Is it a Multiple?

$2^{150}=(2^5)^{30}=32^{30}\equiv 1^{30}=1 \pmod{31}$ ; the answer is $\boxed{Yes}$

It can also be solved using Euler's theorem as follows:

$\begin{aligned} 2^{150} - 1 & \equiv 2^{\color{#3D99F6} 150 \bmod \phi(31)} - 1 \text{ (mod 31)} & \small \color{#3D99F6} \text{Since }\gcd(2,31) = 1 \text{, Euler's theorem applies.} \\ & \equiv 2^{\color{#3D99F6} 150 \bmod 30} - 1 \text{ (mod 31)} & \small \color{#3D99F6} \text{Euler's totient function }\phi(31) = 31-1 = 30 \\ & \equiv 2^{\color{#3D99F6} 0} - 1 \text{ (mod 31)} \\ & \equiv \boxed 0 \text{ (mod 31)} \end{aligned}$

Yes , $31 \mid 2^{150}-1$

$2^{150}\equiv 1\pmod{31}$

$\Rightarrow (2^{30})^5\equiv 1\pmod{31}$ and $2^{30}\equiv 1\pmod{31}$ by Fermat’s Little Theorem.

$\Rightarrow (2^{30})^5\equiv 1\pmod{31} = 1^5\equiv 1\pmod{31} = 2^{150}\equiv 1\pmod{31}$

$\Rightarrow 31 \vert 2^{150}-1$

$\Rightarrow \boxed{Yes}$ .