Hello everyone, My codeforces account has been hacked. The hacker has changed the email and old password so i cant change the password and because i had the "Remeber me" tick marked i am still logged on in my Laptop . The hacker has changes the email to some temporary email "[email protected]" MikeMirzayanov Una_Shem

the Hacker, If u are reading this blog.Bro what will do with my codeforces profile. Trampling on someone else's hardwork. Please give me the password.

As for proofs: 1) i have first of all attached the burner mail which the hacker has put. 2) Screenshot of my official my in which i have recieved the notifications for all previous contests and pings for messages, which i didnt get for this contest 3) a weird ip has logged in my codeforces profile from russia or somewhere as you can see my regular ip is India. 4) a seperate mail showing my email name to confirm the mails are for this username only

IF SOMEONE CAN HELP.PLESE HELP.. I am not able to attach images so i have attached links. Someone please help

https://imgur.com/a/zlt3OPB

• GCD of a set of numbers can be thought as a blue-print of those numbers. If u keep adding the GCD you can make all numbers that belong in that set.

• Every common divisor of a and b is a divisor of gcd(a,b).

• Gcd(a,b) where both a and b are non-zero, can also be defined as the smallest positive integer d which can be a solution/which can be expressed as a linear combination of a and b in the form d=a*p + b*q, where both p and q are integers.

• Gcd(a, 0) = |a|, for a ≠ 0, since any number is a divisor of 0, and the greatest divisor of a is |a|.

• If 'a' divides b*c and gcd(a,b)=d , then a/d divides c.

• If m is a non-negative integer, then gcd(m⋅a, m⋅b) = m⋅gcd(a, b).It also follows from this property that if gcd(a,b)=g, then a/g and b/g should be coprime. Try to derive it yourslef.

• If m is any integer gcd(a,b)=gcd(a+m*b,b).

• The GCD: gcd(a, b) = gcd(b, a%b).

• If m is a positive common divisor of a and b, then gcd(a/m, b/m) = gcd(a, b)/m.

• GCD is a multiplicative function. That is if a1 and a2 are coprime gcd(a1*a2,b)=gcd(a1,b)*gcd(a2,b). -1. In particular, recalling that GCD is a positive integer valued function we obtain that gcd(a, b⋅c) = 1 if and only if gcd(a, b) = 1 and gcd(a, c) = 1. if the gcd is one then they need not be coprime to distribute the gcd, morever each gcd invidually should also be 1.

• The GCD is a commutative function: gcd(a, b) = gcd(b, a).

• The GCD is an associative function: gcd(a, gcd(b, c)) = gcd(gcd(a, b), c). Thus gcd(a, b, c, ...) can be used to denote the GCD of multiple arguments.

• gcd(a, b) is closely related to the least common multiple lcm(a, b): we have gcd(a, b)⋅lcm(a, b) = |a⋅b|.

• The following versions of distributivity hold true: gcd(a, lcm(b, c)) = lcm(gcd(a, b), gcd(a, c)) lcm(a, gcd(b, c)) = gcd(lcm(a, b), lcm(a, c)).

• If we have the unique prime factorizations of a = p1^e1*p2^e2 ⋅⋅⋅ pm^em and b = p1^f1*p2^f2 ⋅⋅⋅ pm^fm where ei ≥ 0 and fi ≥ 0, then the GCD of a and b is gcd(a,b) = p1^min(e1,f1) p2^min(e2,f2) ⋅⋅⋅ pm^min(em,fm).

• In a Cartesian coordinate system, gcd(a, b) can be interpreted as the number of segments between points with integral coordinates on the straight line segment joining the points (0, 0) and (a, b).

• For non-negative integers a and b, where a and b are not both zero, provable by considering the Euclidean algorithm in base n it simple states that: gcd(n^a-1,n^b-1)=n^gcd(a,b)-1

If u want an informal proof think of numbers in base 2 .We are calculating gcd's of number which contains all continuous 1 in their binary representations.. For ex: 001111 000011 their gcd can be the greatest common length which in this case is 2 thus the gcd becomes:000011 .Think of numbers in terms of length maybe you get the idea.

• An identity involving Euler's totient function: Gcd(a,b)=∑φ(k) where k are all common divisors of 'a' and 'b'