Question #142682
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?
1
Expert's answer
2020-11-08T18:45:50-0500

Let x=x= the value of divisor, x>8.x>8. Then for some integers mm and nn


3187=mx+8,3319=nx+83187=mx+8, 3319=nx+8

33193187=(nm)x3319-3187=(n-m)x

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

132=22311132=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,>8, then the smallest possible value of the divisor is 11.



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