Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $N>1$ be a positive integer. What will be the Arithmetic mean of positive integers less than $N$ and co-prime with $N$?

Getting no idea how to proceed!

share|improve this question
add comment

2 Answers

up vote 6 down vote accepted

(The way the problem reads at this time is "What will be the Arithmetic mean of positive integers less than $N$ and co-prime with $N$?" (I mention this in case there was some other version of the question and it got edited.))

$n$ is coprime to $N$ if and only if $N-n$ is coprime to $N$.

The average of $n$ and $N-n$ is $N/2$.

The average of the averages of all such cases is the average of a bunch of numbers each equal to $N/2$.

The average of all numbers in $\{1,\dots,N\}$ that are coprime to $N$ is therefore $N/2$.

Later edit: Here's an example. The numbers in $\{1,\dots,20\}$ that are coprime to $20$ are $1,3,7,9,11,13,17,19$.

They come in these pairs: \begin{align} 1, & 19 \text{ (The average of these two is $10$.)} \\ 3, & 17 \text{ (The average of these two is $10$.)} \\ 7, & 13 \text{ (The average of these two is $10$.)} \\ 9, & 11 \text{ (The average of these two is $10$.)} \end{align} Now take the average of all those $10$s. The average is $10$.

share|improve this answer
add comment

HINT:

As $(a,n)=(n-a,n),$

Let $$S=\sum_{1\le a\le n,(a,n)=1}a,$$

then $$S=\sum_{1\le a\le n,(a,n)=1}(n-a)$$

$$\implies2S=\phi(n)(a+n-a)=n\phi(n)$$ where $\phi(n)$ is the Euler Totient function, the number of positive integers $<n$ and co-prime to $n$

The Arithmetic mean will be $\frac S{\phi(n)}$

share|improve this answer
    
It looks to me as if you've made the problem far more complicated than it is and your answer is wrong. The way the problem reads at this time is "What will be the Arithmetic mean of positive integers less than $N$ and co-prime with $N$?" (I mention this in case you read some other version of the question and it got edited.) –  Michael Hardy Jul 25 '13 at 15:51
    
@MichaelHardy, what is wrong here? I just changed the case of $N$ from upper to lower? Here the mean will be $\frac{n\phi(n)}{2\phi(n)}=\frac n2$ –  lab bhattacharjee Jul 25 '13 at 15:53
    
See my answer. You've made it far more complicated than it is. –  Michael Hardy Jul 25 '13 at 15:54
    
@MichaelHardy, but what is wrong here???? –  lab bhattacharjee Jul 25 '13 at 15:55
1  
To be even more clear - he counts each pair $(a, n-a)$ twice. Once for $a,$ once for $(n-a).$ This gives the total count of $\phi(n).$ That's why he has $2S$ instead of just $S.$ Regardless, this discussion wouldn't have arisen had lab written his answer in a less confusing manner. I've already up-voted yours and just didn't want people to incorrectly assume this answer is incorrect. –  lyj Jul 25 '13 at 16:22
show 16 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.