<s>[INST] How find $\sum_{0 \leq k: \leq 100 \ gcd \left( k, 100 \right) =1} f \left( \frac{k}{100} \right)$ if $f \left( x \right) = \frac{9^x}{3+9^x}$ ? How find of $\sum_{0 \leq k: \leq 100 \ gcd \left( k, 100 \right) =1} f \left( \frac{k}{100} \right)$ if $f \left( x \right) = \frac{9^x}{3+9^x}$ ? 
 [/INST] We can prove $(k,100)=(100-k,100)$
Now prove $f(x)+f(1-x)=1$  
Set $x=\dfrac k{100}$
Now $\phi(100)=\phi(25)\cdot\phi(4)=40$
If $\displaystyle S=\sum_{0\le k\le 100;\left( k, 100 \right) =1} f \left( \frac k{100} \right),$ 
$\displaystyle  S+S=\sum_{0\le k\le 100;\left( k, 100 \right) =1}\left[f \left( \frac k{100} \right)+f \left( \frac{100-k}{100} \right)\right]=\sum_{0\le k\le 100;\left( k, 100 \right) =1}1=40$
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