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

## December 2022

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In the Math Corner of my December 2022 Diary I mentioned that that 2023 is a Harshad Number: that is, a number that is divisible by the sum of its digits. Then I added that:

The number 2023 is also divisible by the sum of the squares of its digits. So far as I know there is no family name for such numbers. I hereby baptise them the Derbyshire numbers. Quick now: How many Derbyshire numbers are there less than 100?

I then added a more formal brainteaser that I'd spotted on Twitter:

Brainteaser:  Which is greater, 1000^1001 or 1001^1000? (The caret of course indicates exponentiation.)

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•  Solutions

The Derbyshire numbers less than 100 are easily enumerated. They are 1, 10, 20, and 50.

If we put those into the OEIS search box we get their sequence A034087.

Sure enough there is no family name listed for these numbers. In all fairness, though, since Amiram Eldar (presumably this one, a cosmologist at the Hebrew University of Jerusalem) has gone to the trouble of listing the first ten thousand of these numbers, they should properly be called Eldar Numbers.

Our new year 2023 bears the 59th Eldar number. The last Eldar-Numbered year before 2023 (i.e. the 58th Eldar Number) was 2010; the next one will be 2040. The 21st century sees four Eldar-numbered years altogether: those three plus the year 2000.

That's quite a wealth for centuries in the general zone of 21. The frequencies of Eldar numbers per century for the 1,700s, 1,800s, 1,900s, 2,000s, 2,100s, 2,200s, and 2,300s go 1, 0, 0, 4, 2, 4, 2 for an average 1.86.

What are the per-century frequencies for higher-numbered centuries? Well, Eldar's table ends with the 10,000th Eldar Number, 1315838; the frequencies per century for the last seven complete centuries up there (i.e. the 1,314,900s thru 1,315,500s) go 1, 0, 0, 1, 0, 1, 3; average 0.86.

I'm sure someone — likely Dr Eldar — has figured out the law governing those averages as the Prime Number Theorem governs the frequency of primes in the neighborhood of N, but so far I haven't located the right paper. If nobody has figured out the law, here's a chance for fame and fortune … or what counts for fame and fortune in the world of math.

For the brainteaser one's guess on first eyeballing it is that 1000^1001 is bigger just because exponents are more powerful — "exponent" … ""power" … see? — than the number being exponentiated. That intuition turns out to be correct.

You just have to work out the number of digits in the expansion of each. For the first one that's easy: 1000^1001 is (10^3)^1001, which is 10^3003, a number of 3,004 digits (just as 10^2 is a 3-digit number).

To get the number of digits in the expansion of 1001^1000, rewrite it as ((1 + 1/1000) * 1000)^1000. That's equivalent to (1 + 1/1000)^1000 * 10^3000.

Now you need to remember that the limit of (1 + 1/n)^n as n tends to infinity is Euler's number e, whose value is 2.718 and change. (You can see the first million digits of e here.) One thousand isn't infinity but it's big enough to ensure that the first two digits of (1 + 1/1000)^1000, at least, agree with e's. This is the case for all n > 73. In fact n = 1000 gets you the first three digits of e, if the Windows 10 calculator can be believed.

So 1001^1000 is 2.7-something times 10^3000. That's a number of 3001 digits beginning 2,7 then continuing for another 2,999 digits.

Three thousand and four digits beats three thousand and one, so 1000^1001 is indeed bigger than 1001^1000 — about 368 times as big.