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riddle puzzle

Puzzle with banknotes

Question

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.

Edited by riddle puzzle

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I think I can do better:

Spoiler

01:   100                                   pay 10 with 100
02:   50 20 20                              pay  5 with  50
03:   20 20 20 20  5                        pay  3 with   5
04:   20 20 20 20  1  1                     pay  5 with  20
05:   20 20 20 10  5  1  1                  pay  3 with   5
06:   20 20 20 10  1  1  1  1               pay  8 with  10
07:   20 20 20  1  1  1  1  1  1            pay  9 with  20
08:   20 20 10  1  1  1  1  1  1  1         pay  8 with  10
09:   20 20  1  1  1  1  1  1  1  1  1      pay 18 with  20
10:   20  1  1  1  1  1  1  1  1  1  1  1   pay 18 with  20

But not sure it is the max.

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I think I have a much better answer.

01: 100                                                                                                                                                    pay 3 with 100

02: 50 20 20 10 5 2                                                                                                                                pay 3 with 5

03: 50 20 20 10 2 1 1                                                                                                                  

On 5/17/2020 at 9:42 PM, harey said:

I think I can do better:

  Hide contents

01:   100                                   pay 10 with 100
02:   50 20 20                              pay  5 with  50
03:   20 20 20 20  5                        pay  3 with   5
04:   20 20 20 20  1  1                     pay  5 with  20
05:   20 20 20 10  5  1  1                  pay  3 with   5
06:   20 20 20 10  1  1  1  1               pay  8 with  10
07:   20 20 20  1  1  1  1  1  1            pay  9 with  20
08:   20 20 10  1  1  1  1  1  1  1         pay  8 with  10
09:   20 20  1  1  1  1  1  1  1  1  1      pay 18 with  20
10:   20  1  1  1  1  1  1  1  1  1  1  1   pay 18 with  20

But not sure it is the max.

@harey  Rule 3 in the puzzle stipulated

 "..........always gives  the nearest bank note (of which he had at the moment)....". 

You have broken this rule at shop 02, by tendering a note of denomination 50. You MUST have tendered note of denomination 20 intead!! 

Edited by Ramakant
typo

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And what about this:

Spoiler

1: 100                                 pay 10 with 100
2:  50 20 20                           pay  9 with  20
3:  50 20 10  1                        pay  8 with  10
4:  50 20  1  1  1                     pay  9 with  20
5:  50 10  1  1  1  1                  pay  8 with  10
6:  50  1  1  1  1  1  1               pay 10 with  50
7:  20 20  1  1  1  1  1  1            pay  9 with  20
8:  20 10  1  1  1  1  1  1  1         pay  8 with  10
9:  20  1  1  1  1  1  1  1  1  1      pay 18 with  20
     1  1  1  1  1  1  1  1  1  1  1

 

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7 hours ago, harey said:

And what about this:

  Hide contents

1: 100                                 pay 10 with 100
2:  50 20 20                           pay  9 with  20
3:  50 20 10  1                        pay  8 with  10
4:  50 20  1  1  1                     pay  9 with  20
5:  50 10  1  1  1  1                  pay  8 with  10
6:  50  1  1  1  1  1  1               pay 10 with  50
7:  20 20  1  1  1  1  1  1            pay  9 with  20
8:  20 10  1  1  1  1  1  1  1         pay  8 with  10
9:  20  1  1  1  1  1  1  1  1  1      pay 18 with  20
     1  1  1  1  1  1  1  1  1  1  1

 

I think I can improve your solution to 10 toys instead of 9, You are welcome to improve better!

But your solution as well as mine both are subject to a big IF - the acceptance of  my note below.

 

1:  100                                                                                cash 100              pay 5 with 100      get 50 20 20 5

2:  50 20 20 5                                                                    cash 95                pay 3 with  5          get 1 1

3:  50 20 20 1 1                                                                 cash 92                pay 5 with 20         get 10 5

4:  50 20 10 5  1 1                                                             cash 87                pay 3 with 5           get 1 1

5:  50 20 10 1 1 1 1                                                           cash 84                pay 8 with 10         get 1 1

6:  50 20 1 1 1 1 1 1                                                          cash 76                pay 9 with 20         get 10 1

7:  50 10 1 1 1 1 1 1 1                                                       cash 67                pay 8 with 10         get  1 1

8:  50  1 1 1 1 1 1 1 1 1                                                     cash 59                pay 10 with 50       get  20 20

9:  20 20 1 1 1 1 1 1 1 1 1                                                 cash 49                pay 18 with 20       get 1 1

10: 20 1 1 1 1 1 1 1 1 1 1 1                                               cash 31                pay 18 with 20        get 1 1

Balance all small change .........                                       cash 13                END

With the best utilization of 87%!! 

 

Notes:

Problem-setter's solution (of 8 toys) seems to interpret the rule 4 as "...............return the change with as less banknotes as possible,............ at the end it turns out that each seller always return at least two banknotes to George". (BOTH OF THESE HIGHLIGHTED CONDITIONS MUST BE COMPLIED IN TOGETHER AND HENCE COMPLIED ACCORDINGLY).

Whereas in your as well as mine (of 9 toys & 10 toys respectively), the interpretation of rule 4 is as " .......... always return  at least two bank notes. Therefore, if at any stage return of 1 banknote alone becomes possible, ignore such option and simply stick to return 2 banknotes instead....." 

I have interpreted so (with more liberty than the problem-setter), especially because the word "always" has NOT been used by the problem-setter in the first part of the highlighted conditions . Word "always" has been used by him in second part of the highlighted condition only.

Edited by Ramakant
clarification of the logic

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