Răspuns:
1; 16.
Explicație pas cu pas:
[tex]S=\frac{1}{1*7}+\frac{1}{7*13}+\frac{1}{13*19}+...+\frac{1}{91*97} =\frac{1}{6}(\frac{1}{1} -\frac{1}{7})+\frac{1}{6}(\frac{1}{7}-\frac{1}{13})+\frac{1}{6}(\frac{1}{13}-\frac{1}{19})+...+\frac{1}{6}(\frac{1}{91}-\frac{1}{97})=\frac{1}{6}*( \frac{1}{1} -\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+...+\frac{1}{91}-\frac{1}{97})=\frac{1}{6}*(\frac{1}{1}-\frac{1}{97})=\frac{1}{6}*\frac{96}{97}=\frac{16}{97}.\\97*S=97*\frac{16}{97}=16.[/tex]
Deci, divizorii improprii ai numarului 97S sunt 1 si 16.
