integers less than 2010 that have exactly 3 factors I have stumbled upon this problem during a past paper sheet and was stumped at how someone might actually do this at all by hand (no calculators and time-constrained) 
I am aware that these sorts of numbers have factors a specific form after messing around, which is:
n=  $1$ x $p$ x $p^2$ where $p$ denotes a prime
However, that would require me working out the number of primes whose squares are less than 2010, identifying from the 44 normal perfect square numbers ( $x^2$ < 2010);  which by hand seems to be incredibly difficult and time taking. This leads me to believe that I'm doing something completely wrong in my approach. Could someone please tell me what I'm doing wrong/fix my mistake. Thanks.
 A: Counting primes less than 44 is perfectly doable by hand inside 30 seconds at most using the sieve of Eratosthenes.  Indeed, they are
2,3,5,7,11,13,17,19,23,29,31,37,41,43
and you can leave them as $n=p^2$ form instead of multiplying them out (although it isn't difficult).  Indeed, I remember memorising all primes less than 200 and recite all squares less than 10000 as primary school exam questions when I was at that age about 30 years ago.
A: Your approach is absolutely fine. All the required numbers are of the form $p^2$ with $p$ being prime and we just have to search for all prime not larger than $44$, i.e. $2$, $3$, $5$, $7$, $11$, $13$, $17$, $19$, $23$, $29$, $31$, $37$, $41$ and $43$.
We can use the Eratosthenes Sieve Method to find all such primes. First list all numbers from $2$ to $44$. Then delete all multiples of the first number, i.e. $2$, except $2$ itself. We circle the number $2$, which is confirmed a prime. Now the smallest number left is $3$. We then delete all multiples of $3$ other than $3$ itself. Circle $3$ to indicate that it is a prime. The smallest integer left is now $5$. Do the same thing for $5$ and repeat the procedure until we pass the square root of $\sqrt{44}$. As $\sqrt{44}$ is less than $7$, the procedure for $5$ is actually the last one. Circle all the remaining integers. They are all prime.
