10 logical and mathematical problems for neural network programmers

Since each lock is initially closed, we need to understand how many times the state of each lock will be changed in order to determine which locks will be open after the 100th pass.

Passages are carried out as follows:

1. In the first pass, the hacker switches each lock, so that locks 1, 2, 3, …, 100 are now open because they were switched once.

2. In the second pass, he switches every second lock (locks 2, 4, 6, …, 100), changing their state to the opposite, i.e. closed ones become open, and vice versa.

3. In the third pass, every third lock is switched (locks 3, 6, 9, …), again changing their states.

This process continues until the 100th pass, when the hacker switches only the 100th lock.

The key observation here is that the lock switches in those passages that are dividers of its number. For example, the 12th lock will be switched in the 1st, 2nd, 3rd, 4th, 6th and 12th passes.

The lock will eventually remain open if it is toggled an odd number of times. This will happen if the castle has an odd number of divisors. However, most numbers have an even number of divisors. For example, the number 12 has divisors 1, 2, 3, 4, 6, 12, of which there are six, and it is an even number.

The exception is perfect squares because they have one divisor that is paired with itself (the same number). For example, the number 9 has divisors 1, 3, 9, and the number 3 appears only once in the list. The squares 1, 4, 9, …, 100^2 are the numbers 1, 16, 81, …, 10000. Exact squares have an odd number of divisors.

So only the square numbers labeled 1^2, 2^2, …, 10^2 will remain open because they are switched an odd number of times. As a result, exactly 10 locks will be opened – these are the squares of numbers from 1 to 10.