Solving Fractional Part Sum S(n): a,b,n Natural Non-Null Numbers

[redount2k9]

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!

 

[Simon Bridge]

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".

 

[redount2k9]

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 learned in middle school.
Thanks!
Note: This is the source of the problem http://www.viitoriolimpici.ro/uploads/attach_data/112/45/26//4e02c08p03.pdf

 

[Simon Bridge]

You didn't understand.

I had a feeling...

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 learned in middle school.
Thanks!

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]

 

[redount2k9]

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!

 

[Simon Bridge]

Cool - link to the solution?

 

[redount2k9]

http://uk.answers.yahoo.com/questio...H4icEYwazKIX;_ylv=3?qid=20121228015407AA8frcq

 

[Simon Bridge]

Thanks :)