<s>[INST] Let $\frac{a}{b} = 1-\frac{1}{2}+\frac{1}{3}-\dots+\frac{1}{67},$ such that $\gcd(a,b)=1$. Show that $101\mid a.$ 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?
 [/INST] $\frac{a}{b}=\frac{1}{34}+\frac{1}{35}+...+\frac{1}{67}$
$\frac{a}{b}=\sum_{i=34}^{50} \frac{1}{i}+\frac{1}{101-i}$
$\frac{a}{b}=\sum_{i=34}^{50} \frac{101-i+i}{(i)(101-i)}$
$\frac{a}{b}=\sum_{i=34}^{50} \frac{101}{(i)(101-i)}$
$\frac{a}{b}=101.\left(\sum_{i=34}^{50} \frac{1}{(i)(101-i)}\right)$
therefore $101\mid a$.
thanks,Carl Schildkraut
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