# Thread: Help with problem solving

1. ## Help with problem solving

So, I'm trying to improve my problem solving skills.

One problem was that you have 10 bags and 1,000 dollars. Arrange the 10 bags so that if I ask for any amount of dollars, it can be given without opening any of the bags.

I struggled for a while and peaked at the answer.

1,2,4,8,16,32,64,128,256,489

How would you go about solving it? I just want to see the thought process in some hope I can improve my own.

Thanks for any help.

2. ## Re: Help with problem solving

Originally Posted by mark090480
So, I'm trying to improve my problem solving skills.

One problem was that you have 10 bags and 1,000 dollars. Arrange the 10 bags so that if I ask for any amount of dollars, it can be given without opening any of the bags.

I struggled for a while and peaked at the answer.

1,2,4,8,16,32,64,128,256,489

How would you go about solving it? I just want to see the thought process in some hope I can improve my own.

Thanks for any help.
are you aware of the binary number system?
any number can be expressed in the binary number system.
suppose some guy asked for a total amount of $10. then first express 10 in binary. it is $(1001)_2$. So $10=2^3+2=8+2$. if someone had asked for any other price again the same procedure follows. 3. ## Re: Help with problem solving "without opening any of the bags"??? I thought for a moment you mignt have meant "without emptying any of the bags" but that can't be it either. I suspect that the problem was to distribute the money so that you can make any amount of money by emptying bags- that is, any bag you take money from, you must take all the money, not just some of the money in any one bag. I would probably do that by a step by step method- obviously one bag will have to contain exactly$1 so that I can take out $1. And I have to be able to take out$2 so I will have to have a second bag containing $2. Of course, with those two bags alone I can take out$3 so I do not need a bag with $3 in it- but I do need one with$4. With "1", "2", and "4", I can make 5= 4+ 1, 6= 4+ 2 and 7= 4+ 3+ 1, but not $8 so I need a bag with$8 in it. With those I can make 9= 8+ 1, 10= 8+ 2, 11= 8+ 2+ 1, 12= 8+ 4, 13= 8+ 4+ 1, 15= 8+ 4+ 2+ 1, but not 16, so I add a bag with $16. In a sense, I am writing those numbers in "base 2": every number can be written use "bits" of 0 or 1 (empty a given bag or not) so each of my bags can be represented by a place in the binary system- powers of 2. I can do this until $2^8= 256$ at which point I have put tex]1+ 2+ 4+ 8+ 16+ 32+ 64+ 128+ 256= 2^9- 1= 512- 1=$511 into bags. At that point, I should but 512 into the tenth bag but I only have 1000- 511= \$489 left to put there.

(Clearly, I don't type fast enough!)

4. ## Re: Help with problem solving

OK tnx.

HallsofIvy: That is pretty much how I started but then doubted my self as I thought I could not make any number doing that and tried some pretty random stuff. Yes I knew of the binary system but couldn't see how to apply it without knowing the answer.

abhishekkgp:

That for the 10 does not work, 8+2, what if I want 6?

5. ## Re: Help with problem solving

Originally Posted by mark090480
OK tnx.

HallsofIvy: That is pretty much how I started but then doubted my self as I thought I could not make any number doing that and tried some pretty random stuff. Yes I knew of the binary system but couldn't see how to apply it without knowing the answer.

abhishekkgp:

That for the 10 does not work, 8+2, what if I want 6?
$6=4+2=(110)_2$. why this does not work for 10?