Final Round

a. The sequence starts as follows. $$\begin{array}{rlrlrlrl}{a}_1& =\frac{10}{11},& a_2& =\frac{12}{14}=\frac{6}{7},& a_3& =\frac{8}{10}=\frac{4}{5},& a_4& =\frac{6}{8}=\frac{3}{4},\\ a_5& =\frac{5}{7},& a_6& =\frac{7}{10},& a_7& =\frac{9}{13}& & \end{array}$$ It seems that the last simplification occurred at $a_4$. With induction to $n$, we will prove that there is no simplification for all $n\ge 5$. At the same time, we will prove that $a_n=\frac{1+2(n-3)}{1+3(n-3)}$ for all $n\ge 5$. For $n=5$, the statement is true, because $a_5=\frac{5}{7}=\frac{1+2(5-3)}{1+3(5-3)}$ and this fraction $\frac{5}{7}$ cannot be simplified further. Now suppose the statement is true for $n=k-1$. Consider $n=k$. Because there has been no simplification for $a_{k-1}$, the numerator of $a_{k-1}$ equals $1+2(k-4)$ and the denominator equals $1+3(k-4)$. Then the number $a_k$ is defined as $\frac{1+2(k-4)+2}{1+3(k-4)+3}=\frac{1+2(k-3)}{1+3(k-3)}$. We will argue by contradiction that there is no simplification here. Namely, suppose there is an integer $d>1$ such that both $1+2(k-3)$ and $1+3(k-3)$ are divisible by $d$. In particular, $3\cdot (1+2(k-3))-2\cdot (1+3(k-3))=1$ will also be divisible by $d$. This gives a contradiction, and the proof by induction is finished. $\square$

b. We will show that there must be a simplification at some point. Indeed, suppose there is no simplification. Just like in part (a), we can show by induction that $a_n=\frac{97+2n}{97+3n}$. In particular, we see that $a_{97}$ is not a simplified fraction, because both the numerator and denominator are divisible by 97, and that is a contradiction. $\square$

c. You can use $c=7$ or $c=27$, for example. Then we get the sequences $$\frac{7}{8},\frac{9}{11},\frac{11}{14},\frac{13}{17},\frac{15}{20}=\frac{3}{4}$$ and $$\frac{27}{28},\frac{29}{31},\frac{31}{34},\frac{33}{37},\frac{35}{40}=\frac{7}{8}.$$ $\square$