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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
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....
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)
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...
Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...