Wikipedia:Reference desk/Archives/Mathematics/2024 February 1

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February 1

Fair 20 sided die by corner...

For a 20 sided die, the average of the face values is 10.5, and let the vertices have the value that is the sum of the faces around that corner, Is it possible to setup a 20 sided die so that half of them sum to 52 and half sum to 53? (this could be equivalently stated by giving values to the 20 verticies of a Dodecahedron and giving the faces the value that is the sum of the corners.Naraht (talk) 04:21, 1 February 2024 (UTC)

I don't have a script written up for it, but there is a note I'd like to make that might be able to help find if this the case. In an icosahedron (what I presume you mean by the 20-sided polyhedron), because faces are triangular, if two faces are adjacent, switching their values only affects the values of their two non-adjacent vertices. This means that you could probably simulate some kind of value "flow" between these kinds of vertices in an attempt to equalize their values. GalacticShoe (talk) 06:45, 1 February 2024 (UTC)

Note that the polyhedron formed from connecting each vertex to the 5 other vertices which are "two triangles away" (i.e. the vertices you can flow from/to by switching face values) is none other than the great icosahedron. GalacticShoe (talk) 06:59, 1 February 2024 (UTC)

The small stellated dodecahedron has the same edges. —Tamfang (talk) 00:45, 3 February 2024 (UTC)

A plausible constraint one might consider in an attempt to reduce the search space is to require the sum of the vertex values of each of the 6 pairs of diametrically opposite vertices to equal 105. (This is only – potentially – helpful if there is a solution.)  --Lambiam 11:11, 1 February 2024 (UTC)

Even more restrictive: require the sum of the face values of each of the 10 pairs of diametrically opposite faces to equal 21. This implies the restriction on sums of vertex values.  --Lambiam 18:34, 1 February 2024 (UTC)

• Yes, you can. Here's one arrangement, for example. --Amble (talk) 05:01, 2 February 2024 (UTC)

     /\  /\  /\  /\  /\
    / 1\/20\/ 2\/10\/19\
   ---------------------
    \ 5/\11/\15/\12/\ 9/\
     \/16\/ 4\/13\/ 3\/18\
      ---------------------
      \ 8/\14/\ 7/\17/\ 6/
       \/  \/  \/  \/  \/

• Thank you!Naraht (talk) 03:20, 3 February 2024 (UTC)

I looked at a related(?) question: [1] —Tamfang (talk) 20:26, 5 February 2024 (UTC)

Powers of two

Is the thing that all powers of two will eventually contain long sequence of zeros at the end? Some calculator softwares indeed make this. --40bus (talk) 16:27, 1 February 2024 (UTC)

No. Numbers ending in 0 are always multiples of 5. All powers of 2 end in 2, 4, 6, or 8. The cycle is 2-4-8-6-2-4-8-6-2-4-8-6-2-4-8-6... Georgia guy (talk) 17:07, 1 February 2024 (UTC)

@40bus: It sounds like a rounding problem. 2⁶⁴ = 18446744073709551616. This may for example be rounded to 1.8446744×10¹⁹, maybe written 1.8446744e+19 if it's written at all and not hidden internally in the software. This rounded value is equal to 18446744000000000000. PrimeHunter (talk) 17:29, 1 February 2024 (UTC)

It depends though; Python computes all digits for whole numbers, and gives 2¹⁰⁰⁰ as 1071...(about 300 digits)...9376. --RDBury (talk) 17:46, 1 February 2024 (UTC)

In binary, which is the base-2 numeral system, powers of two have a specific pattern.

For example:

2^0=1 (binary: 1)

2^1=2 (binary: 10)

2^2=4 (binary: 100)

2^3=8 (binary: 1000)

2^4=16 (binary: 10000)

So Some calculator software may take advantage of this to provide faster computation for powers of two. Harvici (talk) 03:05, 2 February 2024 (UTC)