Problem 4. Consider the collinear points $M_{1}, M_{2}, M_{3}, \ldots, M_{2015}$ in this order, such that $M_{1} M_{2}=2$ $\mathrm{cm}, M_{2} M_{3}=2 M_{1} M_{2}, M_{3} M_{4}=2 M_{2} M_{3}, \ldots, M_{2014} M_{2015}=2 M_{2013} M_{2014}$.

a) Calculate the length of the segment $\left[M_{1} M_{200}\right]$.

b) Compare the lengths of the segments $\left[M_{1} M_{200}\right]$ and $\left[M_{200} M_{300}\right]$.

c) Prove that for any natural numbers $a, b, c, d$, and $1 \leq a<b \leq c<d \leq 2015$, the segments $\left[M_{a} M_{b}\right]$ and $\left[M_{c} M_{d}\right]$ have different lengths.

Dorin Popa, Slatina
## Solution and marking scheme

a) $M_{2} M_{3}=2^{2}, M_{3} M_{4}=2^{3}, M_{2014} M_{2015}=2^{2014}$
$M_{1} M_{200}=2^{100}-2$
b) $M_{200} M_{300}=2^{300}-2^{200}$
$2^{200}-2<2^{300}-2^{200} \Rightarrow M_{1} M_{200}<M_{200} M_{300}$
c) Assuming there exist natural numbers $a, b, c, d$, with $1 \leq a<b \leq c<d \leq 2015$ such that $M_{a} M_{b}=M_{c} M_{d}$, it follows that $2^{b}-2^{a}=2^{d}-2^{c}$
Dividing by $2^{a}$ results in $2^{b-a}-1=2^{d-a}-2^{c-a}$, which is impossible, since the left side is an odd number, while the right side is an even number, so the assumption made is false