Hello,
I have one Math puzzle to which I think I found the solution but wanted to check your opinion on it. The puzzle is the following:
George got a birthday present which was 1 banknote of 100 (the currency does not matter. Lets name it "currency" just for convinience). Valid banknotes in the currency are: 100, 50, 20, 10, 5, 2 and 1.
So he decided to spent some of the money in the shopping mall (he did not take any other banknotes with him. Just the one banknote of 100). At the end of the shopping it turns out that:
1. In each and every shop he bought just one item and therefore just one payment was made.
2. The price of the item was a whole integer (no decimal points are allowed)
3. For each item he never had the exact sum that's why he always gives the nearest banknote (of which he had at the moment).
4. The sellers on the other hand always have enough money in different banknotes that's why they return the change with as less banknotes as possible. However, at the end it turns out that each seller aways return at least two banknotes to George.
The question is:
What is the maximum number of items that George can buy with these restrictions?
I have bolded the important parts in my opinion, so they pop out.
The approach that I took was the following:
1. In each shop the way I tried to figure out the price of the item is to have a change with as less as possible banknotes as change and with as much as higher value of the banknotes in the change
2. And as less as the price can be.
So the solution:
1. First Shop one buying of item that costs - 10
- The payment was made with the available banknote - 100
- The change was - 50 20 20
- Available banknotes for George - 50 20 20 = 90
2. Second Shop one buying of item that costs - 9
- The payment was made with the available banknote - 20
- The change was - 10 1
- Available banknotes for George - 50 20 10 1 = 81
3. Third Shop one buying of item that costs - 7
- The payment was made with the available banknote - 10
- The change was - 2 1
- Available banknotes for George - 50 20 2 1 1 = 74
4. Fourth Shop one buying of item that costs - 9
- The payment was made with the available banknote - 20
- The change was - 10 1
- Available banknotes for George - 50 10 2 1 1 1 = 65
5. Fifth Shop one buying of item that costs - 7
- The payment was made with the available banknote - 10
- The change was - 2 1
- Available banknotes for George - 50 2 2 1 1 1 1 = 58
6. Sixth Shop one buying of item that costs - 9
- The payment was made with the available banknote - 50
- The change was - 20 20 1
- Available banknotes for George - 20 20 2 2 1 1 1 1 1 = 49
7. Seventh Shop one buying of item that costs - 13
- The payment was made with the available banknote - 20
- The change was - 5 2
- Available banknotes for George - 20 5 2 2 2 1 1 1 1 1 = 36
8. Eight Shop one buying of item that costs - 17
- The payment was made with the available banknote - 20
- The change was - 2 1
- Available banknotes for George - 5 2 2 2 2 1 1 1 1 1 1 = 19
I tried different approaches and this gives me the highest number of items - 8. For the last three we have several options but they cannot give more than 8.
Is this the correct answer and if no how can you approach the problem to have more buyings. I found the rule that sellers should return at least two banknotes the most restrictive.