Hannah G.

asked • 05/16/18

Bagging dimes

3 people have 139 dimes. They want to bag the coins in such a way that they can pay any amount between 1 and 139 dimes by handing over a combination of bags without looking inside them. What is the minimum number of bags they need?

Paul M.

tutor
I think the answer is 6 bags each with 2n coins, but I'm not sure.  If you don't get a more certain answer, go with this one!
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05/16/18

Karen V.

I'm not certain about this, but I think it would be 8 bags in total (ignoring that there are 3 people, which I think is irrelevant). The first bag would have 1 (2^0) dime in it, the second would have 2 (2^1) dimes, and so on, with the eighth bag having 128 (2^7) dimes in it.
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05/16/18

Mark M.

How could they hand over 3 dimes?
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05/16/18

Hannah G.

Hi thanks for the help but I think I figured it out and ended with 11 bags. Because I don't think you can go over 139 so including 128 would go end with a combination greater than 139 I ended with bags full of 1, 1, 2, 2, 4, 4, 5, 8, 16, 32, 64 thoughts? And yes I do not think the people have to do with the problem
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05/16/18

1 Expert Answer

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Alexander B. answered • 05/18/18

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5 (7)

PhD in Engineering with 20 yrs of Math and Science Teaching Experience

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It seems like 10 bags will satisfy the condition with coins allocation: 1,2,3,6,9,18,19,27,54. The logic behind is applying base 3 numeric system 1,2,3,6,9,18,27,54  (will cover all combinations from 1 to 120) and adding the last bag with 19 coins to extend the range to 139.
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