Show that some problems have more than one answer by giving at least two answers to the following problem. What change would a restaurant cashier give someone for $1 if he used no more than four of any coin and no coin smaller than a nickel or bigger than a quarter?
Try to come up with 2 combinations of nickels, dimes, and quarters, such that no more than 4 of each are used.
For example: 4 quarters
There can be no pennies and no dimes, becasue they are smaller than a nickel. Also no Kennedy half dollars. That leaves quarters and nickels to make change with. There could be 0, 1, 2 or 3 nickels and 0, 1, 2 or 3 quarters.
See what change you could make with those combinations. There are a lot more than two possiblities.
To find multiple answers to the problem, we need to consider different combinations of coins that satisfy the given conditions.
Let's start by listing the available coins within the constraints:
- Nickel (5 cents)
- Dime (10 cents)
- Quarter (25 cents)
Now, let's find two different solutions by combining these coins:
Solution 1:
1 Quarter (25 cents) + 3 Nickels (3 * 5 cents) = 40 cents
The cashier can give 1 quarter and 3 nickels as change, totaling 40 cents which is less than $1.
Solution 2:
2 Dimes (2 * 10 cents) + 2 Nickels (2 * 5 cents) = 30 cents
The cashier can give 2 dimes and 2 nickels as change, totaling 30 cents which is also less than $1.
Both of these solutions satisfy the given constraints: using no more than four of any coin, and no coin smaller than a nickel or bigger than a quarter. Therefore, they provide valid answers to the problem, demonstrating that there is more than one possible combination of coins for a given amount of change.