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Further developments of the 2-triplewatch

By on 31/12/2016

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

pitagora

 

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.

image04

Which had led to the 2-triplewatch below.

image03

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

image06

Which led to the new 2-triplewatch below.

image05

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

image01

With which I generated the further 2-triplewatch below.

image00

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?

About Roberto Natalini

2 Comments

  1. Pingback: Carnevale della matematica #105 | backup del Post

  2. 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 🙂

    Reply

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