The concept of n-triplewatch

By on 21/11/2016

With this post we start the collaboration with the Italian blog “Pitagora e dintorni” by Flavio Ubaldini a.k.a. Dioniso Dionisi, by republishing the English version of some of the post appeared in the blog. Here we find some thoughts about strange watches. The original post, in Italian, can be found here, and here there are some more remarks.

by Flavio Ubaldini



Some weeks ago, on social media, I saw several times the image below which shows a watch where all the hours are displayed by only using the digit 9 exactly 3 times.

Wait a minute! Didn’t we say that the digit 9 should be repeated exactly 3 times? What about one o’ clock? And what about five o’ clock? Is that correct? The questions above were raised during a social media discussion with
Gianni Amati. That’s how we started to play this game.

The rule is that we want to express all the integers between 1 and 12 by only using the same digit n correctly combined by any mathematical symbol exactly 3 times. The objective being to produce what we called an n-triplewatch

After some discussion, we managed to create some general rules. Here we adopted the notation $n/10=.n$,

1 = $\sqrt{n}\times\sqrt{n}/n$
2 = $(n+n)/ n$
3 =$\sqrt{(n-.n)/.n}$
4 = ?
5 = $n/(.n+.n)$
6 =$(\sqrt{(n-.n)/.n})!$
7 = ?
8 = ?

9 = $(n-.n)/.n$
10 =$\sqrt{n}\times\sqrt{n}/.n$
12 = ?

Moreover, with these rules and with some additional work, we created the 4-triplewatch.


Later, Gianni Amati also answered my question
: is it possible to create an n-triplewatch in the case in which n is not a perfect square? Yes, it is possible! Gianni found the missing formulas and we created the 3-triplewatch.


 Eventually, we also tried to produce the 2-triplewatch and came to the conclusion that the 7 is missing. At least, if we consider some restrictions. As a matter of fact, I found a formula for the 7 but only with the usage of the ceiling function and the cotangent: [cot 2]/(2×2)


However, if you use the ceiling function, everything becomes easier. It’s a bit like cheating. Our conjecture is that it is impossible to create a 2-triplewatch because it is impossible to find a 2-triple formula for the 7. Can anyone find a 2-triple formula for the 7? If no one can find it, as the cases are finite, one might want to try the proof of the impossibility to find a 2-tripleformula for the 7.
What about the other triplewatches? 

If you accept the challenge to produce other n-triplewatches or other rules, please let me know.

About Roberto Natalini


  1. Isak

    21/11/2016 at 23:37

    I’ve uploaded one here, it’s a 1-triplewatch using standard operators in c++

  2. Flavio Ubaldini

    23/11/2016 at 11:27

    Hi Isak. Thanks for your C++ version of the 1-triplewatch. I like it!

  3. Flavio Ubaldini

    06/12/2016 at 16:13

    A student at Politecnico di Torino sent me a new 2-triple formula for the 7

    Here you’ll be able to find a new version of the 2-triplewatch with that new formula.

  4. Shaun Stevens

    16/12/2016 at 12:47

    If you are allowing . then why not allow a \dot as a recurring decimal symbol, so you could have

    7 = 2/.\dot{2} – 2

  5. Flavio Ubaldini

    21/12/2016 at 22:31

    Thanks Shaun Stevens. That’s an even simpler formula. If you don’t mind tomorrow I’ll publish a new post with the new version of the 2-triplewatch with that new formula.

  6. Flavio Ubaldini

    23/12/2016 at 17:34

    Here is a new version of the 2-triplewatch with the new formula that Professor Stevens suggested.

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