Ken's Puzzle of the Week

Rods in a Mobile

The diagram is not drawn to scale.  Every horizontal length is an integer.  If all three horizontal rods are the same length, what is that length when:

  1. The weights from left to right are 1,2,3,4,5?
  2. The weights from left to right are 5,4,3,2,1?
What configuration of the weights 1,2,3,4,5 leads to the longest/shortest rod-length?

Extension: What is the lightest configuration of unique weights for a rod length of 3?  4?  5?

Source: Original.


Solution Solutions were received from Denis Borris, K Sengupta, Mark Rickert, and Joseph DeVincentis.  Joseph had a very thorough analysis:

When the weights are 1,2,3,4,5: The split of weight across the top bar is 3 vs. 12, so that bar must be split into lengths in the ratio of 4 to 1 and so the whole bar is a multiple of 5. For the left bar, the split is 2 to 1 and the bar must be a multiple of 3 also. So if the bar is of length 15, can the weights 3, 4, and 5 be distributed properly? We need 5x + 4y = 3(15-x) = 45 - 3x, or 8x + 4y = 45 with y<x integers. Clearly, the quantity on the left is a multiple of 4 and this is impossible. For this to work, we must have a bar of a length which is a multiple of 4, and the shortest possible is 60. Then 2x + y = 45, and many solutions are possible such as x=20, y=5.

When the weights are 5,4,3,2,1: The split of weight across the top bar is 9 vs. 6, so the top bar is split in the ratio 2 to 3 and must have a length that is a multiple of 5. The left bar is split in the ratio 4 to 5, and must have length a multiple of 9. If the bar is of length 45, the weights on the right bar must be arranged so that x + 2y = 3(45-x) = 135 - 3x, or 4x + 2y = 135, and the bar must be an even length. Then 2x + y = 135 has many solutions on the bar of length 90.

For any ordering of the weights 1 through 5: It is clearly impossible to get a bar of length less than 15. If the top bar is split in a 2:1 ratio, the left side must have two weights that add to 5, and these will require a bar length that is a multiple of 5. If the top bar is split in a 3:2 or 4:1 ratio, the left bar can have weights 1 and 2, 1 and 5, 2 and 4, or 4 and 5, but in all cases the bar length must be a multiple of 3. And for any other split of the weights across the top bar, the top bar alone is forced to be a multiple of 15. The first case above almost worked for a length of 15, and swapping 3 and 4 yields the equation 5x + 3y = 4(15-x) = 60 - 4x, or 9x + 3y = 60, 3x + y = 20. This has the solution x=6, y=2, with 9 units of the bar's length on the left side. So 1,2,4,3,5 is one ordering that allows the minimum bar length of 15.

The top bar holds weights that add to 15, and can be forced to be a multiple of 15 units long if the weights are split not into multiples of 3 or 5; this means they must be split as 4-11, 7-8, or 8-7, so either 1,3, 3,4, 2,5, or 3,5 are in A,B. Each of these forces the bar to be a multiple of A+B. I am going to skip the other configurations as they are at a distinct disadvantage. So now we have to consider the possible balances of C vs. D and E for all permutations of 245, 125, 134, and 124.

For 245 (60 already forced)
245: 2x16 = 4x3 + 5x4 fits on a bar of 20 (overall length still 60)
254: 2x15 = 5x2 + 4x5 fits on 20 (overall 60)
425: 4x3 = 2x1 + 5x2 fits in 5 (overall 60)
452: 4x5 = 5x2 + 2x5 fits in 10 (overall 60)
524: 5x2 = 2x1 + 4x2 fits in 4 (overall 60)
542: 5x2 = 4x1 + 2x3 fits in 5 (overall 60)

