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Euler's Remainder Theorem [#permalink]
04 Oct 2013, 22:47

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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

Fermat's Little Theorem [#permalink]
04 Oct 2013, 22:57

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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....

Chinese Remainder Theorem [#permalink]
04 Oct 2013, 23:06

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Let’s understand this theorem with an example:

Q.5) – Rahul has certain number of cricket balls with him. If he divides them into 4 equal groups, 2 are left over. If he divides them into 7 equal groups, 6 are left over. If he divides them into 9 equal groups, 7 are left over. What is the smallest number of cricket balls could Rahul have?

Let N be the number of cricket balls.

N = 2(mod4) ————–> equation 1

N = 6(mod7) ————–> equation 2 &

N = 7(mod9) ————–> equation 3.

From N=2(mod4) we get, N=4a+2

Substituting this in equation 2, we get the following equation:

4a + 2 = 6(mod7)

Therefore, 4a = 4(mod7)

Hence, 2 x 4a = 2 x 4(mod7)

This gives us a = 1(mod7)

Hence a = 7b+1.

Plugging this back to N=4a+2, we get….

N = 28b + 6

Substituting this to equation 2;

28b + 6 = 7(mod9)

28b = 1(mod9)

Therefore, b=1(mod9)

Hence b = 9c + 1.

Substituting this back to equation N=28b+6;

N = 28(9c+1) + 6

N = 252c + 34

The smallest positive value of N is obtained by setting c=0.

It gives us N = 34

TB – All prime numbers greater than 3 can be expressed as 6K+1 or 6K-1, this is another important result. You would be using this result a lot when it comes to number system problems...