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 Post subject: Conrguence (integers modulo m)Posted: Thu, 26 Apr 2012 17:39:22 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
If i have two equations ,

11a+b(mod26)=17
&
22a+b(mod26)=22

How do i begin to solve for a and b?

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 Post subject: Re: Conrguence (integers modulo m)Posted: Thu, 26 Apr 2012 20:27:29 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12098
Location: Austin, TX
DgrayMan wrote:
If i have two equations ,

11a+b(mod26)=17
&
22a+b(mod26)=22

How do i begin to solve for a and b?

Use the Chinese remainder theorem.

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 Post subject: Re: Conrguence (integers modulo m)Posted: Thu, 26 Apr 2012 22:37:38 UTC
 S.O.S. Oldtimer

Joined: Sat, 21 Jan 2012 03:59:22 UTC
Posts: 182
at first i thought you were messing with me lol

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 Post subject: Re: Conrguence (integers modulo m)Posted: Fri, 27 Apr 2012 05:25:17 UTC
 Moderator

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6006
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
DgrayMan wrote:
If i have two equations ,

11a+b(mod26)=17
&
22a+b(mod26)=22

How do i begin to solve for a and b?

Use the Chinese remainder theorem.

I assume the OP means etc. Why apply CRT here instead of eliminating immediately?

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 Post subject: Re: Conrguence (integers modulo m)Posted: Fri, 4 May 2012 17:52:27 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Mon, 19 May 2003 19:55:19 UTC
Posts: 7949
Location: Lexington, MA
Hello, DgrayMan!

Quote:

. .

. . . . . . . . . . . . . . . . . . . .

.**

n . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

How do we find that "19"?
Here is a primitive method . . .

We want to find so that: .

That is: .

. . Then: .

Since is an integer, must be a multiple of 11: .

We have: .

Since is an integer, must be a multiple of 4: .

We have: .

By inspection, we find

Substitute into (b): .

Substitute into (a): .

And that is how I found the multiplicative inverse of 11, mod 26.

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 Post subject: Re: Conrguence (integers modulo m)Posted: Sat, 5 May 2012 05:39:56 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6006
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
Soroban wrote:
And that is how I found the multiplicative inverse of 11, mod 26.

Why do it the hard way when you could just use Bezout/Euclid?

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