Hi everyone! How to solve this: S(n) = { (a+b)/n } + { (2a+b)/n } + { (3a+b)/n } + ... + { (na+b)/n } where {x} represents fractional part of x. a,b,n are natural non-null numbers and (a,n)=1. I don`t need only an answer, i need a good solution. Thanks!
Welcome to PF; Have I understood you... $$S(n)=\frac{a+b}{n}+\frac{2a+b}{n}+\cdots +\frac{(n-1)a+b}{n}+\frac{na+b}{n}$$ ... ... if this is what you intended, then it looks straight forward to simplify: notice that each term is over a common denominator ... you should be able to see what to do from there. Note: I don't know what you mean by "{x} is the fractional part of x" or "(a,n)=1".
You didn't understand. We say that a number x = {x} + [x] http://en.wikipedia.org/wiki/Fractional_part (a,n) = 1 that means the greatest common factor Ex: (2, 3)=1 (23, 29)=1 (14, 19)=1 Hope this helps but I think you know physics better because this notions are learnt in middle school. Thanks! Note: This is the source of the problem http://www.viitoriolimpici.ro/uploads/attach_data/112/45/26//4e02c08p03.pdf
I had a feeling... Thanks for the detailed description. You did provide the context with "fractional part" but I didn't get it because this was not taught that way, with those words, in NZ when I went to "middle school" (though I may have missed that class due to dodging dinosaurs and contemplating the possibilities of this new-fangled "wheel" thingy.) Mind you - (a,b) for "greatest common divisor" (factor - whatever) would be an older and ifaik uncommon notation - it is more usual to see "gcd(a,b)" instead. This sort of thing makes international forums more fun :D So... $$\sum_{i=1}^n \left \{ \frac{ia+b}{n} \right \}=\sum_{i=1}^n \frac{ia+b}{n}-\sum_{i=1}^n \left \lfloor \frac{ia+b}{n} \right \rfloor$$ It occurs to me that the properties of the floor function may help here? ........ Note: This is the source of the problem http://www.viitoriolimpici.ro/uploads/attach_data/112/45/26//4e02c08p03.pdf[/QUOTE]
Thanks anyway man, a very good math guy helped me on other forum with a modulo n solve. It was the best I`ve ever seen. It is solved now, but thanks!