Ramakant
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Posts posted by Ramakant
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7 hours ago, harey said:
And what about this:
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 1I 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.
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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:
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 20But 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!!
Rearrange this equation Using only move of one stick.L+VIII=X
in New Logic/Math Puzzles
Posted
Take the horizontal stick of L to the right of equal sign, revised equation becomes 1+8=9