Nov 23 - The Scales Problem

 

Assuming the intent is that the merchant uses the scales once per order, the question is then, what four numbers can be added together, in various combinations, to compose the numbers bewteen 1 and 40. This is impossible, since:

one of the numbers has to be 1g to meet the smallest possible amount

one has to be 2g for the next smallest number

these can be added to get 3g

then we need a 4g weight

now we can make 5, 6, and 7

then we need an 8g weight, which means the highest sum is 15, a far cry from 40.

Alternatively, the puzzle doesn’t specify how the merchant weighs the quantities. In theory then the weights could weigh 1g, 2g, 3g, and 4g, and the merchant could just weigh any herb out in this quantity, up to 10g at a time. For example, if someone wanted 20g of something, the merchant could weight out 10g of it 2 times, or 4g of it 5 times, etc. The weights could then be any values so long as one of them is 1g, to guarantee odd numbers and the smallest amount.

Another possible alternative involves the use of subtraction. If for example I had two weights, one of 18g and the other of 19g, I could measure 1g by weighing out 19g, then replacing the 19g weight with the 18g and removing herbs from the other pan until they balanced, which would be 1g. Are there 4 different numbers bewteen 1 and 40 that with addition and subtraction bewteen them (using each number only once) we can generate every number between 1 and 40?

If so, let’s call them w, x, y, and z. Now, we need to create 40 unique permutations of these numbers using addition and subtraction, to represent every number bewteen 1 and 40.

 

 

Let w > x > y > z, then we get the following table

Permutation #	Permutation	Permutation #	Permutation
1	W	21	W+X-Y
2	X	22	W+X-Z
3	Y	23	W+Y-Z
4	Z	24	X+Y-Z
5	W+X	25	W-X-Y
6	W+Y	26	W-X-Z
7	W+Z	27	W-Y-Z
8	X+Y	28	X-Y-Z
9	X+Z	29	W+Y-X
10	Y+Z	30	W+Y-Z
11	W-X	31	W+Z-X
12	W-Y	32	W+Z-Y
13	W-Z	33	W+X+Y+Z
14	X-Y	34	W-X-Y-Z
15	X-Z	35	W+X-Y-Z
16	Y-Z	36	W+X+Y-Z
17	W+X+Y	37	W+X-Y+Z
18	W+X+Z	38	W-X+Y-Z
19	W+Y+Z	39	W-X-Y+Z
20	X+Y+Z	40	W-X+Y+Z

 

Which may be possible if we can choose correct values of each number.

Given our initial restraints, we have the following conditions to satisfy all the permutations

w + x + y + z = 40 for #33

w > x + y + z for #34

x > y + z for #28

Then w + x + y is the second largest possible sum, 39, which tells us z = 1

From here I spent about 2 hours trying to find a combination of numbers, but a pattern never revealed itself to me. I tried a simpler case, what numbers would I need to create the numbers from 1 to 10, and got two answers: 1, 2, 7 and 1, 3, 6. I couldn’t find a pattern there either, so unfortunately I had to surrender to this problem and look up the solution, which is 1, 3, 9, and 27.

Given that I couldn’t solve it I don’t think it needs an extension, but a contraction to a series of smaller ranges building up to 1-40 could help students find a pattern in advance of this problem.