Millicent likes to pay for anything under a couple of dollars with exact change from her coin purse, where she keeps an ample supply of
pennies, nickels, dimes, and quarters. She always uses the smallest number of coins necessary.
One day at a farmer's market she saw a tray of nectarines for sale, at quite a good price for each one, and decided to get one. She picked out
six coins to pay for it.
But before she could do so she saw another that looked so perfectly ripe, she decided to take it home, too. She returned her six coins back in the purse and
picked out enough to pay for two nectarines, which again was exactly six coins.
Then before she paid for the two nectarines, she saw a third one she liked and decided to get all three. Once again, she picked out exactly six coins,
and this time gave them to the vendor.
How much did each nectarine cost?
Solution:
The answer is that each nectarine costs 24 cents.
Construct a table of all combinations of six coins that uses the smallest number of coins necessary.
For example, you would not use 0 Quarters, 3 Dimes, 0 Nickels, and 3 Pennies because you could have used 1 Nickel and 1 Quarter in place of the 3 Dimes (using 5 coins instead of 6).
The table shows the cost of a nectarine if she bought 1 nectarine, 2 nectarines, or 3 nectarines for each of the coin combinations. You need to find one price that occurs in each of the last three columns.
| Q | D | N | P | = | 1 | 2 | 3 |
|---|---|---|---|---|---|---|---|
| 0 | 1 | 1 | 4 | = | .19 | - | - |
| 0 | 2 | 0 | 4 | = | .24 | .12 | .08 |
| 1 | 0 | 1 | 4 | = | .34 | .17 | - |
| 1 | 1 | 0 | 4 | = | .39 | - | .13 |
| 1 | 1 | 1 | 3 | = | .43 | - | - |
| 1 | 2 | 0 | 3 | = | .48 | .24 | .16 |
| 2 | 0 | 0 | 4 | = | .54 | .27 | .18 |
| 2 | 0 | 1 | 3 | = | .58 | .29 | - |
| 2 | 1 | 0 | 3 | = | .63 | - | .21 |
| 2 | 1 | 1 | 2 | = | .67 | - | - |
| 2 | 2 | 0 | 2 | = | .72 | .36 | .24 |
| 3 | 0 | 0 | 3 | = | .78 | .39 | .26 |
| 3 | 0 | 1 | 2 | = | .82 | .41 | - |
| 3 | 1 | 0 | 2 | = | .87 | - | .29 |
| 3 | 1 | 1 | 1 | = | .91 | - | - |
| 3 | 2 | 0 | 1 | = | .96 | .48 | .32 |
| 4 | 0 | 0 | 2 | = | 1.02 | .51 | .34 |
| 4 | 0 | 1 | 1 | = | 1.07 | - | - |
| 4 | 1 | 0 | 1 | = | 1.11 | - | .37 |
| 4 | 1 | 1 | 0 | = | 1.15 | - | - |
| 4 | 2 | 0 | 0 | = | 1.20 | .60 | .40 |
| 5 | 0 | 0 | 1 | = | 1.26 | .63 | .42 |
| 5 | 0 | 1 | 0 | = | 1.30 | .65 | - |
| 5 | 1 | 0 | 0 | = | 1.35 | - | .45 |
| 6 | 0 | 0 | 0 | = | 1.50 | .75 | .50 |
As you can see from the table, only one number occurs in each of the last three columns: 24 cents.
At first, Millicent had 2 dimes and 4 pennies (24 cents) to pay for 1 nectarine.
Then she had 1 quarter, 2 dimes, and 3 pennies (48 cents) to pay for 2 nectarines.
Finally, she had 2 quarters, 2 dimes, and 2 pennies (72 cents) to pay for 3 nectarines.
Seth Cohen noticed that this continues to work beyond the scope of the puzzle:
1 nectarine = 24 cents = 2 dimes, 4 pennies
2 nectarines = 48 cents = 1 quarter, 2 dimes, 3 pennies
3 nectarines = 72 cents = 2 quarters, 2 dimes, 2 pennies
4 nectarines = 96 cents = 3 quarters, 2 dimes, 1 penny
5 nectarines = 120 cents = 4 quarters, 2 dimes
The reason this works is that each time, we add 25 cents minus 1 cent: we replace one penny with one quarter.
Correctly solved by:
| 1. K. Sengupta | Calcutta, India |
| 2. Kamal Lohia |
Holy Angel School, Hisar, Haryana, India |
| 3. Ivy Joseph | Pune, Maharashtra, India |
| 4. Seth Cohen | Concord, New Hampshire, USA |
| 5. Colin (Yowie) Bowey | Beechworth, Victoria, Australia |
| 6. Kelly Stubblefield | Mobile, Alabama, USA |