For 125 (105 already forced)
When 1 and 5 are on the ends of the bar, the bar must always be of even length since these two weights must both be at odd distances or both at even distances from the fulcrum to satisfy 2-parity of the total torque. That said:
125: 1x12 = 2x1 + 5x2 fits in 14 (overall 210)
152: 1x25 = 5x1 + 2x10 fits in 35 (overall 105)
215: 2x11 = 1x2 + 5x4 fits in 15 (overall 105)
251: 2x4 = 5x1 + 1x3 fits in 7 (overall 105)
512: 5x1 = 1x1 + 2x2 fits in 3 (overall 105)
521: 5x2 = 2x1 + 1x8 fits in 10 (overall 210)

For 134 (105 already forced)
When 1 and 3 are on the ends of the bar, we have 1x = 4y + 3(b-x) or 1x + 4y = 3(b-x) where b is the length of the bar. This yields 4x = 4y + 3b or 4x + 4y = 3b, so the bar must be a multiple of 4 in these cases.
134: 1x18 = 3x2 + 4x3 fits in 21 (overall 105)
143: 1x10 = 4x1 + 3x2 fits in 12 (overall 420)
314: 3x3 = 1x1 + 4x2 fits in 5 (overall 105)
341: 3x2 = 4x1 + 1x2 fits in 4 (overall 420)
413: 4x7 = 1x4 + 3x8 fits in 15 (overall 105)
431: 4x2 = 3x1 + 1x5 fits in 7 (overall 105)

For 124 (120 already forced)
124: 1x10 = 2x1 + 4x2 fits in 12 (overall 120)
142: 1x8 = 4x1 + 2x2 fits in 10 (overall 120)
214: 2x7 = 1x2 + 4x3 fits in 10 (overall 120)
241: 2x3 = 4x1 + 1x2 fits in 5 (overall 120)
412: 4x2 = 1x2 + 2x3 fits in 5 (overall 120)
421: 4x1 = 2x1 + 1x2 fits in 3 (overall 120)

So the longest bars required are 420 for combinations such as 25143.

For the bar length 3 and any weights, we have 2A=B (or equivalently A = 2B), C=D+2E, and 2A+2B=C+D+E or A+B=2C+2D+2E. For either case of the 3rd equation, we have a multiple of 3 on the left, and 2D+3E (or twice this) on the right, forcing D to be a multiple of 3. Suppose D is 3. For E=1, we have C=5, and the right side adds to 9, so we need A+B = 2(C+D+E) = 18, A=12 and B=6 for a total weight of 27. If E=2, then C=7, and the right side adds to 12, but we cannot have the left side add to 6 because it would split into 2 and 4 and reuse the 2. If it adds to 24 we get a larger weight. For any other D=3 case, or any D=6 or larger case, the right side will add to more than 12 and so the whole thing will add to more than 18. But since the left side splits 2-to-1, the sum of those weights is a multiple of 3, and since the top bar splits 2-to-1, the sum of all the weights is a multiple of 9, so 27 is the smallest possible sum of weights.

For bar length 4, the left bar splits in a 3-to-1 ratio and the top bar splits 3-to-1 or evenly. The sum of the left bar is a multiple of 4, so the sum of all the weights is a multiple of 16 (when the top bar splits 3-to-1) or a multiple of 8 (when it splits evenly). For the right side there are 3 cases: C = D + 3E, C = 2D + 3E, and 2C = D + 2E. For the overall balance, C+D+E must be a multiple of 4.

For the last case of the right side, 2C = D + 2E, so D is even and C+D+E = 3C - E.  This can be satisfied at minimum weight with C=3,D=4,E=1, and this works with an evenyl split top bar and A,B = 2,6 at the minimum possible weight of 16.

For bars of length 5, the left and top bars each split in a 4-1 or 3-2 ratio. The right bar splits in various ways but must add to a multiple of 5. So A+B is a multiple of 5 and the whole set of weights adds to a multiple of 25. With both 3-2 splits, this is easily satisfied with C=3 (at distance 3), D=5 (at distance 1), E=2 (at distance 2), and A,B = 6,9.
 


Mail to Ken