There is only one 3-digit number that is both a square and a cube (presumably, of an integer). What is that number?At the end of the class, I asked if anyone had solved it. One girl raised her hand. I asked her for the answer, and she said, "\(729\)." I asked how she got that, and she said that she tried a few random numbers. I was intrigued, so I asked her for more details. Turns out the numbers she chose weren't so random. "I knew that 10 squared is a hundred, so I started there," she said. I remarked that this method, which I call "plug and chug", is used by computers to solve problems that we couldn't possibly do in a lifetime of chugging, and that thinking in these (algorithmic) ways can be useful for solving many types of problems, even without a computer.

Intriguing, indeed. I sat down after class and tried to kludge together a short Pascal (the only programming language I know well enough to do this stuff) program to solve this problem. When a student who stayed after class asked about it, I started to realize that I had the wrong initial approach to the problem, and I showed him another approach on the board. This was the algebraic/graphing approach, where you state the problem in algebraic terms, and use a graph to at least constrain the possible solution space:

Let \(y\) equal an integer, and \(y^{2}\) its square. And let \(x\) equal another integer, and \(x^{3}\) its cube. Then the problem is to find a 3-digit number that equals \(y^{2}\) and \(x^{3}\). Put another way:

\(y^{2} = x^{3}\), or \(y = x^{3/2}\), where \(100≤y^{2}≤999\) and
\(100≤x^{3}≤999\)

I used the wonderful Desmos graphing calculator
to visualize this space:

The dark gray rectangle in the middle of the output above is the constrained area of the solution to the problem, and the answer can be gotten by finding the point in that section on the curve that corresponds to integers \(x\) and \(y\). This is somewhat difficult to do with Desmos, but I already knew the answer \((9,27)\), and I did find this point using Desmos' scan functionality.

On the way home from the class, I did some mental figuring, and came to the conclusion that the answer to this puzzle could be gotten by systematically going through the integers 1 through wherever you wanted to stop, and take their 6th power, and these would be the numbers that were cubes as well as squares of some integer. The thinking went something like this: the answer to the puzzle from the book (3-digit number that is a cube and a square) is \(729\), which is \(9\) cubed and \(27\) squared. Well, \(9 = 3^{2}\) and \(27 = 3^{3}\), and \(9^{3} = 27^{2} = 729\), so \((3^{2})^{3} = (3^{3})^{2}\), which is an exponential identity example \(((x^{a})^{b} = (x^{b})^{a} = x^{ab})\). A little mental calculation gave the 2-digit cube/square number as \(64 = 2^{6}\), which backed up the hypothesis. So, to get the numbers which are both cubes and sqaures (of integers, by definition), all I needed to do was to write a Pascal program that listed the numbers which were 6th powers of each of the integers from 1 to whatever number I wanted to go up to.

When I got home, I started writing the program, and with a lot of looking up for syntax, and adding the counting of digits to make it a more interesting problem discussion for the class, I came up with a program which produced the desired output.

But after thinking about it for a while, I decided that Javascript would be a better solution, since this language was made for HTML and browsers, and I've done a fair amount of programming in it for another venture. Here's the result:

The digit-count is in parentheses, as you might have surmised. So, as the problem statement above says, there is only one 3-digit number that is both a cube and a square. It turns out there is another problem in the puzzle book just before the one I gave asking for the 2-digit number that is a cube and a square, and we can see, as the book notes, that the answer is \(64\). There is also only one 4-digit solution \((4096 = 4^{6} = 2^{12})\), two 5-digit solutions, three 6-digit solutions, five 7-digit solutions, and so on. I'm trying to see whether there is any pattern to the number of digit solutions, but haven't discovered anything obvious yet. One pretty cool pattern to note is that the ones digit (farthest to the right) of the solutions go in the repeating pattern \(1, 4, 9, 6, 5, 6, 9, 4, 1, 0\). If you include \(0\) in the integer list, this becomes a palindrome: \(01496569410\). Cool!

And this is only one of the hundreds of puzzles included in this short book of puzzles! Imagine how much fun we'll have solving some of the others!!

The dark gray rectangle in the middle of the output above is the constrained area of the solution to the problem, and the answer can be gotten by finding the point in that section on the curve that corresponds to integers \(x\) and \(y\). This is somewhat difficult to do with Desmos, but I already knew the answer \((9,27)\), and I did find this point using Desmos' scan functionality.

On the way home from the class, I did some mental figuring, and came to the conclusion that the answer to this puzzle could be gotten by systematically going through the integers 1 through wherever you wanted to stop, and take their 6th power, and these would be the numbers that were cubes as well as squares of some integer. The thinking went something like this: the answer to the puzzle from the book (3-digit number that is a cube and a square) is \(729\), which is \(9\) cubed and \(27\) squared. Well, \(9 = 3^{2}\) and \(27 = 3^{3}\), and \(9^{3} = 27^{2} = 729\), so \((3^{2})^{3} = (3^{3})^{2}\), which is an exponential identity example \(((x^{a})^{b} = (x^{b})^{a} = x^{ab})\). A little mental calculation gave the 2-digit cube/square number as \(64 = 2^{6}\), which backed up the hypothesis. So, to get the numbers which are both cubes and sqaures (of integers, by definition), all I needed to do was to write a Pascal program that listed the numbers which were 6th powers of each of the integers from 1 to whatever number I wanted to go up to.

When I got home, I started writing the program, and with a lot of looking up for syntax, and adding the counting of digits to make it a more interesting problem discussion for the class, I came up with a program which produced the desired output.

But after thinking about it for a while, I decided that Javascript would be a better solution, since this language was made for HTML and browsers, and I've done a fair amount of programming in it for another venture. Here's the result:

The digit-count is in parentheses, as you might have surmised. So, as the problem statement above says, there is only one 3-digit number that is both a cube and a square. It turns out there is another problem in the puzzle book just before the one I gave asking for the 2-digit number that is a cube and a square, and we can see, as the book notes, that the answer is \(64\). There is also only one 4-digit solution \((4096 = 4^{6} = 2^{12})\), two 5-digit solutions, three 6-digit solutions, five 7-digit solutions, and so on. I'm trying to see whether there is any pattern to the number of digit solutions, but haven't discovered anything obvious yet. One pretty cool pattern to note is that the ones digit (farthest to the right) of the solutions go in the repeating pattern \(1, 4, 9, 6, 5, 6, 9, 4, 1, 0\). If you include \(0\) in the integer list, this becomes a palindrome: \(01496569410\). Cool!

And this is only one of the hundreds of puzzles included in this short book of puzzles! Imagine how much fun we'll have solving some of the others!!