How Can We Weigh Anything?

Sometimes in math we can intuitively understand the answer to something, know that it’s correct, and yet can’t lay out the proper reasoning in mathematical terminology. It’s a frustrating feeling to know that you have what normally comes at the end of the process, and yet have nothing to show for it. I think this problem captures the essence of this feeling quite well:

A few years ago, a king’s math professor purchased a small farm, a place where he could unwind on the weekends and grow some fruit and vegetables. The farm came with all kinds of tools and various implements, including a large balance scale. Next to the scale was an old (pre-metric) 40 lbs stone with some initials carved into it: apparently, a previous owner had used it to weight 40 lbs of feed.

One morning, while cleaning out his barn, the professor dropped the stone and it broke into four pieces.

The professor was a bit sad about his carelessness, as he had liked that curious old stone. But he soon discovered something interesting: He could use the four pieces of the broken stone to weigh items on the balance scale – as long as these items were in one pound increments, between 1 and 40 pounds.

How much did each of the four pieces weigh?

I’m going to outline my take on the problem in this post, so (as always) don’t scroll beyond the picture if you want to do it yourself.


If you think about this long enough, you may soon come to the idea that the weights could be broken into 1, 3, 9, and 27. This sort of comes about intuitively through our understanding of the numeric system, which takes different powers of a certain base to express any intermediate number. Here, it’s a little different because we can subtract any of the weights, but it all comes around to the same logic. Now, how do we prove this? More importantly, how do we show how we can arrive at the answer without using it to prove its own validity?

Firstly, we know we can measure 40 different weights (1 – 40), but they can also be negative (on the other side of the balance), and we can have 0 as well. So we have 40 + 40 + 1 = 81 different weights we can measure. Now how can we make those combinations: of our 4 weights, each can be used on either side or not be used at all, so each can be used in 3 different ways. This comes around to 34 = 81 different combinations, the same exact number we just calculated. So what does this tell us? If we need to measure out 81 different combinations, and can create 81, that means that an distinct combination of the weights will result in a distinct integer weight (in mathematics this is called being a sharp condition).

Let’s assume our weights: a ≤ b ≤ c ≤ d. Now we know that each distinct combination will result in a distinct integer output, so we know that all of them combined will equal 40:

a + b + c + d = 40

The next smallest combination would be b + c + d, which would have to equal 39:

b + c + d = 39

Following this logic, we can go down next biggest combinations, which have to be:

b + c + d – a = 38

a + c + d = 37

As we form new equations, we can solve them simultaneously to arrive at our answer Only 4 simultaneous equations are needed for 4 variables, however we do need to use one more beyond these (these four will not give discrete values for c and d due to the common term c + d throughout).  The smallest equation which doesn’t contain +c and +d is:

a + b + d = 31

Solving all these equations simultaneously, we find our original answer of:

a=1, b=3, c=9, d=27

The way in math makes you prove something you already know is one of its interesting facets in my opinion, and I don’t think I could’ve found a better problem to display that. This aspect of math does however does make it the subject of many jokes:

A mathematician, a physicist, and an engineer are riding a train through Scotland. The engineer looks out the window, sees a black sheep, and exclaims, “Hey! They’ve got black sheep in Scotland!” The physicist looks out the window and corrects the engineer, “Strictly speaking, all we know is that there’s at least one black sheep in Scotland.” The mathematician looks out the window and corrects the physicist, ” Strictly speaking, all we know is that is that at least one side of one sheep is black in Scotland.”


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