On the first day of the year, the temple was inaugurated with the lighting of all the bulbs... There were 365 bulbs and each one had a push button. Pressing a button, its bulb would turn on if it was off and would turn off if it was on...
On the 2nd day of the year, the 2nd, 4th, 6th, 8th, 10th, ... buttons were pressed.
On the 3rd day of the year, the 3rd, 6th, 9th, 12th, 15th, ... buttons were pressed.
On the 4th day of the year, the 4th, 8th, 12th, 16th, 20th, ... buttons were pressed.
On the 5th day of the year, the 5th, 10th, 15th, 20th, 25th, ... buttons were pressed.
And so every day of the year...
On the last day of the year, how many bulbs and which ones will be on?
This is a challenge of Nina Guindilla's solved by Pepe Chapuzas:
On the last day of the year, a light bulb will be on if its button is pressed an odd number of times. This happens only if the button number is a perfect square... because a perfect square has an odd number of divisors... because its divisors are paired except its square root... because the perfect square can be writen as a product of two different divisors or as a product of two equal divisors (its square root)... because only perfect squares have their square roots as divisors...
Pepe Chapuzas was out of breath... but continued...
So, the 19 bulbs turned on will be the numbers...
1, 4, 9, 16, 25, 36, 49, 64, 91, 100, 121, 144, 169, 196, 225, 256, 289, 324 and 361.
If the year was leap we'd have the same light bulbs turned on... because the following perfect square is 400 > 366 .
Finally Pepe asked:
What is the smallest natural number N which has exactly D divisors?
Nina Guindilla made this table:
What a chaotic sequence!