Tag Archives: Square

Square-Cube Ages

The article that you are about to read in this page was written by the staff of the MATHCOUNTS Foundation and published in their Winter 1995 edition of their newsletter MATHCOUNTS NEWS. It is is reprinted here by permission.

Mild Turns Wild

Every once in a while, tucked deep within a MATHCOUNTS School Handbook or MATHCOUNTS competition, is a problem that seems somewhat mild, but after all
is said and done, could make a person go wild!

Last year my age was a perfect square. Next year it will be a perfect cube. How old am I?

This particular MATHCOUNTS problem isn’t a difficult one, until another dimension is added, as Terrel Trotter, Jr. of Escuela Americana in San Salvador, El Salvador did. He asked his students, “if we drop the idea of age and use larger numbers, can we find other examples of squares and cubes that have a difference of two?” At first glance, this seems like an easy answer, but it’s anything but.

The problem appeared a few months ago in Trotter’s monthly classroom tabloid, Trotter Math News. Accompanying the problem was a reminder: “problem solving is not merely computing the sum of two fractions or the product of two decimals, but rather doing whatever is necessary to answer a big question, using whatever method that may help, guessing, looking for patterns, using calculators — the whole works.”

That’s great advice, but even with the use of calculators and spreadsheet software, the students came up empty-handed. When they reported their findings, Trotter confessed, “I don’t know if there is indeed an answer!”

Not ready to accept defeat, Trotter went to the MATHCOUNTS head office in search of some answers.

MATHCOUNTS Curriculum Coordinator Scott Stull recalled, “when the problem was first written, we didn’t have a strong argument that the answer was unique for all integers. However we did convince ourselves that the answer was unique for all reasonable integral ages for a human being.”

To answer Trotter’s question, Stull enlisted the help of Richard Case, P.E., director of strategic development for IBM Corporation and MATHCOUNTS national judge, and also Harold Reiter, Ph.D., professor of mathematics at The University of North Carolina at Charlotte and MATHCOUNTS question writer.

The two initially agreed that another solution seemed possible but how they went about finding the alternate solution was quite different.

Case developed an algorithm to test the first n integers. He reduced the problem to finding solutions of the form {x, y} for the equation y3 – 2 = x2. His algorithm involved taking an integer, cubing it, subtracting two, taking the square root, and evaluating whether or not the result was also an integer. He was able to confirm that the solution {5, 3} is unique for the first 10,201 positive integers (that’s as high as his computer could go).

Reiter’s approach was slightly different. He developed his argument through number theory. Referencing an article by John Stillwell titled, What Are Algebraic Integers and What Are They For?, Reiter found an argument by Euler verifying that the original solution is the only solution:

A rigorous proof of this argument relies on abstract algebra, and establishing the existence of prime factorization for the ring of algebraic integers Z[-2].

However, Stull warned, “this cannot be generalized to say that there is only one solution to any problem of the form y3j = x2.” Take this problem for instance:

Two years ago the age of a certain tortise and that tortise’s child were both perfect squares. In two years both of the ages will be perfect cubes. How old is each tortise?

Now that the first problem has been tamed, what about this one? Is there a solution? If there is, is it unique? A reward is offered to the first team (of four students) who sends in a solution and an argument verifying or disproving the uniqueness of the solution. To collect the bounty, send your solution to MATHCOUNTS, 1420 King St., Alexandria, VA 23314-2794.

Good luck!

P.S. (March 1999) Don’t rush to send in your solution; the problem has already been solved and the reward claimed. But it’s still a nice problem about the mommy and baby tortise, don’t you think?

Happy & Dizzy Numbers


Before we can explain what a happy number is, you have to learn a new idea, called “recurrent operations.” As the word “recur” means “to happen again”, a recurrent operation must mean a mathematical
procedure that is repeated. A very simple example would be the rule “add 5 to the result”. If we started with the number 0 and applied that recurrent operation rule, we would produce the sequence 0, 5, 10, 15, 20, … ; this list is the famous “multiples of five”.

