Here’s a pleasurable little puzzle via Jo Morgan:
It’s a charming problem. I’m charmed, anyway, and I don’t charm easily. (“Immune to charm,” doctors have called me, with a tearful eye.)
Ironically, some experimental button-mashing is actually a pretty good way to get started here. Even easier, you can just use a spreadsheet to generate all 90 possibilities in a few moments, and then pluck out the best answers. Hard to argue that the pencil-and-paper approach is a time-saver.
(Perhaps that’s no surprise. Jo found the problem in a book that’s about not denigrating calculators, but celebrating them! Just as well; I, for one, have spent too many years being altogether too fusty about calculators.)
Still, if you want to render the calculator moot, we can simply push further. No need to stop at 2 digits. Instead, let’s take a 3-digit, 4-digit, or better yet, an n-digit number and divide it by the sum of its own digits.
What’s the largest result we can obtain, and how?
What’s the smallest result we can obtain, and how?
A good high-school math puzzle! I’ll leave you to it.
P.S. You want to go further? OK: compare the smallest result for n + 1 digits to the smallest result for n digits. What happens to their ratio as n grows?
P.P.S. You may object to this problem: “It’s all about summing digits,” you say, “which means it’s contingent on Base Ten. And since there’s nothing special about Base Ten, this is just silliness. It isn’t real mathematics.” Fine then, purist. How do the answers change if we work in Base k?