If it’s not divisible by $2$, could it be divisible by $4$? Is it possible for its smallest factor to be bigger than $8$?

$53$ ends in a $3$ so it’s not divisible by $2$.

$5+3 = 8$ so it’s not divisible by $3$.

Now, do we need to check $4$? If it were divisible by $4$ then it’d also be divisible by $2$, and we’ve already checked $2$, so we know it can’t be divisible by $4$.

It doesn’t end in a $0$ or $5$ so it’s not divisible by $5$.

If it’s not divisible by $2$ or $3$ then it won’t be divisible by $6$, so we don’t need to check for that. In fact, we don’t need to check for divisibility by any composite numbers, only prime numbers.

Moving on to $7$, $7\times7 = 49$ and $7\times8 = 56$, so $53$ isn’t divisible by $7$.

Can we stop checking yet? If it does have any factors (other than $1$ and itself), then we could write it as something $\times$ something, where both those numbers are at least $8$ (because we’ve ruled out all the numbers below $8$ already). But $8\times8 = 64$ which is already bigger than $53$. Or, to look at it another way, $53\div 8\lt 7$, so if we divide $53$ by any number at least as big as $8$ then we’re going to get something less than $7$. But we already know that no number less than $7$ evenly divides $53$. We can stop checking here. $53$ is prime.