TITLE: A fast (and correct) way of finding a common divisor for $1805$, $1991$, $2146$ so that they have the same remainder
QUESTION [6 upvotes]: A question says: 

"Which $n \in \Bbb N$ should the numbers $1991$, $1805$, $2146$ be divided into, so that there will be the same remainders left?"

I arranged them in order: $1805,1991,2146.$ Then did subtraction: $1991-1805=186,$ $2146-1991=155,$ then $186-155=31.$ This last result is the answer. But I am skeptical about my way of finding the answer (and which seems to be a little bit stranger to me), as it wouldn't work in some cases. I would like 
1) to know your thoughts on my method, 2) you to provide some hints in order to show the right way of solving this.

REPLY [4 votes]: Given three numbers $a,b,c$ your are looking for $\gcd (b-a,c-b,c-a)$.  Your approach is a good step and the answer is guaranteed to divide $b-a$ and $c-b$.  You have done one step in Euler's method, but the answer could be smaller.  For example, let the numbers be $1,19,33$.  You would do $19-1=18, 33-19=14, 18-14=4$ but are not done as the $\gcd$ is $2$.  You can now keep going with $14-3 \cdot 4=2$ and you are done because $2$ divides both $18$ and $14$.