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Solution of Codeforces Round #831 (Div. 1 + Div. 2) || Tutorial of Codeforces Round #831 (Div. 1 + Div. 2)

Solution of Codeforces Round #831 (Div. 1 + Div. 2) || Tutorial of Codeforces Round #831 (Div. 1 + Div. 2) Codeforces Round #831 (Div. 1 + Div. 2, based on COMPFEST 14 Final) Editorial 1740A. Factorise N+M Author:  Pyqe Developer:  Pyqe Tutorial 1740A - Factorise N+M There are multiple solutions for this problem. We will discuss two of them. One solution is to choose  m = n m = n . This always guarantees that  m m  is prime, because  n n  is always prime. And we can see that  n + m = n + n = 2 n n + m = n + n = 2 n , which is always not prime, because  n > 1 n > 1  always holds. Another solution is to choose  m = 7 m = 7 . If  n n  is odd, then  n + m n + m  will be an even number greater than  2 2  and therefore not prime. Otherwise  n n  is even. The only even number prime number is  2 2  and it can be verified that  2 + 7 = 9 2 + 7 = 9  is not a prime number. Time ...

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