Of course, there are all kinds of recurrent operation rules in mathematics. Another important rule is “multiply each result by 2″. If we used 1 as our first number, this sequence shows up: 1, 2, 4, 8, 16, 32, … ; this list is the also famous “powers of two”. So, you see it’s really not such a difficult idea now, is it?

However, in order to produce “happy numbers”, we will invent a rule that is just a little bit more complicated. (After all, you
didn’t expect this to be that easy, did you?) Our rule now will be given in two steps: (1) find the squares of the digits of the starting number; then (2) add those squares to get the result that will be used in the repeat part of your work.

Here is an example. Let’s start with 375. We write:

32 + 72 + 52 = 9 + 49 + 25 = 83

Now we repeat the R.O. procedure with 83. This gives us:

82 + 32 = 64 + 9 = 73

Of course, we continue with 73. This will produce 58.


But we can hear you saying: “When do I stop? What’s the point of all this?” That’s the beautiful part of the story. The answer is: when you see something strange happening. The strange thing that tells when a number is happy is simply this: the result of a 1 eventually occurs. Here is an example, starting with the number 23:

4 + 9 = 13; 1 + 9 = 10; and 1 + 0 = 1.

It’s that easy! When you reach a 1, the starting number is called happy. [But don’t ask why it’s happy, instead of sad; that’s just what the books say.]

Once you determine a number is happy, you can say all the intermediate results are also happy. The numbers 13 and 10 must also
be considered as happy, because they too produce a 1.

Can you find some more happy numbers? Yes. If you know a certain number is happy, it’s easy to find many more. How? One way
is to insert a zero or two. Look: above we saw that 23 was happy, right? This means that 203 is also happy; so is 230. A larger example is 2003. See? Now you can make many, many happy numbers, using an old one with as many zeros as you wish.

But that’s the easy way. You want something a bit more challenging, don’t you? Well, that’s your task now — find some more happy numbers without using the “zeros” technique. Okay?

Part II: Dizzy Numbers

The term “dizzy numbers” was invented by me. It is based on an idea that should occur to anyone searching for happy numbers, because often they find themselves “going in circles”, literally, i.e. getting
dizzy. Here’s why:

Recall the number 375 from above? It produced the sequence 83, 73, 58,… But we stopped there in our explanation of the RO procedure. If we had continued, we would have had 89, 145, 42, 20, 4, 16, 37, and then back to 58! Hmm… now that’s strange, isn’t it? We’ve returned to where we were (58) just eight steps earlier; we’ve gone in a circle. We’ve produced an 8-term numerical cycle. Hence, we’re getting a little dizzy. (Get it?)

So we can now define more formally a dizzy number to be one that is either part of that cycle or produces a sequence that enters the cycle eventually (like 375 did).

Now, do you want to hear something really strange? All numbers that are not happy are dizzy! That’s right. No matter how big or small a number may be, if you use the sum-of-the-squares-of-the-digits RO procedure on it, you either reach a 1 or the 8-term cycle. Amazing,
isn’t it?

Now armed with this new knowledge, you are ready to classify any number as happy or dizzy.

Have fun!

For more activities about recurrent operations, go to Kaprekar or Ulam.

Update: (6/24/02)

For additional information about this interesting topic, go to Mathews: Happy Numbers.

The Math Price is Right

Perhaps some of the readers of this page will not appreciate the unique reference being made in the title of this activity to a famous TV game show, called “The Price Is Right“. If you are one of those, here is a brief description of that program, so that the math activity presented below will make sense.

On the program the contestants won nice prizes if they could guess the monetary value of the object in question: TV sets, stereos, or other valuable items. There were often three persons competing for the same prize simultaneously. Each would state his or her best estimate
of the price. The winner was the person whose estimate came the closest without going over! Simple idea, but effective. It depended highly on an individual’s number sense (a hot topic these days in the mathematics literature) and general good sense about the value of material objects
in today’s economy.

