Factor 43361 knowing $\phi(43361)$ 
It is given that the number $43361$ can be written as product of two distinct prime numbers $p_{1}$ and $p_{2}$. Further, assume that there are $42900$ numbers which are less than $43361$ and co-prime to it. Then find $p_{1}+p_{2}$.

A simple search on Google yielded $p_{1}$ and $p_{2}$ to be $131$ and $331$. But what would be the proper way to find it?
 A: As carmichael561's answer shows, it is not neccessary to factor the number to solve the original problem. However, the factorization can be done like this:
It is given that 43361 is the product of two primes, and that there are 42900 numbers less than 43361 that are co-prime to it. I.e. that $ \phi(43361) = 42900 $ where $ \phi$ is Euler's totient function.
Furthermore, we know that $ \phi(p) = p-1 $ when $ p $ is a prime, and that $\phi(n) = \phi(p_1)\phi(p_2) $ when $n$ is a product of the two primes $p_1$ and $p_2$.
Using this, together with factoring of 42900 gives
$$ 43361 = p_1 p_2 $$
$$ (p_1-1)(p_2-1) = 42900 = 2 \cdot 2 \cdot 3 \cdot 5 \cdot 5 \cdot 11 \cdot 13 $$
Since 43361 end with 1, both $p_1$ and $p_2$ must end with 1, and thus both $p_1-1$, and $p_2-1$ must be multiples of 10. Then we have that $p_1 = q_1 \cdot 10 + 1$ and $p_2 = q_2 \cdot q_3 \cdot 10 + 1$ where $q_1, q_2,$ and $q_3$ are the remaining prime factors 3, 11, and 13.
Since $ 11 \cdot 10 + 1 = 111$ is not prime, $q_1$ can't be 11. Similarly, since $11 \cdot 13 \cdot 10 + 1 = 1431$ is not prime, $q_2$ and $q_3$ cant't be 11 and 13, and then $q_1$ cant be 3 either. The only posibility is then that $q_1$ is 13, and thus 
$$ 43361 = (13 \cdot 10 + 1)(3 \cdot 11 \cdot 10 + 1) = 131 \cdot 331$$  
A: Let $n=43361$. You're given that $n=p_1p_2$ is a product of two distinct primes, and also that $\phi(n)=42900$, where $\phi$ is Euler's totient function. Since $\phi$ is multiplicative, it follows that
$$ 42900=\phi(n)=(p_1-1)(p_2-1)=n-(p_1+p_2)+1=43362-(p_1+p_2)$$
and so
$$ p_1+p_2=43362-42900=462.$$
A: Here is a method that I used, it helps you if you don't know the Euler's totient thing.
Since 42900 numbers before 43361 are divisible by neither p1 nor p2 we can tell that 462(inclusive of 43361) numbers are divisible by either p1 or p2 (with 43361 as an exception which is divisible by both).
let us say x numbers are divisible by p1 and y numbers by p2.
then, taking multiples of p1,
                        *p1, 2p1, 3p1, .....43361 is the series*    with a total of x 
                                                                    numbers in it.

we know from arithmetic progression that,
An= A+ (N-1)D
                                43361= p1+(x-1)p1

solving, we get                     x= 43361/p1
similarly,
                     *p2, 2p2, 3p2, .....43361 is the series*    with a total of y 
                                                                 numbers in it.
                                43361= p2+(y-1)p2

solving, we get                      y= 43361/p2
since x+y= 462,
                             43361/p1 + 43361/p2 = 462

                              43361(p1+p2/p1.p2) = 462

                              43361(p1+p2/43361) = 462

                                

Solving, p1+p2= 462
