What is the unique three-digit positive integer satisfying 100x = 11 (mod \(997\))
+1622 Since x is a 3 digit integer, then the equation is:
\(100x = 11 + 997k\)
The smallest value of k is 11, or else x could not be a 3 digit integer satisfying the restrictions.
100x must end with two 0's, so 997k must end with 89.
Since 7 is the only number that can multiply the last digit of 997 to make the new number end with a 9.
After some testing we get k = 37.
Here is our new equation:
100x = 11 + 997 * 37
100x = 36900
x = 369.
