Category Archives: Simple Operations

Activities with Simple Operations

Pandigital Diversions

In this page, WTM will present some interesting activities similar to those that appear in another page (Digital Diversions). The difference, however, will be that here all ten digits — 0, 1, 2,…,9 — will be used. This is the reason the word “pandigital” is used in the title. (“pan” is a prefix that means “all, every one“.)

Activity #1
Our first discovery activity involves another popular “P” word: palindrome. We begin with a palindrome of 9 digits. Next, you need to double that number and carefully examine the result. If you know how to multiply by 2, do it that way. If not, just add the palindrome to itself, like so:

+   673454376

Either way is okay.

Now, look at that final result. Wow! Isn’t that strange?

Let’s try the same idea with these numbers.

1. 682454286
2. 763454367
3. 764353467
4. 862454268
5. 892151298
6. 971353179
7. 981252189
8. 982151289

Activity #2

Now, we have a unique situation this time. The three palindromes listed here are the only ones that behave in our pandigital way upon performing the doubling process twice. Yes, double each number, then double that result again.

1. 481262184
2. 672393276
3. 673454376

(Note: If you don’t like to double, then double again, what can you do instead?)

Activity #3
It is rather appropriate that for this 3rd activity in the series that our little factor should be 3 itself. So try multiplying the palindromes below by 3 to see what happens. Be careful.

1. 345282543
2. 345828543
                   3. 543282345  (note the digits)
                   4. 567828765  (note the digits)
5. 782353287
6. 984393489
7. 935131539

Activity #4
Are you ready for a new twist to our marvelous multiplying math idea? This time we’ll use two palindromes – a big one, of course, like before, but also a small one. In fact, our small one is the smallest, non-trivial palindrome of all: 11.

So, what are the pandigital products of these beautiful numbers multiplied by 11?

1. 189414981
2. 369252963
3. 567252765
4. 589010985
                  5. 765414567 (note the digits)

Activity #5
Who’s Absent?

Something interesting occurs if we change our rules of our basic game a little bit. Observe the following:

2765115672 is a nice palindrome of ten-digit size. (Remember: that’s not ten different digits.) If we multiply it by 3, we get 8295347016, which is a nice pandigital number. All ten digits are present.

But what if we had written down our palindrome hastily, omitting one of the pair of 1’s that are in the middle? It could happen, you know. We do make mistakes from time to time, whether we’d to admit it or not. What affect would this produce on our product? Let’s see.

276515672 x 3 = 829547016

Ta-dah! An interesting value, don’t you think? We have a 9-digit number composed of 9 different digits! And look closer – who’s missing? Who is absent? Why, the digit 3, of course. Now that’s truly interesting.

But is it a once-in-a-lifetime event? Well, lucky for us numerophiles, it isn’t. In fact, there’s another one nearby. This time the palindrome to be shortened by removing a central digit of 1 is 2753113572.

First, try multiplying it by 3 to see the true pandigital product. Then multiply the shorter version by 3 again. Who’s absent this time? (Hint: it’s not 3.)
And our luck continues if other numbers besides 3 are used as our factor with a reduced size palindrome. Here’s a small listing of some dual-use palindromes with their accompanying factor:

1. 12 x 87399378
 2. 12 x 147111741
 3. 12 x 251383152
4. 13 x 82533528
5. 15 x 95311359
 6. 15 x 275131572
7. 17 x 92977929
     8. 27 x 172585271 (*)

Remember: always remove the middle digit, or one of a middle pair of identical digits.
(* Special note: the palindrome here is not just any old palindrome. It also happens to be a prime number. Hence, it’s called a palprime, for short.)

Activity #6
Some Division Trivia

The number statement given in red in the previous activity provides us with a new angle to pursue. Using the idea of inverse operations, we can change the multiplication into a division like so:

276515672 x 3 = 829547016

829547016 ÷ 3 = 276515672

We might notice that the left side, the division, now consists of all ten digits, right? Hence, it is a pandigital division statement. This nice idea should prompt any good number hunter to look for more such cases.

