancels to:

$\displaystyle

x=1 \cdot 3^{-1} \bmod 8$ (not sure if I can cancel like that)

So$\displaystyle

a=3^{-1}

$

$\displaystyle

3a=1 \bmod 8$

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

$\displaystyle

a=3 \bmod 8

$

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.

- Jul 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, $\displaystyle 6x=2 \bmod 8$

Going back to the definition of the congruence, we have : $\displaystyle 6x-2=8k$, where $\displaystyle k \in \mathbb{Z}$.

That is to say $\displaystyle 3x-1=4k$.

So this is the same as $\displaystyle 3x=1 \bmod 4$.

When there is a common factor in $\displaystyle a=b \bmod n$, divide a,b, and n by this factor.

---------------------------

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