HMMT November, 2014

Solution:

Answer: 15 Let $d=\operatorname{gcd}(a, b)$ for some $a, b \in \mathbb{Z}^{+}$.
Then, we can write $d=a x-b y$, where $x, y \in \mathbb{Z}^{+}$, and
$$\begin{aligned} & 2^{a}-1 \mid 2^{a x}-1 \\ & 2^{b}-1 \mid 2^{b y}-1 \end{aligned}$$
Multiplying the right-hand side of (2) by $2^{d}$, we get,
$$2^{b}-1 \mid 2^{a x}-2^{d}$$
Thus, $\operatorname{gcd}\left(2^{a}-1,2^{b}-1\right)=2^{d}-1=2^{\operatorname{gcd}(a, b)}-1$.
Using $a=16$ and $b=36$, we get
$$\operatorname{gcd}\left(2^{16}-1,2^{36}-1\right)=2^{\operatorname{gcd}(16,36)}-1=2^{4}-1=15$$