A combination lock X number of positions. To open the lock, you move to a certain number in the clockwise direction, then to a number in the counterclockwise direction, and finally to a third number in the clockwise direction. Consecutive numbers in the combination cannot be the same. If there are 1500 students at a high school, what is the smallest number for the number of positions (X), so that each student has a unique combination?
How I did it:
For the lock, how it goes is that if it has X positions, you have X numbers to choose from to go clockwise. In the counterclockwise position, you have X-1 positions to go to because of the consecutive numbers rule. For the third movement clockwise, you can also go X-1 because you can't move to the second number.
Therefore, the number of combinations is (X) * (X-1) * (X-1). So we can just test numbers until we get to 1500 or greater. I tested a couple of numbers and got the answer of 13 positions.
Thus, 13 * 12 * 12 = 1872.
My textbook says the answer is 12 positions, but how can that be?
12 * 11 * 11 = 1452, which is less than 1500. There wouldn't be enough locks for the 1500 students.