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@Above WTH ! I think it's a fake solve

Delta0001 wrote: @Above WTH ! I think it's a fake solve

Yes, your solution is wrong as the question mention non profit integer weight of control weights

Choose powers of $3$. Then, see that $1,3,9,27$ can be adjusted to obtain any weight in the balance pan. SO, the least possible of control weights is $4$.

khan.academy wrote: Choose powers of $3$. Then, see that $1,3,9,27$ can be adjusted to obtain any weight in the balance pan. SO, the least possible of control weights is $4$. Wrong, you cannot measure any number with 2 in its base-3 representation (like 2) this way. Remember, the control weights and measuring weights are on different pans.

meet18 wrote: Delta0001 wrote: @Above WTH ! I think it's a fake solve

Yes, your solution is wrong as the question mention non profit integer weight of control weights What do you mean by non-profit ?

Delta0001 wrote: What do you mean by non-profit ? A more reasonable question is: Where does the question mention this? Apparently it is a combination of a horrible typo of 'integer' and an even more horrible autocorrect. You have my word, this combination is deadly. Sidenote: His/her question (assuming my correction is what the user meant) is legitimate. I really fail to understand your solution. But what I did notice is that $12$ is an upper bound on the minimum. Just break $1,2,4,8,16,32$ into any two non-integral positive numbers in any way you want. Lower bound on minimum is $6$ because if there are just $5$ of them, then we can make at most $32$ different weights using them while we need at least $40$. My current plan on reducing the number required is We used $4$ weights to form $16$ and $32$. Can we make it $3$? And so on...... It does seem to have this problem that if we form $32$, we might make $16$ impossible. It maybe does restrict us with other cases but not with this particular one: if we form $32$, we definitely don't need $16$.

Upper bound:7. Use these set of weights: $0.5,0.5,1.5,2.5,5.5,10.5,21.5$ Proof of lower bound: Asymptotic argument. First of all as $1$ is achievable there must be two non-integer weights summing up to $1$. As $2$ is possible, there must be a weight less than or equal to $2$. As $4$ is possible, we can now see that there must be a weight less than or equal to $4$. Similarly there are two more weights at least which are less than $8$ and less than $16$ respectively. As all these weights can have maximum sum $31$ a seventh weight is required to get $40$ as in the example shown above.

Starting with \(0.5,0.5\), we can greedily add the smallest possible half-integer necessary to get a new number to reach Mathotsav's sequence of \(0.5, 0.5, 1.5, 2.5, 5.5, 10.5, 21.5\). This allows the representation of all integers \([0;42]\). This is easily shown to be optimal - a sequence of length 6 could never be sufficient, since at least half of the \(2^{6}=64\) subsets must sum to a non-integer, as both a subset containing the first element and one identical to it except for the first element can't be integers. Note: the doubles of these numbers seem to be sequence A001045 in OEIS.

I will offer another example of 7 weights: 1/3, 2/3, 4/3, 8/3, 16/3, 32/3, 64/3. (Obviously none of them are integers.) Since this sequence can form any k/3 where k varies from 1 - 127 and 40=120/3 shows its correctness.

Manuel_von_Norlanski wrote: I will offer another example of 7 weights: 1/3, 2/3, 4/3, 8/3, 16/3, 32/3, 64/3. (Obviously none of them are integers.) Since this sequence can form any k/3 where k varies from 1 - 127 and 40=120/3 shows its correctness. I wonder whether there exists a function of X to give the greatest "Weighable" value of X weights (under the circumstances of this problem) and guess that it is the greatest integer not greater than (2^X-1)/3.