We’re happy to report that many do exist, and they come in different categories according to the number of digits in the dividend and the divisor. With the help of our collaborator, Jean-Claude Rosa, leaving the palindromic quotients as an exercise for you, we present the following summary:

A. abcdefghi ÷ j

the smallest: 128943760 ÷ 5

the largest: 862504731 ÷ 9

number of solutions:146

B. abcdefgh ÷/ ij

the smallest: 12905376 ÷ 48

the largest  :98346501 ÷ 27

number of solutions:140

C. abcdefg ÷ hij

the smallest: 1293864 ÷ 507

the largest: 9857016 ÷ 234

number of solutions: 95

D.  abcdef ÷ ghij

the smallest: 105468 ÷ 2397

the largest: 972630 ÷ 4815

number of solutions:118

E. abcde ÷ fghij

the smallest: 26970 ÷ 13485

the largest: 98760 ÷ 12345

number of solutions: 94

Activity #7
Pandigitals and Pi

Every school boy and girl, at least in my school days long ago, learned about the famous number “pi”. Of course, the value of this fascinating number is approximately 3.14, but we were often taught about a fractional approximation of 22/7, or the mixed number version 3 1/7. This leads us to wonder if we could connect our topic of pandigitality with pi. Here is what we have come up with so far.

First, let’s examine 22/7, the so-called improper fraction form. Using all ten digits, we can show these three representations:

 49302     56034     62370
--------  --------  --------
 15687     17829     19845

(In fact, these are the only ways possible with all ten digits.)

Turning now to the mixed number form, we are faced with a slightly different situation. The whole number part, 3, is fixed and can’t be changed, whereas the fraction, 1/7, can take many forms.

This means we need to find ways to write 3 n/d, where n & d are composed of the remaining nine digits. Unfortunately, such is not possible; there do not exist two numbers of the form abcd and efghi, such that

                             abcd         1
                         3  ------- =  3 ---
                             efghi        7

Hence, we must be content with cases that use fewer digits. The only one that uses 8 digits in the fraction is

3 -------

(Who’s absent?)

The smallest case is 3 2/14, which is interesting in its own right as it is made from the first four counting numbers (a.k.a. the natural numbers). There are five other cases where the numerator is a single digit. That means, fractions of the form a/bc.

As we increase the number of distinct digits, we can state these following observations:

  • There are 15 cases where the numerator is a 2-digit number. We leave it as an exercise to you, dear reader, to find those fractions.
  • There are 3 cases where the fraction takes the form abc/def.the smallest is 104 / 728the largest is 108 / 756Again we will let you have the pleasure of finding the middle one.
  • There are 15 cases where the fraction takes the form of abc/defg.the smallest is 208 / 1456the largest is 972 / 6804Since the region to search now is considerably larger – 200 to 999 – we won’t ask you to find them, unless you really want to.

Well, what do you think about pi now?


The numerical data used in some of the activities above come from Jean-Claude Rosa, a mathematician and school math teacher from France. A summary of his work appears as WONplate 114 in the World!OfNumbers.

Others come from the work of Patrick De Geest and likewise appear in his World!OfNumbers

Digital Diversions

A Digital Diversion

Perform the following steps as indicated to see an interesting result.

  1. Form the smallest possible 4-place number using the four largest digits.
  2. Add that number to itself. The result is sum #1.
  3. Take sum #1 and add it to itself. The result is sum #2.
  4. Finally, take sum #2 and add it to itself. The result is sum #3.
  5. While all 3 sums are interesting & share a common property, sum #3 is probably the most unique of all, especially when compared with the original 4-place number.

What is that strange aspect?

Do you care for another similar trick? Then try this:

  1. Form the largest possible 4-place number using the four largest digits.
  2. Divide that number by 4.
  3. Multiply the quotient just obtained by 5.
  4. Observe the resulting product carefully.

Aren’t numbers marvelous?

