largest number that leaves same remainder while dividing 5958, 5430 and 5814? 
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. $\ddot \smile$ 
 A: Since $5958 \equiv 5430 \mod{x}$, then $x | 5958 -5430 = 528$.  Likewise $x | 
5958 - 5814 = 144$ and $x | 5814 - 5430 = 384$.  So $x$ is a common divisor of
$528,$ $144,$ and $384$.  So you want $x=\gcd(528, 144, 384) = 48$.
A: Let the number leaving the same remainder be $a$ and the remainder be $b$.
So, $\gcd(5958-a,5430-a,5814-a)=b$
Note that 
$(1)$ $$\gcd(x,y,z)=\gcd{\gcd(x,y)\gcd(y,z)}$$
$(2)$ $$\gcd(x,y) \mid x-y$$
A: 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.
A: The only "small" solutions (below 10000000) seem to be ${1,2,3,4,6,8,12,16,24,48}$. 
Of course, you can add $lcm({5958, 5430, 5814}) = 1741612770 $ to a solution and get another, bigger one.
The set above was found by letting do the computer what a computer can do best: Computing stuff. Here is the Haskell Code I used:

[x |  x <- [1..870806385 + 49], 5958 `mod` x == 5430 `mod` x, 5430 `mod` x == 5814 `mod` x]

