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Fractional part sum 
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#1
Dec2912, 10:30 AM

P: 12

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 nonnull numbers and (a,n)=1. I don`t need only an answer, i need a good solution. Thanks! 


#2
Dec2912, 10:46 PM

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Welcome to PF;
Have I understood you... $$S(n)=\frac{a+b}{n}+\frac{2a+b}{n}+\cdots +\frac{(n1)a+b}{n}+\frac{na+b}{n}$$ ... ... 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". 


#3
Dec3012, 02:58 AM

P: 12

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/upload...4e02c08p03.pdf 


#4
Jan113, 10:32 PM

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Fractional part sum
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 newfangled "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/upload...4e02c08p03.pdf[/QUOTE] 


#5
Jan213, 02:33 AM

P: 12

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!



#7
Jan313, 01:36 AM

P: 12



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