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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.

Thanks.

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    $\begingroup$ It looks like you have the correct answer and I agree with your reasoning. Textbooks are written by humans, and we are all prone to mistakes on occasion. You have probably just found a typo in the solutions. $\endgroup$ – JMoravitz Sep 10 '15 at 0:43

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