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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

ganesh
2005-08-23 14:23:29

Yes, seven bags is right, but I would have put the Dollars in the bags this way:-
1, 2, 4, 8, 16, 32 and 8.
Your solution is correct too!

mikau
2005-08-23 09:47:25

I didn't say nothing is hard. Rocks are hard, skateboarding is hard. But math isn't hard. Its just unfamiliar.

Hey, do you think I got the above problem right?

MathsIsFun
2005-08-23 09:30:25

It was great, too.

And I liked your "nothing is hard, it is just unfamiliar" post too.

sad sad sad

mikau
2005-08-23 09:02:35

I'm just ticked cause I lost my introduction post. :-(

MathsIsFun
2005-08-23 08:56:50

I knew you would have to sad

Only because my last backup was just before you registered.

I am in the process of upgrading to the latest version of the forum software to (hopefully) avoid further hacks.

mikau
2005-08-23 08:45:35

weird. I needed to reregister my name.  Thats me. Just so you know.

mikau
2005-08-23 08:40:23

while the forum was hacked, someone asked this question:

hey can anyone help me out?

problem:

Divide $71 in whole $ increments into a number of bags so that I can ask for any amount between $1 and $71, and you can give me the proper amount by giving me a certain number of therse bags without opening them.  What is the minimum number of bags you will require?

please help if you can,
thanks

lets see, well if you have to divide them into a group of equal parts, then you'd have to divde it by 71. Anything more then one each and you would not be able to ask for one.

If you can divide it into groups of unequal parts, each bag containing a certain number of whole dollars, then it gets slightly more interesting. If you put 10 in one bag, and put the remaining 61 into individual bags, then you will be able to make the numbers between 1 and 10, by using the additional ones. Therefore I suppose if you put extra in one bag, you need enough singles to cover the numbers from 1 to that point. Lets say we put 36 in one bag and put the remaining 35 into individual bags. For numbers less then 36 you can use the individual bags, for 36 you can use the big bag, any more and you can add the big bag to any number of small bags to make any number from 37 to 71. Now we reduced 71 to 36 bags, lets see if we can the 35 small bags to less bags. Again lets take the midpoint. If 18 in one bag and the remaining 17 into individual bags, we can make the numbers less then 18 by using the individual small bags, we can make 18 using the big bag, and any number between 18 and 36 by using the big bag and adding the individual small bags. So now we've got one big bag with 36, another big bag with 18, and 17 litte bags. If my reasoning proccess is correct, it looks like we are dividing the number by 2 using integer division, then giving the bigger bag the remainder, and the repeat.

So it should be  36, 18, 9, 4, 2, 1, 1. Hope I worked that out right. Kind of a confusing algorithm. I'm not used to doing integer division. lol. Anyway it looks like the answer is 7 bags.

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