If 6x = 7(mod 11), how would we know/get to x = 3(mod 11), my minds gone blank on how to get to this.
And also, if 5x = 4(mod13), how would we know/get to x = 6(mod 13)
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Modular Arithmetic Q watch
- Thread Starter
- 21-01-2010 18:47
- 21-01-2010 19:18
I'm fairly sure there's a principle of modular arithmetic that could solve that with a single line but I can't find my notebook atm.
So we're taking the long way
6x = 7 (mod 11) => 6x = 18 (mod 11)
By definition 6x-18=11k => 6(x-3)=11k (which means k is divisible by 6)
=> x-3=11k/6=11p (p=k/6) => x = 3 (mod 11)
- 22-01-2010 10:20
- 22-01-2010 15:01
If the numbers are small, trial and error is really about as good as it gets.
If the numbers are larger, e.g. 1234x = 5678 mod 7919, use the extended Euclidean algorithm.