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April 14th, 2010, 06:38 PM   #1
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application to ax=b (mod m)

describe all solutions of the given congruence.

22x = 5 (mod 15)

proof: the gcd(22,15) =1 and 1 is a divisor of 15. Thus by a corollary, there is solutions.

the multiplicative inverse of 22 in Z_15 is 13.
So (22*13)x = 15 *13 (mod 15) => x = 65 mod 15 => x = 5 mod 15
Hence the reminder is 5. Solutions in Z_15 is 5+15Z?

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April 18th, 2010, 10:42 PM   #2
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Re: application to ax=b (mod m)

Hello tinynerdi,

Quote:
Originally Posted by tinynerdi
15 *13 (mod 15) => x = 65 mod 15
How come, you mustn't use 15*13=195?

Hoempa
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April 27th, 2010, 10:08 AM   #3
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Re: application to ax=b (mod m)

Typo; should have been 5 not 15.
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