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Balance scale - how many weights do you need?
By Anonymous on Monday, April 2, 2001 - 06:02 am:

This is a challenge, Hope you can provide some help in possible explanations and solutions.

1)Given a balance scale and an unlimited supply of weights of amounts m and n, what is the smallest amount that you can weigh when you are allowed to put weights on both sides of the scale?

2)Suppose you are allowed to put weights on one side of a balance scale and the object to be weighed on the other side. How many weights do you need to to weigh objects of all integer weights from 1 to 30?

3)Suppose you are allowed to put weights on both sides of a balance scale and the object to be weighed on either side. How many weights do you need to weigh objects of all integer weights from 1 to 80?

Thanks in advance for all of your help and explanations!

By Brad Rodgers (P1930) on Tuesday, April 3, 2001 - 01:28 am:

I'll forward a guess for 2: 5. Weights 1,2,4,8, and 16. With these weights, the scenario is exactly analogous to the binary number system. In case your not familiar with the binary number system, you can see that this would cover all the numbers by seeing that if we were to add a weight of 1 to one side, we could sort of 'carry over' weights similar to the way we carry over numbers in the base ten system to get all numbers. In case that didn't make sense, you can see that this will work by just trying to get 1 through 30 from those numbers. This is just an upper bound though - I can't prove that a series of 4 numbers won't work.

For 1, do you want m's weight in terms of n? otherwise I don't see how we can solve it. If it's the case that you want m in terms of n or vice versa, I think it would be the least common multiple of m and n. You can see this by just seeing that you'll need to have a number of n's equal to a number of m's.

I haven't been able to find an answer for 3 yet, and once again, these are all just lower bounds, I haven't been able to prove a thing yet.

Hope this helps,

Brad

By Brad Rodgers (P1930) on Tuesday, April 3, 2001 - 02:05 am:

Actually, I may be able to prove that that is the minimum solution for 2. The proof is fairly simple. First, note that 1 must be in the solution. Next, see that as we cannot have two 1's, we need a 2, then, as we can't have two 2's or two 1's, we need a 4. And so on. I've assumed the (I hope) trivial step of saying that doubles will do no good. This is obviously true as if we had a double of a number, we might as well replace the double with 2×(the number being doubled), and it will lend itself to more additions being found. It will lend itself to more additions being found as anytime we use it, we would've had to use two numbers before.

Hopefully that was at least grammatically coherent...

Brad

By Brad Rodgers (P1930) on Tuesday, April 3, 2001 - 06:16 pm:

I've yet to go through and prove my answer for number 3, but I think that the anwer will be once again five. I'm not sure if this is schoolwork (if it is, I've already done too much), so I'll just give you a hint (a rather large one): think about the trinary number system.

Brad

By Brad Rodgers (P1930) on Tuesday, April 3, 2001 - 06:22 pm:

And, of course if this isn't schoolwork, I'd be glad to guide you through the entire problem.

Brad

By Rachelle Najman (T4009) on Wednesday, April 4, 2001 - 06:13 am:

Hi Brad,
Please guide me through this problem, this isn't schoolwork. Thanks so much!!

By Brad Rodgers (P1930) on Wednesday, April 4, 2001 - 06:23 pm:

Okay, the weights we need are 1,3,9,27, and 81. The way to show that this works is once again fairly simple. First, note that 1,3,and 9 produce all integers up to 13. Now, it follows that when we have 3,9, and 27, we can produce all the integers from 3 to 39, but in multiples of 3. Now, we still have the 1, so that means for each answer we can get we can add or subtract the 1. This shows that we can find sums for all integers from 3-39. A similar argument holds for the trio 9, 27, and 81. This shows that all integers from 1-80 can be formed in some manner. But, once again, I can't prove that there isn't a combination of 4 that will work just yet.

Brad