Tuesday, June 3, 2008

CoPrimes, Euler's Number and Fermat's theorem

Co-Primes are very important , when it comes to finding remainders.
First things first, What is a Co-Prime ?
2 numbers are said to be co-primes/relatively prime if their HCF is 1.
That is the 2 numbers have no other common factor apart from the number.

Co-Primes by themselves serve no purpose.You need to use them with
Fermat's theorem / Chinese Remainder Theorem etc.

Fermat's theorem states that
Remainder (x ^ ( Euler's Number of N) )/N ) = 1
Here X and N have to Co-Primes.

Now , what is Euler's Number?
Euler's number of a particular number is the number of co-primes less than
the number. The total number of co-primes that a certain number can have
is infinite. So we need to determine the number of coprimes less than the number.
Euler's number for a prime number P is P-1. As all numbers less than that
prime number will coprimes of the prime number. As in 1,2,3,4,5,6 will all be
coprimes of 7.

For other numbers :
Euler's Number of a number X = X(1-1/a)(1-1/b)....
Here a and b are the prime factors of the number.
We look at the number of the prime numbers that are factors of the given number.
We ignore the number of times the prime number occurs.
So say the number is 18.
18 has 2 and 3 as factors.
So Eulers No of 18 = 18(1-1/2)(1-1/3)
= 6

So if we get a sum like ,
Find the remainder when (11 ^ 97 )/7
1)See if 11 and 7 are relatively prime to each other
2)If they are relatively prime, Find the Euler's number of the denominator.
3)Express power in terms of the euler's number.
For eg, in this case the Euler's number for 7 is 6.
So 97 = 6k+1.
We know that if we can factorize the numerator in a division, the end remainder
will be the product of the remainders of each individual factor.
So this becomes (11 ^ 6)/7.....(16 times)* (11 ^ 1)/7
Now as per Fermat's theorem,
(11 ^ 6) /7 is 1.
So the answer is the remainder when (11^1) is divided by 7.
This can be used when the numerator can be factorized.
This can be used when the numerator and the denominator are co-primes/relatively prime.
Now The question arises what is they are not coprimes.
If 2 numbers are not co-primes,it means they have a common factor.
See if you can remove the common factor from both the numerator and denominator.

In case that can't be done , we will have to use other methods to find the remainder.

6 comments:

Unknown said...

Thanks a ton!!! I cant tell u how helpful this was!!! I had been searchign for this from the past few days.. and ididnt understand it anywhere else.. U really explained soo soo well! Thank you again! and God bless you :)

Prasanth Guruprasad said...

can you tell me how you would find the remainder of(11^97)/77

Unknown said...

=11/11(11^96/7)
=11/11[(11^(6*16)]/7)
(because we know that the Euler no for 7 is 6)
but the remainder of [11^(16*6)] /7
is 1 ie 11^96=7k+1
=11/11[{(7k)+1}/7] where k is some constant
=11/77 viz 11

meenu said...

thnx...d language was simple nd easy to undrstnd.....

Unknown said...

Thank you sooo much.. It was very helpful :)

Unknown said...

Very good explanation..Bravo!
Thnx a lot!