FINAL ROUND

a) Let $n=a{q}_1+r_1=b{q}_2+r_2$, $a,b\in \mathbb{N}$, $a<b$, $r_1\ne 0$, $r_2\ne 0$. Suppose that $n=r_1+r_2$. Then $n=a{q}_1+r_1=b{q}_2+r_2=r_1+r_2$, ($q_1\ge 0$, $q_2\ge 0$) and $r_1<a$, $r_2<b$, so $b{q}_2=r_1<a\le b$. The inequality $b{q}_2<b$ holds only if $q_2=0$. But then $r_1=b{q}_2=b\cdot 0=0$, a contradiction.

b) Let, by condition, $$n=29q_1+r_1=39q_2+r_2=59q_3+r_3=r_1+r_2+r_3,$$ $r_1<29$, $r_2<39$, $r_3<59$. From these inequalities it follows that $59q_3=r_1+r_2\le 28+38=66$, so $q_3=1$. Then $$r_1+r_2=59.(1)$$ Further, $39q_2=r_1+r_3\le 28+58=86$, so $q_2\le 2$. Consider two cases: 1) $q_2=1$. Then $$r_1+r_3=39.(2)$$ $98=59+39=r_1+r_2+r_1+r_3=n+r_1=29q_1+2r_1,$ i.e., $29q_1+2r_1=98$, so $q_1$ is even and $29q_1<98$, hence $q_1=2$. Then $r_1=\frac{1}{2}(98-2\cdot 29)=20$. But from (1) it follows that $r_2=59-20=39$, which is impossible since $r_2<39$.

2) $q_2=2$. Then $$r_1+r_3=n-r_2=39q_2=78.(3)$$ Then from (1) and (3) we obtain $29q_1+2r_1=137$, so $q_1$ is odd and $29q_1<137$, hence either $q_1=1$ or $q_1=3$. If $q_1=1$, then $2r_1=108$, i.e., $r_1=54$, a contradiction. If $q_1=3$, then $r_1=25$, $r_2=34$, $r_3=53$, and $n=25+34+53=112$, which satisfies the problem condition.