Let $\frac{a}{b} = 1-\frac{1}{2}+\frac{1}{3}-\dots+\frac{1}{67},$ such that $\gcd(a,b)=1$. Show that $101\mid a.$

Question

Let $\frac{a}{b} = 1-\frac{1}{2}+\frac{1}{3}-\dots+\frac{1}{67},$ such that $\gcd(a,b)=1$. Show that $101\mid a.$

i got this problem in olympiad mathematics .

$\frac{a}{b}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{67}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{66}\right)$

$\frac{a}{b}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{67}-\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{33}\right)$

$\frac{a}{b}=\frac{1}{34}+\frac{1}{35}+...+\frac{1}{67}$

Then what will I do?

Answer 1

Hint:

You're on the right track. In the new sum you have, you can pair up the terms $1/n$ and $1/(101-n)$:

$$\sum_{n=34}^{50} \frac{1}{n}+\frac{1}{101-n}.$$

Can you take it from here?

Answer 2

$\frac{a}{b}=\frac{1}{34}+\frac{1}{35}+...+\frac{1}{67}$

$\frac{a}{b}=\sum_{i=34}^{50} \frac{1}{i}+\frac{1}{101-i}$

$\frac{a}{b}=\sum_{i=34}^{50} \frac{101-i+i}{(i)(101-i)}$

$\frac{a}{b}=\sum_{i=34}^{50} \frac{101}{(i)(101-i)}$

$\frac{a}{b}=101.\left(\sum_{i=34}^{50} \frac{1}{(i)(101-i)}\right)$

therefore $101\mid a$.

thanks,Carl Schildkraut