Recent content by d4rkwarrior

But how do I prove that? What method can I use?

micromass; how can I continue from here? By using mod 4, the sum, S, will be: S \equiv 0, 1, 2, 3 (mod 4). Also, 13 \equiv 1 (mod 4) How do I exclude the possibility of S \equiv 1 (mod 4)?

I am supposed to prove it by congruences, and by the way showing that it is impossible with last digit can't prove it's impossible with two, not even in a similar way. And I have to prove that it is IMPOSSIBLE or possible for certain values of n.

Homework Statement Does there exist an integer n, such that 1+2+3+...+n, ends with the last two digits 13? Homework Equations 1+2+3+...+n = n(n+1)/2 The Attempt at a Solution I reached a conclusion that 1+2+3+...+n \equiv 13 (mod 100). Also the sum has to be greater than 100, but...