Why does $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ have new divisors $59$ and $509$ all of a sudden? I am a noob when it comes to math so please bear with me. Why $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ has $2$ new divisors $59$ and $509$. I mean, all of its divisors are prime factors and by no way $59$ nor $509$ can be reached when multiplying any of its divisors. 
How does the remainder $1$ affect the divisibility of a number?
 A: When you multiply any set of numbers and then sum one, you are guaranteeing that non of the numbers in the product part will divide your result (there will always be a remainder of 1 and that is only divisible by 1, therefore it doesn't have a factor greater than 1 with the product). 
If your set of choice is the first $k$ primes, then non of those can possibly divide the result of the "product + 1", hence new prime factors appear and you can even find the result is a prime of it's own (like $2\times3\times5+1$).
A: Nice question, this is a co-prime triple and no one really knows what is going on with them http://en.wikipedia.org/wiki/Abc_conjecture
A: You can understand this much better if you know the concept of 'co-primes'.
Two numbers are said to be co-primes, if they do not have common factor other than 1. 
Also note that for any positive integer $n$ where $(n\neq1)$, $n$ and $n+1$ are always co-primes as they will not have any common factor.Hence they are co-primes.
Let $a= 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$.
Let $b = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$
As $a$ and $b$ are co-primes, none of the factors which divide $a$ will divide $b$.
Also note that any number of the form $p_1 \cdot p_2 \cdot  \cdot  \cdot  \cdot p_n + 1$,where $p_1,p_2,\cdot\cdot\cdot p_n$ are all the primes till $p_n$, is either a prime or divisible by a prime not included in the list. 
