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Sa se determine numerele naturale \( n\in\mathb{N}* \) si \( r_1<r_2<r_3<...<r_n\in \mathb{N}* \) astfel incat \( \frac{17}{18}=\frac{1}{r_1}+\frac{1}{r_2}+...+\frac{1}{r_n} \) .

\( \frac{1}{r_1} \)\( +\frac{1}{r_2}+......+ \)\( \frac{1}{r_n}=1 \)\( -\frac{1}{18} \).
Avem de exemplu, pentru n=9:
\( \frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{33}+\frac{1}{45}+\frac{1}{385}=1 \)
Dar \( \frac{1}{2}+\frac{1}{2}=1 \). Atunci \( \frac{1}{2}+\frac{1}{2}(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{15}+\frac{1}{33}+\frac{1}{45}+\frac{1}{385})=1 \); \( \frac{1}{2}+\frac{1}{6}+\frac{1}{10}+\frac{1}{14}+\frac{1}{18}+\frac{1}{22}+\frac{1}{30}+\frac{1}{66}+\frac{1}{90}+\frac{1}{770}=1 \).
Deci, o solutie in numere naturale este (2;6;10;14;22;30;66;90;770).