- Snapshots of modern mathematics
- Diderot Mathematical Forum 2013: “Mathematics of Planet Earth”
- Pierre de Fermat and Andrew Wiles in Czech Republic stamps
- Stefan Banach (March 30, 1892 – August 8, 1945)
- Diderot Mathematical Forums
- Guessing the Numbers
- What is mathematics for Ehrhard Behrends
- What is mathematics for Krzysztof Ciesielski
- The Three Ducks Trick
- What is mathematics for Franka Brueckler

# Further developments of the 2-triplewatch

We continue the collaboration with the Italian blog “Pitagora e dintorni” by **Flavio Ubaldini a.k.a. Dioniso Dionisi. **Here we find some new thoughts about strange watches.

**by Flavio Ubaldini**

In the conclusions of previous article, The concept of n-triplewatch, I wrote:

*“our conjecture is that it is impossible to create a 2-triplewatch because it is impossible to find a 2-tripleformula for the 7. Can anyone 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.”*

Well, our conjecture was clearly wrong. First of all, because an American colleague had already created the 2-tripleformula below for 7.

Which had led to the 2-triplewatch below.

Additionally, Lorenzo Folcarelli, a young student at Politecnico di Torino, sent me the new 2-triple formula below for the 7.

Which led to the new 2-triplewatch below.

Finally, professor Shaun Stevens suggested a further 2-triple formula for the 7.

With which I generated the further 2-triplewatch below.

I just wanted to remind you that 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.

We produced some general rules.

1 = √n×√n/n

2 = (n+n)/ n

3 = √((n-.n)/.n)

4 = ?

5 = n/(.n+.n)

6 = (√((n-.n)/.n))!

7 = ?

8 = ?

9 = (n-.n)/.n

10 = √n×√n/.n

11=nn/n

12 = ?

Is anyone able to produce new triplewatches that are not already listed here or, even better, new general rules?

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Uli Seidel

16/03/2017 at 19:24

I suggest the following general rules which make use of the floor function of π:

4 = √n×√n/n + [π]

7 = √n×√n/.n – [π]

8 = nn/n – [π]

12 = (n-.n)/.n + [π]

Maybe that there are other solutions with more elegance