Now for the “Math” Price…

We can turn this basic idea of closest without going over into a math class activity that uses higher level thinking, calculators, and the concept of squaring a number (something very necessary when one enters algebra and advanced math). It goes like this:

The class is told that they’re going to play a game much like the TV show. They will do two things:

1. Choose a number.

2. Multiply it by itself. (This is the squaring idea.
And where the calculator comes in.)

If one’s result is the closest to some pre-set TARGET number, announced before the selection process of Step #1, then the goal has been accomplished.

Initially, only whole numbers would be used, as I am assuming that we are playing this game with say, 4th grade students. So a game may have gone something like this:


1. Bob chooses 21 and Ralph chooses 22.

2. Bob’s square number is 441 whereas Ralph’s is 484.

Hence Ralph is the winner.

NOTE: if Ralph had chosen 23, his square of 529 would have been closer than Bob’s value, but it was over 500, hence could not win.

After play has gone on for some time, and the students are becoming more adept at playing it, it is recommended to start extending the game into other dimensions. One thing that can be done while still working with whole numbers is to use the concept of the “cube” of a number. This merely means that one uses the selected number three times as a factor in the multiplication step.

For example: 1728 is the cube of 12 because

12 × 12 × 12 = 1728

Obviously, larger target numbers need to be selected now. But that’s okay; the computation is not hard due to the use of the calculator. The hard part is the thinking! (Hmmm… but that’s good, too.)

A second thing that can be tried is the use of decimals. Even at the 4th grade level this should cause no great difficulty. We are, after all, talking about money here. And most primary school students are familiar with prices such as $12.95 and the sort. Returning to the squaring version of the game, we can proceed in this way:

Let’s use Bob and Ralph again. In trying to come close to 500 again, Bob might try 22.3, whereas Ralph chooses 22.4. Now when Bob squares his number he gets “497.29“. (Very close.) But poor Ralph! His square of “501.76” went over the target this time. So, he loses. What is nice about this feature of the game is that the squares of numbers in the “tenths” are numbers in the “hundredths“, which merely resemble money amounts. It is also important for students to see a fundamental pattern here, namely,

ab.x2 = cde.yy

[The reader is to understand that my focus is on the “x” and “y” parts;

a number with one decimal place has a square with two decimal places. It’s shocking how many students don’t observe this.]

Finally, the game can be turned into a single-person activity in this way:

How close can you come to a given target number, using the squaring procedure, if you are allowed as many guesses as you wish?

This takes the idea away from its competitive setting and puts

it in a problem solving one. This brings us back to recording our investigations in our old friend, the “T-chart“. Let’s see how it might look for a target of 200.

                         n   |    n2
                       14    |  196  too low
                       15    |  225  too high
                       14.2  | 201.64  too high
                       14.1  | 198.81  too low

It is clear that 14.1 produces the winning value this time. If students can handle it, one could proceed to values of n that have 2 decimals places. The principal change here will be that the squares will have 4 decimal places, that’s all.


Not to be overlooked in this work is that we are actually preparing the student for the concept of “square root” (and “cube root”) which will be confronted in the future, concepts that need careful development prior to their formal use in higher mathematics. If some groundwork is laid in the early years, then things will go more smoothly later on.

Distinct Digit Squares


A. When a number is multiplied by itself, the resulting product is called a SQUARE NUMBER, or simply a SQUARE.

12 × 12 = 144 so 144 is a square number.
35 × 35 = 1225 so 1225 is a square number.
133 × 133 = 17,689 so 17689 is a square number.

B. Sometimes a square is made up of digits that are all different, that is, it has “no repeats”. Such a square is called a distinct-digit square (DDS).

Example: 13 × 13 = 169; there are no repeated digits in 169,
so it is a distinct-digit square.

But 21 × 21 = 441; since the 4 is repeated in 441, this is not
a distinct-digit square.


You are to use your calculator to help you make a list of ten (10) distinct-digit squares. But–one more thing–they must all contain either 5 or 6 digits. That is, they should be “5-place” or “6-place” numbers.

