Coins
What are
the best denominations for coins to have? In the United States we have 1, 5, 10, and
25 cents. We also have 50 cents and $1 coins, but they are not in a general
use. In Europe the new coinage consists of 1, 2, 5, 20 and 50 cents and also 1 and 2
euros. In the past and in different
countries we have had a huge variety of denominations. Remember the Spanish
pieces-of-eight, and the English sixpence.
It has been estimated that about
$600 worth of coins pass through the hands of a typical American every year.
More than $10 billion in coins is sitting in drawers, cans, and bottles in
American homes, and $13.5 billion in coins is in circulation. (Discover
Magazine, October 2003.)
The question of the "best"
choice of denomination for coins has only a small mathematical component. It
depends mainly on:
What we are accustomed to.
The structure of prices. (Many prices end with, $ ...
.99, but this is changed when tax is added.)
Shopping habits.
(Do people make small purchases every day, or do they shop once a week?)
How change is computed. (By a person or automatically.)
Here we ask only a "mini
question" that deals with some mathematical aspects of making change.
Question.
What should be the denominations for
three coins, such that any transaction up to 10¢ would require at most two
coins? This means that either you pay
the exact amount using one or two coins, or you pay
with just one coin and get one coin as change.
Examples.
| Amount: |
Payment or payment minus change: |
| 1 |
1 |
| 2 |
2 |
| 3 |
2 + 1 or 5 - 2 |
| 4 |
2 + 2 or 5 - 1 |
| 5 |
5 |
| 6 |
5 + 1 |
| 7 |
5 + 2 |
| 8 |
impossible |
| 9 |
impossible |
| 10 |
5 + 5 |
1¢, 5¢ and 10¢ are also not good.
| Amount: |
Payment or payment minus change: |
| 1 |
1 |
| 2 |
1 + 1 |
| 3 |
impossible |
| 4 |
5 - 1 |
| 5 |
5 |
| 6 |
5 + 1 |
| 7 |
impossible |
| 8 |
impossible |
| 9 |
10 - 1 |
| 10 |
10 |
After you find a solution, think
about designing the coins (their diameters and thicknesses). Assume that 1 cent
has the volume of a US cent, and that for the other denominations,
the volume is proportional to their values.
Solutions.
There
are three solutions to this problem: 1, 5, 8; 2, 4, 5, and 3, 4, 5.
| Amount: |
(1, 5, 8) |
(2, 4, 5) |
(3, 4, 5) |
| 1 |
1 |
5 - 4 |
5 - 4 or 4 - 3 |
| 2 |
1 + 1 |
2 or 4 - 2 |
5 - 3 |
| 3 |
8 - 5 |
5 - 3 |
3 |
| 4 |
5 - 1 |
4 or 2 + 2 |
4 |
| 5 |
5 |
5 |
5 |
| 6 |
5 + 1 |
4 + 2 |
3 + 3 |
| 7 |
8 - 1 |
5 + 2 |
4 + 3 |
| 8 |
8 |
4 + 4 |
4 + 4 or 5 + 3 |
| 9 |
8 + 1 |
5 + 4 |
5 + 4 |
| 10 |
5 + 5 |
5 + 5 |
5 + 5 |