BetonaBG Posted September 23, 2004 Share Posted September 23, 2004 First of all, I'd like to say Hello to everyone as this is my first post Now the fun part, I spend 2-3h in painful strugle and I was unable to prove the following question. If anyone knows how to solve it it will be largely appreciated Prove for any positive integer n that: 2196^n – 25^n – 180^n + 13^n is divisible by 2004 Have Fun Link to comment Share on other sites More sharing options...
haggy Posted September 25, 2004 Share Posted September 25, 2004 2004 = 167*3*4 These are relatively prime to one another. 2196^n – 25^n – 180^n + 13^n = (13*167+25)^n -25^n -(167+13)^n +13^n Therefore 2196^n – 25^n – 180^n + 13^n = 25^n -25^n -13^n +13^n (mod 167) 2196^n – 25^n – 180^n + 13^n = 0 (mod 167) Likewise 2196^n – 25^n – 180^n + 13^n = 0 (mod 3) 2196^n – 25^n – 180^n + 13^n = 0 (mod 4) Therefore 2196^n – 25^n – 180^n + 13^n = 0 (mod 2004) Link to comment Share on other sites More sharing options...
BetonaBG Posted September 27, 2004 Author Share Posted September 27, 2004 I'll say simple - Thanks but my gratitude pales in comparatively with my words. Link to comment Share on other sites More sharing options...
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