Answer to Question #142682 in Combinatorics | Number Theory for mkami

Question

When dividing each of the numbers 3187 and 3319 by some positive integer number, the remainder happened to be equal to 8. What is the smallest possible value of the divisor?

Answer 1

Let $x=$ the value of divisor, $x>8.$ Then for some integers $m$ and $n$

$$3187=mx+8, 3319=nx+8$$

$$3319-3187=(n-m)x$$

$$(n-m)x=132$$

$$132=2\cdot2\cdot3\cdot11$$

The list of all positive divisors is as follows: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132

Since $>8,$ then the smallest possible value of the divisor is 11.