The Haselbauer-Dickheiser Test can be found at http://matrix67.com/iqtest/.
In this blog post, I will study Question 21 from this test.
The question is about the construction of a square from 24 non-overlapping smaller squares, each with a side of unique length.
Below I present the original question for your convenience:
Please do not read the rest of this article, if you want to attempt to solve this question on your own. The rest of this article describes my attempt at solving this question and you should not read it, unless you want to or you do not mind coming across relevant ideas, spoilers, hints, solutions, and strong opinions concerning this test.
You have been warned and I now consider that you continue to read knowing that what you come across for the rest of this article may forever spoil things for you and/or present strong opinions against this test.
Last warning: please do not read this blog post, unless you are certain that you know what you are doing. If you are not sure, then it would be best if you stopped reading at this point.
OK. If you are here, it means that you want to know my opinion. Well, ok then!
To cut a long story short, my opinion is that the test is highly inappropriate. In this blog post, I will focus on the study of question 21.
Actually, this one of the good questions of the exam. I am saying this because the exam contains some highly inappropriate questions as well. But there is still a problem with this question. As with some other questions in this test, the answer can be found in one of Martin Gardner’s books.
So, why would Haselbauer and Dickheiser copy (not just get inspiration, but blatantly copy) questions from Gardner’s books, questions that are fully answered in those same books, when they claim, and I believe them, that they want the integrity of the test to be high?
I said and I am planning to prove that the Haselbauer-Dickheiser Test is inappropriate (and this for many reasons), but this is ridiculous!
Anyway, back to answering the question. I will provide information about Martin Gardner’s book later on in this blog post. But first, I would like to present my attempt at solving this question, which I did before I read the corresponding chapter of the book.
So here is how I solved this question without reading the corresponding chapter.
I observed the image and I thought that I should tackle the problem beginning from the smallest square. I named each of the 24 squares with a number from 1 to 24, counting in my order of preference that I chose loosely based on the order in which I imagined that my calculations would proceed. Anyway, giving a name to each square is arbitrary, because the actual name does not matter, what matters is that we need to have a way to refer to each square. So, I chose the names that are depicted in the following image:
Please note that image above does not depict the size of the side of each square. It depicts the name of each square. I made this so that I can refer to each square and so that I can calculate the size of the side of each square.
So, imagine that the smallest square is named 1, or rather A1, because I am going to use Excel and this way, I will easily create the spreadsheet so that the size of each square will correspond to the cell that has the name of that square.
Here is the Excel spreadsheet that I created.
And here is the way I worked, what I was seeing when I was working at the solution:
So, I had these two windows side-by-side as I was working with the spreadsheet.
Let me describe the spreadsheet. Next to each cell is another cell that depicts the formula used in the first cell. I only enter numbers in the four white cells. The numbers in the yellow cells are automatically calculated by the corresponding formulas in those cells.
Cell A1 corresponds to the side of square 1, the smallest square. Cell A2 corresponds to the side of square 2, the one on the left of square 1. I do a trial and error analysis, where I guess the values of the sides of squares 1, 2, 14, and 15. And I let the formulas that I entered produce the values of the sides of the other squares.
So, the side of square 3 is on the cell A3, which contains the formula =A1+A2, because the side of square 3 is the sum of the sizes of squares 1 and 2. And so on for the other squares.
So, by guessing the sizes of four squares, I calculate the sizes of the rest of the squares. The values I choose have to be small for the squares 1, 2 and 14, and the value for the square 1 has to be the smallest of the three. Also, by looking at the value that is produced from square 1 and square 2 for square 10, I choose a somewhat smaller value for square 15. This is because the image gives me this hint, if I roughly compare square 15 and square 10.
In order to know if the values I enter by trial and error produce the correct result, I check that the sizes of the all encompassing square are all equal. This is why I have created the column D. Each cell is the resulting calculation of one of the four sizes of the all encompassing square. By entering values in the four white cells in column A, I do not really care about the values produced in the yellow squares of column A. I really care about the four values in column D. They all have to be equal to each other. So this is how I know if I have the correct values in the four white cells in column A.
By trial and error, I found that when A1=1, A2=3, A14=2, and A15=29, all values in column D match, i.e. they are equal to each other and equal to 175. So, this means that I solved the puzzle. So, the all encompassing square has a side equal to 175. And the square in question, square A10, has a side equal to 38. And this is the answer: 38.
The way I solved the puzzle, I was very cautious that the numbers I input might have been coprime, especially A1 and A2. When I first saw this puzzle and understood that I would solve it using trial and error, basing my guesses and beginning from the smallest square and the one next to it, I was extra careful not to assume any relationship between these two numbers. In hindsight, things were not that bad, since A1=1 and A2=3, but I could not have known. So I was extra careful to account for a case like A1=2, A2=5 and so on.
Another thing is that the question should have been stated in a way that acknowledges that there are infinite such rectangles and we are searching for the smallest one. Indeed, any integer multiple of the solution is a solution as well. For example, multiplying by 2, we get A1=2, A2=6, A14=4, A15= 58 which also produces a solution. And multiplying by 3, we get A1=3, A2=9, A14=6, A15= 87 which also produces a solution. And so on to infinity. But the question forgets, or rather assumes that we correctly assume it. Good assumption from our part, bad assumption from the question’s part.
As I said, the answer is depicted in one of Martin Gardner’s books. Martin Gardner’s book titled “The Second Scientific American Book of Mathematical Puzzles and Diversions”, which is the 2nd book out of a 15-book series has the answer to this puzzle. Below, I present a screenshot of the pages 205 and 206 from this book.