Largest Number Squared


If you multiply 142 by itself, what is the product?  _________
If you multiply 781 by itself, what is the product?  _________
Now look at your two answers.
The first one was a 5-place number, and the second one was a
6-place number, right?
(If not, you made a mistake somewhere.  Do the wrong one(s) again.)


You now see than when you multiply a 3-place number by itself, you might get a 5-place or a 6-place product.

Your problem is to use your calculator to find the largest 3-place number that when multiplied by itself gives just a 5-place product.

(Hint: The number is greater than 142.)


Compute these two products:

1022 × 1022 = ________
7803 × 7803 = ________

Do you see that the first product is a 7-place number, and the second one is an 8-place number? (If not, check your work as before.)

This time you are to find the largest 4-place number which when multiplied by itself will still only make a 7-place product.

(HINT: It is greater than 1022.)


Compute these two products:

17 × 17 = _______
83 × 83 = _______

Do you see that the first product is a 3-place number, and the second one is a 4-place number?

This time you are to find the largest 2-place number which when multiplied by itself will still only make a 3-place product.

(HINT: It is greater than 17.)

PROBLEM IVThe Brainbuster

You have done three problems with your calculator that were almost the same. Each time you had to find the largest number which
when multiplied by itself gave a product with an odd number of places,

Now you will be asked to do the whole thing one more time–this is the BRAINBUSTER!

Find the largest five-place number which when multiplied by itself gives only a nine-place product.

But unfortunately, this time your calculator will not be able to help you; a 9-place number is too big for the calculator’s display area.

However, things are not so bad if you will look at the answers you found for the first three problems. There is an important clue there that will tame this tough problem. Do you see it?

The largest 3-place product came from ______;
The largest 5-place product came from ______;
The largest 7-place product came from ______.

Same-Digit Pairs of DDSs


In first section you found several squares that we called DDSs. (Remember: these are squares whose digits are “all different, no repeats”.)

In this section, we will explore something interesting about certain of those DDSs. Look at these squares:

37² = 1369 and 44² = 1936

Both 1369 and 1936 are DDSs, of course. BUT, there is one more thing that is strange: they both contain the same digits, just arranged in a different order.

There are many more cases like this. Before you start the exercise below, make sure you understand this idea by finding the squares for these two numbers: 32 and 49.


In the groups of numbers below, two of them will give DDSs with the same digits, but arranged in a different order. The other numbers also produce DDSs, but do not have the same digits. Find the correct
pair in each group.

  1. 144, 175, 174
  2. 305, 153, 198
  3. 136, 228, 267, 309
  4. 233, 193, 305, 172
  5. 152, 142, 118, 179, 147

Below is given a large group of numbers that will give “same-digit pairs”, like you found above; some will not. Find the numbers that make this type of pair and put them together.

267 281 186 273 224
213 282 286 226 214

Once in a while we can find three or more DDSs that use the same digits. Look at this example:

36² = 1296     54² = 2916     96² = 9216

Do you see that all three squares contain the same digits, only in a different order. Now this is strange indeed! And it does not happen as oiften as was true for the same-digit pairs. But, as we will see, it can happen several times, if we are patient enough to look.

The following eleven numbers will produce DDSs that can be grouped into three same-digit families. Each family will have at least three members in it, maybe more. Can you separate all of them into their proper families?

181 148 154 128
209 203 269 196
191 302 178 .

So far, all of our DDSs have been only 5-place numbers. But the same thing can happen with 6-place DDSs, too. And, would you believe it? There are even more pairs and family-sized groups than you saw before.

Here are several numbers that will produce DDSs pairs or families. Can you separate them as you did before?

324 353 364 375
403 405 445 463
504 509 589 645
661 708 843 905


This piece was written by me and published in The Oregon Mathematics Teacher, Sept. 1978. At that time calculators with a 10-digit display were not the common models available to students at the elementary or middle school levels. So the “Brainbuster” problem above needs to be adjusted to take that into account, or only permit the use of 8-digit models while
doing this activity.