2017/18 AHSMC Part 1 Problem 4

If $a$, $b$ are the first two numbers in the sequence, then the third number in the sequence must be $\dfrac{b}{a}$ (because $a$ times $\dfrac{b}{a}$ equals $b$). Similarly we can find that the sequence of numbers around the circle must be $a$, $b$, $\dfrac{b}{a}$, $\dfrac{1}{a}$, $\dfrac{1}{b}$, $\dfrac{a}{b}$, $a$, $b$, $\cdots$. Thus the sequence repeats after six terms. If there were a shorter sequence of at least three different terms starting $a$, $b$, its length must divide into $6$, so it must have length $3$. This would mean that the fourth term $\dfrac{1}{a}$ would equal the first term $a$, which means that $a = 1$, but then the second and the third terms ($b$ and $\dfrac{b}{a}$) of the sequence are equal, a contradiction. So the only possible length is $6$. The answer is (C).