as students arrive the first day of school, they find 300 pennies arranged on a table each penny is facing heads up and is labled as Penny 1, Penny 2, Penny 3, ect. as each studenty passe the table they are to turn over only the pennies that match their number in line, student one turns over all the pennies, student #2 turns over all pennies divisible by two, student three turns over all pennies divisible by three ect. You are to identify the set of all penny numbners that are heads up after all three hundred students have passed by the table?
This is based on the fact that every perfect square has an
odd number of factors and every other positive integer has
an even number of factors.
To illustrate:
The factors of NON-perfect square 24 are 1,2,3,4,6,8,12,24
Notice there are 8 of them. That's an EVEN number.
The factors of PERFECT square 36 are 1,2,3,4,6,9,12,18,36
There are 9 of them. That's an ODD number.
The reason for that is that the pair of factor (p, N/p) furnishes
two factors of N as long as p does not equal to N/p, but when
p = N/p, that only furnishes 1 factor. And that only occurs when
N = p².
Each penny that corresponds to a perfect square will have been
turned over an odd number of times, making it tails. and every
other penny will have been turned over an even number of times
and they will be heads. So all will be heads except the pennies
corresponding to perfect squares, and they will all be tails.
Edwin