largest number that leaves same remainder while dividing 5958, 5430 and 5814?

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.

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 $x=\gcd(528, 144, 384) = 48$.

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

$(1)$ $$\gcd(x,y,z)=\gcd{\gcd(x,y)\gcd(y,z)}$$ $(2)$ $$\gcd(x,y) \mid x-y$$