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. $\ddot \smile$
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.
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 &lt;- [1..870806385 + 49], 5958 `mod` x == 5430 `mod` x, 5430 `mod` x == 5814 `mod` x]