How can prime numbers be found mentally?

Question

At a careers fair I was given a test to see how good I am at mental maths, And I was given multiple questions, asking whether a number was a prime.

Example question:

Which of these numbers isn't a prime?

$$257,317,287,263$$

Well, my first insentive is to check the 10s digit, so I added y to my answers and divided it by that y value, (Like $\frac{k+3}{3}$ is proved when $k$ is a multiple of $3$)

Turns out, none of the worked, not even the correct answer. (Which is $287$)

Which begs the question, how do you find a prime number mentally?, On a calculator it's as simple as displaying it's prime factors, but even then the computer inside is dividing it by $1,2,3,4,5,\ldots$ until its square root

My only guess is that there is a sequence that can be used to list a few primes, that knocks off $2$ of the candidates...

Edit: My technique was half right, rather than sticking to 3 i should have gone further up the prime numbers up to 20.

Answer 1

For numbers less than $400$, you only need to test whether it has a prime factor less than $20$, that is, $2,3,5,7,11,13,17,19$. Testing for $2,3,5,11$ is easy. For the others, just do the division. It helps to subtract clear multiples of one of those primes. For instance, $317$ is not divisible by $17$ because $300=317-17$ isn't. Similarly, $263$ is not divisible by $13$ because $250=263-13$ isn't (or $3=263-260$ isn't, noting that $260=13\cdot 20$).

Answer 2

$287=7\times 4\times 10+7$, so it is a multiple of 7.

Answer 3

If you have a 3 digit number $\rm ABC$ and $\rm AB$ is divisible by $\rm C$, then $\rm ABC$ is clearly not prime.

$${\rm ABC = AB \cdot 10 + C = C}\cdot(n\cdot 10+1)$$

Answer 4

If you know about quadratic reciprocity, that can sometimes be helpful. In this case, you might notice that $287$ is close to $289 = 17^{2}.$ It follows that if $p$ is a prime divisor of $287,$ then $2$ is a quadratic residue (mod $p$), so that $p \equiv \pm 1$ (mod $8$). The first prime congruent to $1$ (mod $8$) is $17$, so no prime divisor of $287$ is congruent to $1$ (mod $8$). The only prime congruent to $7$ (mod $8$) which is small enough to have a chance of dividing $287$ is $7$ itself, which does indeed work. In this case, the strategy is not quicker than trial and error, but for larger numbers, quadratic reciprocity can prune possibilities.