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?
12 thoughts on “The Anti-Calculator Puzzle”
The largest answer is 00, 000, 0000, etc, when each divided by their sum it is infinite. (If you consider positive zero.)
I can accept zero as positive but I cannot accept it as a two digit whole number. Furthermore, division by zero is not infinite, it is undefined.
This IS a fun one. I did a similar thing with a Mensa puzzle I saw in a book as kid. There exists a non-trivial four-digit number that when multiplied by nine gives the same four-digit number in reverse (i.e. ABCD x 9 = DCBA). Find the number. I took this and proved that such a number exists when multiplying by n-1 (in lieu of 9) base-n for all n>2 (i.e. ABCD x (n-1) = DCBA).
I still have the proof lying around somewhere. I’ll have to digitize it and write a post on my blog. I’ll link it back here when I do.
How do you feel about leaving answers in the comments? I think I finally worked this one out, in general, but it was kind of trickier than I feel it should have been. I wouldn’t mind showing what I’ve done, getting ideas, and learning better techniques, but it’s your site and I don’t feel right just dropping it here.
Go for it!
I did a full solution for maximum and minimum of this little function for all bases k and all number of digits greater than 1.
You can find it at my blog
That’s http://themathdragons.wordpress.com/2019/11/30/a-little-discrete-math/ .
This isn’t my “natural” type of puzzle, but that last paragraph challenged me!
I ended up somewhere that I didn’t expect, so I wrote it up at https://loopspace.mathforge.org/CountingOnMyFingers/DigitSum/