[Guest]

my friend created an interesting curiosity, the unbalanced scale. each arm is of a different length, and the two pans wiegh differently. you can currently tell the left arm is longer, but the right side is heavy.

your task is to determine the ratio between the two arm lengths, and the differance in wieght between the two pans, using any wieghts you see fit, and the minimum number of wieghs.

edit: it's sfe to assume the ratio is not more than 32:1.

Edited May 5, 2011 by phillip1882

[k-man]

Is the left arm a whole number of times longer than the right arm?

[Guest]

jst keep adding 1unit weights to it till they become balanced (stable).

that is the diff in weights of pan. and diff in length can also be found through that.

[Guest]

both are whole numbers, each less than or equal to 32 inchs (or meters or whatever unit of measure you want.)

so you could have the left arm be 15 and the right arm 7.

[Guest]

For the min total weighings, we need some parameters regarding the range of weights possible for the pans. That is needed to determine the min number of weighings for step 1, which is to bring the scales into balance.

Step 1, add or subtract weights to the left side until it balances the right side. Each time weight is added or subtracted, it should be half the amount added or subtracted on the previous weigh. When they balance, the weights on the left side will equal the difference in pan weight multiplied by the ratio between the arm lengths.

Step 2, now that they scales balance, add 1 weight to the right side. Add 16 to the left side. If the left side is heavy, then subtract half that, 8, else add 8. Repeat this, adding half the amount added/subtracted last time if left is heavy or subtract if right is heavy, and no change if they balance. The scales should balance within 5 weighs, with all possible ratios covered from 1 to 31. If they do not balance, then that means there are 31 weights on the left side and the ratio must be the only other option not weighed, 32.

Final, the ratio between the arm lengths is the ratio of the weights added in step 2. The difference in pan weights is the weight from step 1 divided by the ratio in step 2. The total weighings are 5 for step 2 and undetermined for step 1.

That is unless this is a trick question, and the "minimum" weighings is actually 2, where you get lucky and balance the scales with just one try in both steps 1 and 2.

Edited May 5, 2011 by Nana7

[Guest]

Start with 1 & 16 if less go to 1 & 8 if less go to 1 & 4 if less 1 & 2 so 5 would be the least numbers of weights!

[Guest]

Suppose that the scale is balanced => L1/W1 = L2/W2 => W1 = L1*W2/L2

Add weights so that the scale is balanced again => L1/(W1+a) = L2/(W2+b)

Substituting,

L1/(L1*W2/L2+a) = L2/(W2+b)

L1*L2/(L1*W2 + a*L2) = L2/(W2+b)

L1/(L1*W2 + a*L2) = 1/(W2+b)

Cross-multiplying

L1*W2 + b*L1 = L1*W2 + a*L2

L1/L2 = a/b.

The pans can have any 2 weights with the ratio W1/W2 = a/b, thus any difference you desire.

[Guest]

you answered in the spirit of the puzzle.

while neither of you gave a minimum number of wieghs, given how the puzzle is worded this is impossible.

i ment this to be solvable, and hoped you would understand that the max ratio betwen the pans is also ment to be 32:1 with each pan being a whole number less than or equal to 32. i really need to think these things through more before posting the puzzle.