TITLE: Which of the following is not a prime number?
QUESTION [18 upvotes]: Which of the following is not a prime number ?

$a.)\ 911 \ \ \ \ \ \ \ \ \ \ b.)\ 919 \\
\color{green}{c.)\ 943} \ \ \ \ \ \ \ \ \ \ d.)\ 947$
This was asked in my exam and the time given per question was $1-3\ \text{ min}$
If the time was $10\ \text{min}$ for this question I would have solved this comfortably but since the time was less i couldn't solve it with dividing every option with shorter primes   .
I look for a simple and short way.
I have studied maths up to $12$th grade.

REPLY [2 votes]: As others have mentioned, you need to check divisibility only against prime numbers that are smaller than or equal to the square root of the dividend.
And since the square roots of all the dividends are smaller than $31$, you need to check divisibility of each dividend only by $2,3,5,7,11,13,17,19,23,29$.
Checking divisibility against $2,3,5$ is easy, so I'm not going to expand on it.
Checking divisibility against $7$ is easy in this case:

$911$ is not divisible by $7$ since $910$ is
$919$ is not divisible by $7$ since $910$ is
$943$ is not divisible by $7$ since $950$ would have to be
$947$ is not divisible by $7$ since $940$ would have to be

Checking divisibility against $11$ is easy in this case:

$911$ is not divisible by $11$ since $900$ would have to be
$919$ is not divisible by $11$ since $930$ would have to be
$943$ is not divisible by $11$ since $990-44=946$ is
$947$ is not divisible by $11$ since $990-44=946$ is

Checking divisibility against $13$ is easy in this case:

$911$ is not divisible by $13$ since $910$ is
$919$ is not divisible by $13$ since $910$ is
$943$ is not divisible by $13$ since $930$ would have to be
$947$ is not divisible by $13$ since $960$ would have to be

Checking divisibility against $17$ is partially easy in this case:

$911$ - $\color\red{\text{you'll have to do the math}}$
$919$ - $\color\red{\text{you'll have to do the math}}$
$943$ is not divisible by $17$ since $960$ would have to be
$947$ is not divisible by $17$ since $930$ would have to be

Checking divisibility against $19$ is easy in this case:

$911$ is not divisible by $19$ since $930$ would have to be
$919$ is not divisible by $19$ since $900$ would have to be
$943$ is not divisible by $19$ since $950$ is
$947$ is not divisible by $19$ since $950$ is

Checking divisibility against $23$ is easy in this case:

$911$ is not divisible by $23$ since $920$ is
$919$ is not divisible by $23$ since $920$ is
$943$ is divisible by $23$ since $920+23$ is