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?

Related Posts

• 

  A game of free will
  A game of free will
• 

  Fighting the coronavirus with mathematics
  Fighting the coronavirus with mathematics
• 

  International Day of Mathematics on March 14th
  International Day of Mathematics on March 14th
• 

  Lunar Arithmetic
  Lunar Arithmetic

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

This site uses Akismet to reduce spam. Learn how your comment data is processed.