## July 2020

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In the Math Corner of my July Diary I included some brief general remarks about the factorial function. Then I appended this factorial-related brainteaser, which I stole shamelessly from the July 26th "Mind-Benders for the Quarantined" email list run by Dr Peter Winkler at the National Museum of Mathematics. All I can do by way of recompense is urge you to subscribe to Dr Winkler's list: The link is here.

Consider this number, which you might call a super-factorial, or factorial-factorial:

1!×2!×3!×4!×5!× … ×100!

Referring to the individual factorials that make up this number as "terms," so that 2! is a term, 37! is a term, and so on, can you remove one term in such a way that what is left is a perfect square?

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**• Solution**

The solution below is the one given by Dr Winkler, who credits both puzzle and solution to Jeremy Kun, author of
*A
Programmer's Introduction to Mathematics*. "There are several other equally valid ways to get the answer," says Dr Winkler. Indeed
there are. I received several solutions from my readers. Every one of them got the correct answer, but I don't think any two arrived at it by the same
method.

Let's call our big product *N* and observe that it's not far from already being a product of perfect squares. The product of
*N* 's first two terms, for example, is 100 × 99!^{2} (which happens to be a perfect square since
100 = 10^{2}).

In fact we could pair up all the terms and write *N* in the following form:
100 × 99!^{2} × 98 × 97!^{2} × 96 × 95!^{2} × … × 4 × 3!^{2} × 2 × 1!^{2},
which is a perfect square times the number
*M* = 100 × 98 × 96 × … × 4 × 2. But we can
rewrite *M* as
2^{50} × 50 × 49 × 48 × … × 2 × 1 = (2^{25})^{2} × 50!.

So *N* is the product of a lot of perfect squares and the number 50!. It follows that if we remove the 50! term from *N*,
we're down to a perfect square!