15. (i) Express the fraction $23 / 30$ as the sum of three irreducible fractions, whose denominators are pairwise coprime;
(ii) Express $23 / 30$ as the sum of two irreducible fractions, whose denominators are coprime.
15. (i) Let $23 / 30=b_{1} / a_{1}+b_{2} / a_{2}+b_{3} / a_{3}$. $\left(a_{1}, a_{2}\right)=\left(a_{2}, a_{3}\right)=\left(a_{3}, a_{1}\right)=1$, $a_{j} \geqslant 2,\left(a_{j}, b_{j}\right)=1$. Since $30=2 \cdot 3 \cdot 5$, we can set $a_{1}=2, a_{2}=3, a_{3}=5.15 b_{1}+10 b_{2}+$ $6 b_{3}=23.23 / 30=1 / 2-1 / 3+3 / 5$.
(ii) $23 / 30=a_{1} / 5+a_{2} / 6 . a_{1}=3, a_{2}=1$. Both problems can have other solutions, but the three denominators in (i) are unique.