How can one with an accurate balance-scale measure any weight up to let’s say 1600 gram with an error margin of 1 gram, when available are only 7 different counterbalance (weights)


3 Answers 3


In the traditional version of this problem, dealing with exact integer weights, the standard answer is to

use weights increasing in powers of 3: 1, 3, 9, 27, etc.

How much error can this cope with? Answer:

up to 1/2 a unit.


by doubling these: 2,6,18,...,1458 we can get up to 2+6+18+...+1458=2186 units. (In this case, grammes.)


1) First weight is 2g to handle 1-3g.

2) In order to weigh something at 4g (outside error margin of 1g) we can put the 2g weight on it pushing the lower limit to 6g. Now we need a 7g weight which now we can handle from 1-10g (2 + 7 + error of 1)

3) Next weight we want is 21g. For 11-13 we add 7+2, for 14 we add 7, for 15-17 we add 7 and do 21+2, and for 18-20 we add 2. Then we can also do weights above 21 using the 2 and 7 getting us to 31g.

4) Continuing from what we figured in the last step, if we take the total (30) and add it to the next weight we don't know yet (32) we get 62. One higher than that 63g should be the weight we need. 32-34 are good with +30. For 35-36 we add 21+7. For 37-38 we add 21+7 and do 62+2. For 39-41 we add 21+2... etc

Assuming this pattern holds, a weight W(i) = 2 * N + 3 where N is the sum of weights W(1) to W(i-1).

Giving us...

2, 7, 21, 63, 189, 567, 1701 which would actually handle weights up to 2551g with a margin of error of 1g

Edit to add, each successive number after 7 is the preceding number times 3. I'm not as well-versed in math as I used to be so I don't immediately know the reason for that but it lines up with the other answer of powers of 3. This however gets a higher total possible weight you can measure.

  • 1
    $\begingroup$ The difference between this and the other answer is that this answer assumes we will only weigh integer weights, which is why this gets a higher total. $\endgroup$
    – ffao
    Oct 31, 2016 at 22:56

The answer is

We can't


The maximum weight we can calculate is 318g

Think of it this way...

This works a lot like binary...

The key fact I've focused on is that the margin of error is only 1g. If we want a margin of error to 1g then the minimum weight we need is $2g$ (As that can cover 1g and 0g is, well, N/A). With a $2g$ weight we can measure $1g$ to $3g$.

So the maximum we can make with that is $3g$. So the next weight we need is $5g$. With this we can calculate $4g$ to $6g$. We have all number covered so far up to $8g$...

Now the next weight we need is $2g + 5g = 7g + 3g$ (margin of error) $= 10g$ With this we can calculate $9g$ to $11g$.

So that adds up to $17g$ meaning we can get up to $18g$. The next weight should be $20g$ because that can cover $19g$. So now we can cover up to $2 + 5 + 10 + 20 = 37g$. So we can cover up to $38g$, meaning the next one should be $40g$ which can cover $39g$.

The rule for finding the next weight ($W$) is:


where $T$ is the current total the wights add up to.

Now we only have seven weights. So the seven weights using the rule are:

$2g$, $5g$, $10g$, $20g$, $40g$, $80g$ and $160g$

Therefore the maximum weight we can calculate is 318g (317g + 1 margin of error).


As you can see, that is nowhere near 1600g.

  • $\begingroup$ You could do better even without the error margin. $\endgroup$
    – ffao
    Oct 31, 2016 at 22:14

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