modular arithmetic - Find the greatest number that will divide $2327, 2677, 4007$ and $497$ and will leave a remainder of $17, 37 ,47$ and $57$ respectively.

Find the greatest number that will divide $2327, 2677, 4007$ and $497$ and will leave a remainder of $17, 37, 47$ and $57$ respectively.

My attempt:

Since all these numbers leave a remainder,

$\begin{align}2327 -17 &\to 2310 \tag1\\ 2677 -37 &\to 2640 \tag2 \\ 4007 -47 &\to 3960 \tag3\\ 497 -57 &\to 440 \tag4\end{align}$

Now simply by brute force method (took me a couple of tries) I ended up with the result $110$.

$\begin{align} 2640-2310&=330 && \tag 5 \\ 440-330 & = 110 \tag6\end{align}$

This result seems to work for me. I checked the remainders after subtracting and they all comply.

But the question explicitly asks for the the greatest result. How do I know there are no better answers?

And also is there a faster way to solve it? (This was a MCQ and we are supposedly to spend less than 2 min per question and I took well over 10).

Thanks!

Answer

You are correct in your method. $$2327 \equiv 17 \pmod {x} \iff 2310 \equiv 0\pmod x$$ $$2677 \equiv 37 \pmod {x} \iff 2640 \equiv 0\pmod x$$ $$4007 \equiv 47 \pmod {x} \iff 3960 \equiv 0\pmod x$$ $$497 \equiv 57 \pmod {x} \iff 440 \equiv 0\pmod x$$
Thus $$x=\gcd(3960, 2640, 2310, 440)=\gcd((\gcd(3960, 2640), \gcd(2310, 440))$$ Now apply the Euclidean Algorithm, to get, as you got, $110$. So your answer is correct.