## Brainteaser 1795: Dicey columns

**From The Sunday Times, 9th February 1997** [link]

A boy amused himself by building three separate columns of some equal-sized dice. In each column each die was placed squarely on the one below it. He then looked at the tops of the columns, read each as a digit, and then read the digits together to form one long number.

For example, had the tops been:

then he would have read 113.

He was surprised to find that the number which he read equalled the total number of visible spots all around and on top of his three columns of dice. He also multiplied together the three numbers from the tops of the columns and found that the answer equalled the total number of dice which he had used.

What was that total number of visible spots?

For a more challenging Teaser try to find a corresponding situation with more than three columns of dice. There is only one other possibility.

The text of this puzzle is taken from the book *Brainteasers* (2002, edited by Victor Bryant), so may differ from the puzzle originally published in the newspaper.

In fact, the puzzle published in the paper was the *more challenging* variation.

[teaser1795]

## Jim Randell 7:01 am

on14 March 2019 Permalink |Suppose the tops of the three columns show

A, B, C.Opposite faces on a die sum to 7, so each die shows 14 spots on the vertical sides, and the top dice also show

A, B, C, so if there arendice the total number of spots visible is:(where

ABCis an alphametic expression).Also the number of dice is given by:

which gives us the following alphametic expression to solve:

where

A, B, Care limited to be digits from 1 to 6.This can easily be solved using the [[

`SubstitutedExpression()`

]] solver from theenigma.pylibrary.Run:[ @repl.it ]Solution:The total number of visible spots was 133.So there are 9 dice in total (3 columns of 3 dice each).

In the book the puzzle then goes on to suggest:

We can try the same thing with four dice, showing

A, B, C, D:So the solution to this extra puzzle is with 180 dice (4 columns of 45 dice each), showing 2536 visible spots.

There are no solutions for 5 dice, at least not reading the numbers on top of the dice as a base 10 number. If we read them as a base 8 number then there are 5 solutions with 5 dice (and 1 with 4 dice, and 4 with 3 dice). (You can try it by adding a [[

`--base=8`

]] argument to the commands).LikeLike

## Jim Randell 2:43 pm

on14 March 2019 Permalink |Writing:

we can simplify this to:

So we only needs to consider values of

AandBto deriveC:The improvement in run time (if any) is negligible.

However, this gives us a manual way to solve the puzzle.

Cis an integer between 1 and 6, and as the denominator of the expression forChas a divisor of 7 then so must the numerator. In fact it must be the case that 7 divides(11A + B). So for any given value ofAthere is only a single value forBthat is possible, from which we can calculate potential values forC:And only one value for

Agives a viable value forC.LikeLike

## GeoffR 7:29 pm

on8 August 2019 Permalink |For three columns of dice:

For four columns of dice:

LikeLike