177. "Common Divisor". Here is a puzzle that readers often ask me (of course, the specific numbers in it are different). A correspondent of a provincial newspaper reported that many teachers had undermined their health in vain attempts to solve it! Probably, he exaggerated a bit, because the question is actually simple, if you know which angle to approach it from.

It is as follows. Find a number that, when dividing the three numbers 480608, 508811, and 723217, would give the same remainder.
177. When dividing these numbers by the sought number, the remainders are the same. Therefore, if we subtract one number from the other as shown below, the difference will be divisible by the sought number without a remainder.

![](https://cdn.mathpix.com/cropped/2024_05_21_56f37ca27ac3d928838dg-221.jpg?height=163&width=448&top_left_y=2077&top_left_x=813)

The prime divisors of the number 28203 are \(3, 7, 17, 79\), and \(214406 - 2, 23\), 59, 79. The only common divisor of the two differences is 79. Therefore, the sought number is 79, and the common remainder is 51. Simple, isn't it?