have a problem on hand - this one is kinda easy- but am stuck :P

the problem is :

Solve using CRT:

x congruent to 8(mod13)

x congruent to 1(mod 5)

x congruent to 5 (mod 6)

i have gotten most of it. just want to be sure

thanks

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- Apr 9th 2006, 03:39 PMdarren_a1Chinese Remainder Theorem
have a problem on hand - this one is kinda easy- but am stuck :P

the problem is :

Solve using CRT:

x congruent to 8(mod13)

x congruent to 1(mod 5)

x congruent to 5 (mod 6)

i have gotten most of it. just want to be sure

thanks - Apr 9th 2006, 04:07 PMThePerfectHackerQuote:

Originally Posted by**darren_a1**

Also,

thus, we may rely on Chinese Remainder Theorem.

----

By Chinese Remainder Theorem, we know that,

Where,

Also,

Finally,

thus,

thus,

thus,

Thus,

Thus,

- Apr 9th 2006, 04:33 PMdarren_a1
Thanks! That was awesome - this was a problem in m Discrete Math class - so i posted it here! Sorry for trouble caused - am still a bit confused as to how you calculate b1 b2 and b3 :(

Cheers - Apr 10th 2006, 07:28 AMThePerfectHackerQuote:

Originally Posted by**darren_a1**

.

Is your problem based on*how*to solve these congruences.