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. $\ddot \smile$