# Fractional part sum

1. Dec 29, 2012

### 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 dont need only an answer, i need a good solution.

Thanks!

2. Dec 29, 2012

### 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".

3. Dec 30, 2012

### redount2k9

Last edited: Dec 30, 2012
4. Jan 1, 2013

### Simon Bridge

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]

Last edited: Jan 1, 2013
5. Jan 2, 2013

### redount2k9

Thanks anyway man, a very good math guy helped me on other forum with a modulo n solve. It was the best Ive ever seen. It is solved now, but thanks!

6. Jan 2, 2013

### Simon Bridge

Cool - link to the solution?

7. Jan 3, 2013

8. Jan 3, 2013

Thanks :)