| Manifesto | Bookstore | Contents | Amazon store | Term index | What changed? | Contact | Recommend |
Outline Mathematics
|
|
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
Now add up all of your (four) answers, and it always adds up to 49. Why? |
Copyright © 1996-2009 Alexander Bogomolny
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
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 , has faces which are paired - a face and its opposite - into 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 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 possibilities for one die and also possibilities for the other. In all, there are 3×3 = ways to combine the three pairs from one die with the three pairs from the other.
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, and being the numbers on the faces. Following up on the instructions leads to the following sum:
| 6×2 + 1× + 6× + 1× |
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 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 and B-C and . 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 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 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 +
= T×t + T×b + t×B + B×b = T×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) = |
And finally comes the punch line: recollect that the dice are created so that in any pair of the opposite faces the numbers add up to ! Therefore, T + B = 7 and also t + b = 7. And, as every one knows, 7×7 is !
Copyright © 1996-2009 Alexander Bogomolny
| 34222322 |  |
