Weigh-scale Problem (New Solution)

In math class today, we were challenged with a classic problem about measuring weights using an old-fashioned set of scales. While the correct solution was not immediately evident to me, an alternative solution presented itself. I take the time now to show you the original solution, my solution and how I found it.

Problem: A weigh-scale has four 'test weights' that are used to help balance the scales and measure the weight of unknown objects. If these scales can correctly measure the weight of any object whose weight is a whole number from 1 to 40 (1kg, 2kg,...39kg, 40kg.), how much do each of the four 'test weights' weigh?

Solution:

1kg - 3kg - 9kg - 27kg

This is because, the difference between weights is a range twice the amount of the former weight, allowing the any number in between to be reached from either bound.

Proof:

Below is a table values of weights. The first number shows the weight of the unknown object and the second and third sets of numbers show the balancing of numbers.

1: 1 = 1
2: 2 + 1 = 3
3: 3 = 3
4: 4 = 1 + 3
5: 5 + 1 + 3 = 9
6: 6 + 3 = 9
7: 7 + 3 = 1 + 9
8: 8 + 1 = 9
9: 9 = 9
10: 10 = 9 + 1
11: 11 + 1 = 3 + 9
12: 12 = 3 + 9
13: 13 = 1 + 3 + 9
14: 14 + 1 + 3 + 9 = 27
15: 15 + 3 + 9 = 27
16: 16 + 3 + 9 = 1 + 27
17: 17 + 1 + 9 = 27
18: 18 + 9 = 27
19: 19 + 9 = 1 + 27
20: 20 + 1 + 9 = 3 + 27
...
40: 40 = 1 + 3 + 9 + 27

The interesting thing about this solution is that you can carry this process on indefinitely. In the same way that you can reach all numbers with just 1,2,4,8,16, ... etc. you can reach them with 1,3,9,27,81, ... etc. The trick to this puzzle seems so mundane when presented in this fashion, so let's bring in a new idea!

My Solution:

2kg - 6kg - 18kg - 54kg

You'll notice now that my numbers are quite big. In fact, rather coincidentally, they are exactly double. But what is funny, is that I discovered this solution and remained oblivious to the original solution, until some one showed it to me. So let me show you why this one works. Note, this solution works so well, that you can go farther than 40kg, you can go to 80kg!!

Proof:

Reaching the even numbers with my solution is identical to the original solution, although I didn't see that at the time. However, without the number '1', reaching the odd numbers is difficult. Now, two 'weighings' are required. But despite that, reaching the numbers is just as fast except now we can reach 80 with just four 'test weights'!

1: 1 < 2
1 > 0
2: 2 = 2
3: 3 + 2 < 6
3 > 2
4: 4 + 2 = 6
5: 5 < 6
5 + 2 > 6
6: 6 = 6
7: 7 < 2 + 6
7 > 6
8: 8 = 2 + 6
9: 9 + 2 + 6 < 18
9 > 2 + 6
10: 10 + 2 + 6 = 18
11: 11 + 6 < 18
11 + 2 + 6 > 18
12: 12 + 6 = 18
13: 13 + 6 < 2 + 18
13 + 6 > 18
14: 14 + 6 = 2 + 18
15: 15 + 2 < 18
15 + 6 > 2 + 18
16: 16 + 2 = 18
17: 17 < 18
17 + 2 > 18
18: 18 = 18
19: 19 < 2 + 18
19 > 18
20: 20 = 2 + 18
...
27: 27 + 2 + 6 + 18 < 54
27 > 2 + 6 + 18
...
35: 35 + 18 < 54
35 + 2 + 18 > 54
36: 36 + 18 = 54
37: 37 + 18 < 2 + 54
37 + 18 > 54
...
71: 71 < 18 + 54
71 + 2 > 18 + 54
...
79: 79 < 2 + 6 + 18 + 54
79 > 6 + 18 + 54
80: 80 = 2 + 6 + 18 + 54

Just to name a few!

I really like how pretty this looks. It's very easy to do, once you see the pattern play out over enough numbers. Now, I'll give you a little explanation behind how I landed on these numbers.

When I first saw the problem, I knew that the four numbers (weights) should probably add up to 40 (or at least close to). That way, when the biggest number came up, these four numbers could still reach it. But I was struggling to identify (only) four numbers that were large enough to reach 40, but small enough to hit each little number as it came up. Believe it or not, my first quess came along the lines of 1 - 4 - 9 - 20. It seemed to me that there was adequate space between the numbers, but that they could still reach in between. So seemed a good place to start, save for my initial constraint that the four numbers had to add up to 40. These added up to only 34. This lead me to believe that '1' was out! No way could I have four numbers and have '1' as one of them. So I next looked at 2 as my lowest number. This immediately prompted the question, well how would you weight '1'? Hmm... well, I suppose '1' is less that '2' (my lowest number), and since these are weigh-scales, we'd easily be able to determine that '1>0'. Thus, began my rationale behind using inequalities. After '1' was solved, '2' was easy. It was '2=2'! But '3' was hard. If my second weight was '5', then '3+2=5'. But by that reaoning, '6+5+2=13' and '14+13+5+2=34'. So, my four numbers would be 2 - 5 - 13 - 34. Does this work? Actually, it does!! But it can only reach 54 (not 80) and more importantly, before I had even got to working out all four numbers, I had already thought of changing it.

If I can get away with 1<2, then I can get away with 2+3<6 and from that I can do anything. 2+6=8 and so, the next number we could have to deal with is 9. And again, I can get away with 2+6+9<18, so I did so. And from this, I produced way of figuring out the next number. 2+6+18=26, so 2+6+18+27<54 was the next step in the logic. And from that I produced my guiding numbers. And they are 2 - 6 - 18 - 54. And with these, I can tackle all of them to 80. Not 2+6+18+54=80. So, if I only had three weights, I could reach 26. i.e. 2+6+18=26.

So, if you are faced with this problem in the future, you have the skills to overcome it. And what if you are presented with 1-40,000 instead of 1-40?? Can it be done? Is this the most efficient way? What are the implications? Below I have set a question dealing with just that:

How many weights (#) are necessary to reach 40,000kg? How high would those weights reach (kg)?

ANSWER MY QUIZ
(side bar of page)

Hope you enjoyed!