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04 Oct 2013, 23:47
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Euler's Remainder Theorem:
Euler’s theorem states that if p and n are coprime positive integers, then ———> P φ (n) = 1 (mod n), where Φn=n (1-1/a) (1-1/b)….
Before we move any further, let us understand what mod is. Mod is a way of expressing remainder of a number when it is divided by another number. Here φ (n) (Euler’s totient) is defined as all positive integers less than or equal to n that are coprime to n. (Co-prime numbers are those numbers that do not have any factor in common.)
For example ———-> 24=23 x 3
———-> Therefore, we get 24 x (1-1/2) (1-1/3) = 8
———-> which means that there are 8 numbers co-prime to 24.
They are 1, 5, 7, 11, 13, 17, 19, 23.
Let us understand this theorem with an example:
Q.1) – Find the remainder of (7^100) / 66
Answer-
As you can see, 7 and 66 are co-prime to each other.
Therefore, Φ 66 = 66 x (1-1/2) x (1-1/3) x (1-1/11) = 20
So, (7^100) = 1(mod 66)
Please post any doubts...
Euler’s theorem states that if p and n are coprime positive integers, then ———> P φ (n) = 1 (mod n), where Φn=n (1-1/a) (1-1/b)….
Before we move any further, let us understand what mod is. Mod is a way of expressing remainder of a number when it is divided by another number. Here φ (n) (Euler’s totient) is defined as all positive integers less than or equal to n that are coprime to n. (Co-prime numbers are those numbers that do not have any factor in common.)
For example ———-> 24=23 x 3
———-> Therefore, we get 24 x (1-1/2) (1-1/3) = 8
———-> which means that there are 8 numbers co-prime to 24.
They are 1, 5, 7, 11, 13, 17, 19, 23.
Let us understand this theorem with an example:
Q.1) – Find the remainder of (7^100) / 66
Answer-
As you can see, 7 and 66 are co-prime to each other.
Therefore, Φ 66 = 66 x (1-1/2) x (1-1/3) x (1-1/11) = 20
So, (7^100) = 1(mod 66)
Please post any doubts...
04 Oct 2013, 23:57
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Fermat’s theorem is an extension of Euler’s theorem. If, in the above theorem, n is a prime number then, pn-1 =1(mod n)
Consider an example;
Q.2) Find remainder of 741 is divided by 41.
Here, 41 is a prime number.
Therefore, [7 40 x 7 /41] (By Fermat’s theorem)
which is equal to 7.
TB – Wieferich prime: is a prime number p such that p2 divides 2p − 1 – 1 relating with Fermat little theorem, Fermat’s little theorem implies that if p > 2 is prime, then 2p − 1 – 1 is always divisible by p....
Consider an example;
Q.2) Find remainder of 741 is divided by 41.
Here, 41 is a prime number.
Therefore, [7 40 x 7 /41] (By Fermat’s theorem)
which is equal to 7.
TB – Wieferich prime: is a prime number p such that p2 divides 2p − 1 – 1 relating with Fermat little theorem, Fermat’s little theorem implies that if p > 2 is prime, then 2p − 1 – 1 is always divisible by p....
05 Oct 2013, 00:01
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Wilson’s theorem states that a number ‘n’ is prime if and only if (n-1)! + 1 is divisible by n
For example –
Q.3) Find the remainder when 30! Is divided by 31.
Here, 31 is a prime number.
From Wilson’s theorem, we have (31-1)! /31 = – 1
Hence, 30 is the remainder. [whenever there is a negative remainder, subtract it from the divisor and you get the remainder]
Let us have a look another example.
Q.4) Find the remainder when 29! Is divided by 31.
You can write 29!/31 as ———> 30 x 29! / 30 x 31 (multiplying numerator and denominator by 30)
Therefore, 30! / 30 x 31 = 1
We have already found from Wilson theorem 30! /31 is 30. 30 in the denominator cancels out, hence the remainder is 1.
From the foregoing examples, the derivative arrived at from this theorem is, (P-2)! = 1(mod P)..
For example –
Q.3) Find the remainder when 30! Is divided by 31.
Here, 31 is a prime number.
From Wilson’s theorem, we have (31-1)! /31 = – 1
Hence, 30 is the remainder. [whenever there is a negative remainder, subtract it from the divisor and you get the remainder]
Let us have a look another example.
Q.4) Find the remainder when 29! Is divided by 31.
You can write 29!/31 as ———> 30 x 29! / 30 x 31 (multiplying numerator and denominator by 30)
Therefore, 30! / 30 x 31 = 1
We have already found from Wilson theorem 30! /31 is 30. 30 in the denominator cancels out, hence the remainder is 1.
From the foregoing examples, the derivative arrived at from this theorem is, (P-2)! = 1(mod P)..
