$(0,0)$ and pans at $\left(\pm1,0\right)$ , $\left(\pm2,0\right)$ (as shown in the diagram). Find the least number of positive integer weights you need in order to measure any integer weight from 1 to 1000. Note that any number of weights can be placed in any of the 4 pans (even the pan with the object you want to weigh).

You have a balance with fulcrum atClarification: You want to be able to weigh the integer weights from 1 to 1000, but you should not assume that the object you weigh will have an integer weight.

The answer is 5.

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I believe that putting weight in distant-from-fulcrum pan is balanced by putting twice the weight in opposing near-the-fulcrum pan. I know it's well-known in physics, but I believe it should be mentioned in the problem.

Not a complete solution here, I can't solve it yet but probably my work can help a bit. For every n different standard weight, we can put each in pan a, pan b, pan c, pan d, or not put the weight. This will cause 5^n possibility. Note that not putting anything at all doesn't contribute to weighing, so there are 5^n - 1 cases remaining. Note that each possibility has complete opposite, and that the case of having standard weighs in opposing position are actually similar. now we can observe (5^n - 1)/2 different cases. We're still assuming the perfect condition that each cases can get balanced by two different non-standard weight (because the non-standard weight can be put in distant pan or near pan. Now with n standard weight in this perfect condition, we can weigh 5^n - 1 non-standard weight.

However, we also can tell that to measure certain integer weight we don't have to get the perfect balance. For example, we can measure a non-standard 1 kg weight just by knowing that it's lighter than a 2 kg standard weight. Now if we assume the perfect condition such that all of those 5^n - 1 different condition can be used to balance 5^n - 1 different even weights, we can assume that we actually can measure 2 (5^n -1) different integer weights. In this case, n= 4 should be enough to measure 1000 first integer weights. I have a hard time deciding the case where it can work, but if it doesn't work, then obviously the problem is extreme double-counting.

For n=5, we can use 2, 10, 50, 250, and 1250 to measure up to 3124 different weights