## Outline Mathematics

Arithmetic Word Problems

# A "Math trick" with two dice

In introductory algebra or advanced arithmetic classes pupils often face the following problem:

Take 2 dice. Dice are made up so that opposite faces add up to 7. Make sure that 6 is opposite 1, the 5 is opposite 2, and the 4 is opposite 3. Roll the dice and then

- multiply the top two numbers
- multiply the bottom two numbers
- multiply the top of one die by the bottom of the other
- and now multiply the other top and bottom

Now add up all of your (four) answers, and it always adds up to 49. Why?

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### Solution

Take 2 dice. Dice are made up so that opposite faces add up to 7. Make sure that 6 is opposite 1, the 5 is opposite 2, and the 4 is opposite 3. Roll the dice and then

- multiply the top two numbers
- multiply the bottom two numbers
- multiply the top of one die by the bottom of the other
- and now multiply the other top and bottom

Now add up all of your (four) answers, and it always adds up to 49. Why?

(In the text below, some words are omitted. These have been underlined. Click just above the line. See what happens.)

First note that each die, a cube,cube,square,triangle,circle,pyramid, has six,four,five,six,seven,eight faces which are paired - a face and its opposite - into three,three,four,six pairs. Let's call a pair of faces *relevant* to a roll of a die, if this happens to be the top/bottom pair of faces after the roll. Thus, for a roll, one of the three possible pairs is relevant.

When two dice are thrown, there are two relevant pairs: one for each die. An important question is, How many different combinations of two relevant pairs are there? There are three,three,four,six possibilities for one die and also three,three,four,six possibilities for the other. In all, there are

Let's take an example. Assume one die rolled 6 on the top, the other 2, which means that the first die's relevant pair is 6/1, that of the second 2/5, 1,1,2,3 and 5,4,5,6 being the numbers on the bottom,side,adjacent,bottom faces. Following up on the instructions leads to the following sum:

6×2 + 1×5,3,4,5 + 6×5,3,4,5 + 1×2,1,2,3

which is the sum 12 + 5 + 30 + 2 = 49. Which we should have expected, right? Now, if the first die showed 1 at the top, the relevant pair would be the same and the calculations would be just a little different:

1×2 + 6×5 + 1×5 + 6×2

with the sum 2 + 30 + 5 + 12 = 49. A simple rearrangement of the sum before: the four,four,five,six products are the same, their order in the sum is slightly different.

What is the point of all this pair counting? Well, be patient, you'll see the light shortly. Denote the pairs (of faces) A, B, C. The possible combinations are:

A-A, A-B, A-C, B-A, B-B, B-C, C-A, C-B, C-C.

The pairs A-B and B-A lead to the same calculations (please verify this), as are the pairs A-C and C-A,A-B,C-B,C-A and B-C and C-B,C-A,A-B,C-B. The conclusion is that, in order to solve the problem, it is only necessary to carry out the calculations for six pairs:

A-A, A-B, A-C, B-B, B-C, C-C.

Not a big task actually and a good exercise, too. To sum up: in order to solve the problem, all one has to do is verify that in six,two,four,six,eight cases indicated above the result is always 49.

This is one solution. A more advanced (one can say *algebraic*) solution goes like that. Assume T and B are the top and the bottom numbers on one die, and their lower case counterparts t and b,t and b,t and 2,t and 3,t and t are the top and the bottom numbers on the second die. The instructions lead to the following sum:

T×t + B×b + T×b + t×B.

This sum can be modified using the commutative law (both for addition and multiplication):

T×t + T×b + B×b + t×B,B×b + t×B,T×b + t×B,B×b + t×T =

T×t + T×b + t×B + B×b =

T×t + T×b + B×t + B×b,B×b + t×T,B×t + B×b,B×t + T×b

Now, the distributive law enters the fray twice:

T×t + T×b + B×t + B×b =

T×(t + b) + B×(t + b) =

(T + B)×(t + b),(T + b)×(t + B),(t + B)×(t + b),(T + B)×(t + b)

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