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Intern  Joined: 04 Oct 2013
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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

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)

Intern  Joined: 04 Oct 2013
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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....
Intern  Joined: 04 Oct 2013
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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)..
Intern  Joined: 04 Oct 2013
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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...
Intern  Joined: 04 Oct 2013
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Q) If n is a positive integer, what is the remainder when [7(8n+3) + 2] is divided by 5?

This problem can be easily solved with concept of cyclicity.

You should realize that unit digit of all numbers raised to powers start repeating itself.

For example, the cyclicity of 7 is 4. This means that the unit digits repeat after on every 4th power. See for yourself below:

71 = 7 75 =7 79=7

42 = 9 76= 9 710=9

73 = 3 77 =3 711=3

74 = 1 78 =1

In this question, consider n=1.

You are looking at unit digit of 7^11 which will be 3 from concept of cyclicity.

Therefore, last digit will be 5 for [7(8n+3) + 2]. Hence it’s completely divisible by 5.
Intern  Joined: 04 Oct 2013
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Remainder can never be negative; its minimum value can only be 0.

Consider an example of -30 / 7. Here, remainder is 5.

It would not be (-28 – 2 / 7), but [(-35+5)/7]

When you divide, you will get remainder of -2. Since remainder can never be negative, we subtract it from quotient, here 7 – 2 = 5.

Negative remainder is useful when you are trying to solve a problem with higher power.

Consider an example…….

Q) Find the remainder for 7^7^7 is divided 32

32 can be factored into 8 and 4,

Therefore, we will divide the question into two parts:

Remainder [7^7^7/8] x Remainder [7^7^7/4]

which gives: Remainder [(-1) ^7^7/8] x Remainder [3^7^7/4] [(-1) raised to odd power is -1 and raised to even power is 1]

Remainder [(-1)/8] x Remainder [(-1)/4]…. which gives

———-> 7 x 3 = 21
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_________________ Re: Euler's Remainder Theorem   [#permalink] 30 Dec 2018, 11:46
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