Sniper Joe wrote:

How many different combinations of coins must the machine be programmed to accept?

Whoever wrote this must not know programming; it only needs to know the different coins and their values -_-

Technically, I think the answer is infinate, because there's nothing that says you can't put in more than 50 cents.

If you want it exact... hmmm.

I can't think of a mathmatical way to solve that beside going through all the possibilities. What was the last thing your class learned?

It's easy, if the book doesn't want you to hit 50c exactly. Just write "infinate" for your answer;)

The answer they want is 10:

2 quarters
1 quarter +
5 nickels
3 nickels + 1 dime
1 nickel + 2 dimes
10 nickels
8 nickels + 1 dime
...
5 dimes

The correct answer is rather harder to ascertain. Most vending machines will let you go over the max amount with a single coin, but no further. This machine has to be programmed so that it will accept up to 70¢, but nothing over 50¢. That is, if you add two dimes and then two quarters, it needs to take the last quarter.

That means it must have all the above plus:

2 quarters +
2 dimes
1 dime + 2 nickels
4 nickels
3 nickels
1 dime + 1 nickel
1 dime
2 nickels
1 nickel
1 quarter +
4 dimes + 1 nickel
3 dimes + 3 nickels
2 dimes + 5 nickels
1 dime + 7 nickels
9 nickels
4 dimes
3 dimes + 2 nickels
2 dimes + 4 nickels
1 dime + 6 nickels
8 nickels
3 dimes + 1 nickel
2 dimes + 3 nickels
1 dime + 5 nickels
7 nickels
3 dimes
2 dimes + 2 nickels
1 dime + 4 nickels
6 nickels
5 dimes + 1 nickel
4 dimes + 3 nickels
3 dimes + 5 nickels
2 dimes + 7 nickels
1 dime + 9 nickels

So the most correct answer is 31, but it's obviously not the simple one they want.

Lummox JR

Ah, if only I had paid attention in Discrete Math