TITLE: largest number that leaves same remainder while dividing 5958, 5430 and 5814?
QUESTION [8 upvotes]: Largest number that leaves same remainder while dividing 5958, 5430 and 5814 ?


$$5958 \equiv 5430 \equiv 5814 \pmod x$$
 $$3\times 17\times 19 \equiv 5\times 181\equiv 3\times 331\pmod x$$
$$969 \equiv 905\equiv 993\pmod x$$
After a bit of playing with the calculator,
I think the answer is $48$ but I don't know how to prove it. 
Sorry if the answer is too obvious, I am still trying to wrap my head modular arithmetic and not very successful yet.It would be great help if anybody would give me some hints on how to proceed ahead. Thanks. $\ddot \smile$

REPLY [21 votes]: Okay, first we set it up even though we have no idea where we are going.
$5958 \equiv n \mod x$
$5430 \equiv n \mod x$
$5814 \equiv n \mod x$.
Then we noodle a bit to make it smaller and more pallatable.
$5958 - 5814 = 144 \equiv 0 \mod x$
$5814 - 5430 = 384 \equiv 0 \mod x$
and 
$5958 - 5430 = 528 \equiv 0 \mod x$
Well, now it's very clear that $x$ is a common divisor and all we have to do is find the $x = \gcd(144,384, 528)$
$x = \gcd(144 = 2^43^2, 384 = 2^7*3, 528 = 2^4*3*11) = 2^4*3 = 48$.
And we're done.