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Here is the question, from the 2013 Division A mathletes competition in Illinois. I'm doing this for practice for my test and I've seen this come up more than once:

Find the largest integer that divides $300$, $417$, and $764$ with remainders $R_1$, $R_2$, and $R_3$, respectively, such that $R_2=R_1+3$ and $R_3=R_2+5$.

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So you want \begin{align} 117=417-300= 3 \mod n\\ 347=764-417 = 5\mod n \end{align} where $n<300$. Hence $n \mid 114$ and $n \mid 342$. Since \begin{align} 114= 2\cdot 3\cdot 19 \ \ \text{ and } \ \ 342= 2\cdot 3^2\cdot 19 \end{align} then $n=2\cdot 3 \cdot 19=114$ is the greatest common divisor.

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  • $\begingroup$ Thank you so much! I did not know about that operation entirely. Totally opened up the problem! $\endgroup$ – PhilTheLawyer Apr 14 '17 at 5:24

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