# »  Solutions to puzzles in my VDARE.com monthly Diary

## 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 = 102).

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 250 × 50 × 49 × 48 × … × 2 × 1 = (225)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!