Otdfbt

I'm not not schooled in math. I'm 50 years old and I only have about a grade 8 level. But I do enjoy math and heard a question in the show "Growing Pains of a Teenage Genius" that interested me. So please forgive me. I do not speak "math."

The question has been posted here already, but I don't think the correct answer was given, and since I'm new, I haven't earned the points to be able to comment on that post. So I've started my own post.

The question is, if you have 1000 pennies lined up in a row, all heads up, and you turn over every second penny, then every third penny, then every fourth penny, etc. all the way until you turn over the thousandth and last penny, which ones will be heads up?

I've figured out that the answer is that the square numbers will be heads up. It is only the square numbers that will be flipped an even number of times to land them in the position they started out in. But I don't know why that is.

What is it about square numbers that they are the only ones that get flipped an even number of times through the process of flipping every 2nd, 3rd, 4th,...etc, penny?

I thought it must have something to do with factoring since the primes will only get flipped once, but the widening gap between each succession of flips is a bit complicated to visualize, and I don't know how to work that with factoring square numbers.

Is there something special about factoring square numbers that's applicable here?

How do you visualize this problem mathematically?

Each penny will be flipped a number of times equal to the number of divisors it has (including or not including $1$ based on the specific wording of the problem).

Supposing that $d$ is a divisor of $n$, i.e. that there is some $k$ such that $n = dtimes k$, then $k$ is also a divisor. In the event that $k$ is different than $d$ then it will be counted separately than $d$ when counting the total number of divisors. In this way, every single divisor $d$ of $n$ that we wish to count will have a corresponding different divisor $k=fracnd$.

All except the circumstance where $n$ happens to be a square number $n=r^2$ in which case you have $r$ is a divisor and the corresponding paired divisor $fracnr$ is again equal to $r$ and so is not distinct and need not be counted a second time.

Let's look at $12$ for an example.

$12$ has the divisors $colorred1,colorblue2,colorpurple3,colorpurple4,colorblue6,colorred12$. Note how the numbers with matching colors are paired together and multiply together to give $12$.

Now, let's look at a square number as an example like $16$.

$16$ has the divisors $colorred1,colorblue2,colorpurple4,colorblue8,colorred16$. Notice here again we have the numbers with matching color multiply together to get $16$. However, in the center since $16$ is square you only have one number of that color, not two, again because the corresponding divisor associated with it happens to be the same number. This pattern continues for all numbers. Every square number has an odd number of divisors and every non-square number has an even number of divisors and it is for this reason that the only pennies left turned heads up will be the ones at the square number positions.