Which of the following is not a prime number?

the smaller prime factor has to be less that $\sqrt{961} = 31$

Don't start at $2$ and work your way up. Start at $29$ and work your way down.

Four fails at $29$. In all four cases, the first digit of the quotient was $3$.

Two fails at $23$.

Then $943 = 23 \times 41$.

or

If they taught you the difference of two squares method...

\begin{array}{|c|rrrr|} \hline 31^2 & 961 & 961 & 961 & 961\\ n & 911 & 919 & 943 & 947\\ \hline 31^2 - n & 50 & 42 & 18 & 14\\ +(2\cdot 31 + 1) & +63 & +63 & +63 & +63\\ \hline 32^2 - n & 113 & 105 & \color{red}{81} \\ \hline \end{array}

So \begin{align} 32^2 - 943 &= 9^2\\ 32^2 - 9^2 &= 943\\ (32-9)(32+9) &= 943\\ 23 \times 41 &= 943 \end{align}

Since you don't know what you're looking for, the most sensible strategy is to check the numbers to see if they're divisible by $2, 3, 5, 7, \dots, 29$ and stop when you find one that is. Obviously, this involves a lot of individual checks, so the problem is how to go fast. For $2, 3, 5, 11$ there are well-known criteria for divisibility, and you quickly determine that none of the four numbers are multiples of these.

Since the four numbers are all close together, it's easiest to simply list the multiples of $7, 13, 17, \dots$ that are close to the numbers.

The multiples of $7$ are: $910, 917, 924, 931, 938, 945, \dots$.

The multiples of $13$ are: $910, 923, 936, \dots$

The multiples of $17$ are: $901, 918, 935, \dots$

The multiples of $19$ are: $950, 931, 912, \dots$

The multiples of $23$ are: $920 - 23, 920, \mathbf{943}$.