## My Problem With 7

My reposted problem of a couple days ago, about building all the digits of a clock face using exactly three 9’s and simple arithmetic combinations of them, caught in my mind, as these things will sometimes do. The original page missed out on a couple ways of using exactly three 9’s to make a 1, but it’s easy to do. The first thing to wonder about was how big a number could we make using exactly three 9’s? There must be some limit; it’d be absurd to think that we could make absolutely any positive integer with so primitive a tool set — surely 19,686 is out of the realm of attainability — but where is it?

Not 13, at least: it’s equal to $9 + \sqrt{9} + .\bar{9}$, since the infinitely long repeating decimal $.\bar{9}$ is equal to one. Infinitely long repeating decimals may take a little work to understand, but it’s no worse than using radicals or factorials. How about 14? That seems be obviously 9 + 5, which since we need three 9’s to make a 5 seems out of the range, or 18 – 4, which doesn’t help much. Maybe 14 is the first digit we miss, in this parallel to the Chicken McNuggets problem. 15 is attainable easily; that’s $9 + 9 - \sqrt{9}$; 16 less obviously so. 17 is a snap, using repeating digits again. 18 looks like we might have to make do with using just two 9’s, but those square roots come to save us again; replace one nine with the square root of 9 times the square root of 9. It’s an inefficient way of writing it, but if we wanted compact and efficient writing we wouldn’t be doing this problem at all, would we?

But the other question is: do we have to do it with 9’s? Can we use other digits as a base and produce all the digits of a clock face? Some numbers are pure gimmies: whatever digit we use as the base — call it N — we get 1 for free by taking, among other forms, $N^{N - N}$ which is 1 for any number. We also get 2 for free: that’s $\frac{N + N}{N}$. The original number we get also, as $\sqrt[N]{N^N}$. And we get 11, by $\frac{NN}{N}$, where the numerator there is really N in the tens place followed by N in the ones place, which isn’t properly an arithmetic operation but is close enough for our needs.

It seems to me that among the single-digit numbers, 1, 4, and 9 have privileged positions. They’re perfect squares, so we can use the square root of them to produce integers again. If an integer isn’t a perfect square, then its square root is irrational, and it’s a pain to get from an irrational back to a rational number other than zero.

Using 1 as a base seems like a bad start: we can get 1 for free. 2 is, among other forms, $1^1 + 1$. 3 is just 1 + 1 + 1. 4 is … well, I’m stuck there. I’d take suggestions.

How about 4, then, as a base? Once again we get 1 (and for that matter 0) for free. Two we can write as $\sqrt{4}\frac{\sqrt{4}}{\sqrt{4}}$. Three is a little less obviously wasteful: $\sqrt{4} + \frac{\sqrt{4}}{\sqrt{4}}$, and four is another of our freebies. Five is almost a repeat: $4 + \frac{\sqrt{4}}{\sqrt{4}}$. This trick keeps paying off, too, for six: $4 + \frac{4}{\sqrt{4}}$, or we can strike out for a little originality and do $\sqrt{4} + \sqrt{4} + \sqrt{4}$.

It’s seven that stumps me. $4 + \sqrt{4}$ seems like a good start, and we just need to get one more to get there, but I don’t see a way of producing that stray 1. The repeating-decimal 9 is equivalent to 1 and comes in handy that way. A repeating-decimal 4, though, isn’t anything but four-ninths, leaving us an ocean away from the whole numbers we wanted. We have the same problem trying from the other way, trying to take one away from 4 plus 4.

There are some more functions we might throw in while keeping to the spirit of the puzzle: logarithms and trigonometric functions, for example. These probably won’t, though; with a tiny handful of exceptions they’ll just turn integers into irrational numbers, and we don’t want that.

Possibly I’ve just not thought about this long enough, or fiddled with combinations enough. There are a lot of possible combinations, even if we count things like the fusion that gives us 11 as anomalies. But if you see a way to get a 7 out of exactly three 4’s, you’ve reached it before I have.