Another Digital Diversion

It is a well known, and easily proven fact, that using 4 distinct digits, such as 3, 7, 2, 4, and so on, you can form 24 different 4-place numbers. The same principle would apply using 4 different letters of the alphabet, namely you could form 24 different “words”. The words need not make sense or even be pronounceable. Likewise, .you could arrange 4 persons in four chairs in a row in 24 different ways. Such an arrangement is called a permutation of the 4 items under consideration.

Now to continue in the line of reasoning of our previous digital activity…

Take the digits 2, 3, 7, & 9, and form all 24 possible 4-place numbers. It is our goal now to find which of those numbers will yield for us the remaining five digits (1, 4, 5, 6, & 8) when they are doubled.

Extra 1: While you still have your list of 24 candidates handy, try this with those that did not work out: multiply by 5 in order to get products formed by the same five digits (1, 4, 5, 6, & 8).

Extra 2: Now replace the 3 above with a 6 and repeat the above doubling & quintupling process on those 24 permutations. That is, use 2, 6, 7, & 9 to yield results containing only the digits of 1, 3, 4, 5, & 8.

A Digital Diversion a la Kaprekar

A little known piece of number trivia concerns a mathematician from India, named D. R. Kaprekar. He discovered that if you use the digits 1, 4, 6, & 7 to make the largest and smallest 4-place numbers, that the difference between them is a number that is composed of those same four digits. This is easily shown as follows:

7641 – 1467 = 6174

The number, 6174, is therefore called Kaprekar’s Constant. Much has been written about this idea in various websites*, math books & periodicals.

However, your task here is (1) to list all 24 permutations of those digits as 4-place numbers; then (2) use some other number(s) to multiply them by in order to obtain the remaining five digits (2, 3, 5, 8, & 9). [Hint: the factors you need are less than 10.]

Extra: three of the permutations will yield good results when multiplied by 12, 16, & 53.

Note: use of a spreadsheet is recommended for this problem.

[*For my website, go to Kaprekar. There are other links there as well.]

DD’s Two-by-Two

By now, you should be good at listing the 24 permutations of 4 distinct digits to form 4-place numbers. Also you should be pretty good at finding which permutation(s) yield the remaining unused digits to form the final result.

So without further ado, we will give you nine sets of 4 digits to work with. Find which permutations produce the other five digits in their products when multiplied by 8.

(If you think about it for a moment, that’s actually what you were doing in the problem task at the beginning of this collection of activities. That is, doubling something three times in sequence is equivalent to multiplying by 8: N * 2 * 2 * 2 = N * 8.]

Here are the sets. Each set will produce 2 good products. Sorta like Noah putting the animals in the ark 2-by-2, isn’t it?

            {1, 2, 3, 7}    {1, 4, 5, 9}    {1, 4, 6, 9}

            {1, 5, 6, 9}    {2, 3, 5, 9}    {3, 4, 8, 9}

            {3, 5, 8, 9}    {4, 5, 8, 9}    {4, 6, 7, 9}

Kaprekar Revisited

Above in another activity you were introduced to the set we call Kaprekar’s Digits. This time we will insert the digit 0 and call the set the augmented Kaprekar set.: {0, 1, 4, 6, 7}. You will be multiplying by numbers formed from that set: e.g. 10746, 41067, etc. As before, the set for the products will be {2, 3, 5, 8, 9}.

Now the number of permutations for 5 things is 120, which is a bit boring to list out. Therefore, we will change our approach this time. Your task will be in the form of a traditional matching quiz. This means, we will list the permutations that work in the left column and the other factor in the right column. Once a match has been made, those numbers can be set aside and considered as completed.

There is another difference this time. The second factor will not be a whole number. Instead it will be non-integral, greater than 1. An improper fraction if you prefer. This proved to be necessary due to the nature of the digits used.

So let’s go. Match them up!

permutation factor
67014 8/3
70146 6/5
67140 9/8
71460 16/7
74610 17/15
41067 9/4
14607 4/3
61470 17/4
17460 5/4
47016 4/3
41706 7/3
14076 19/4
46710 8/5
17460 5/4

Pandigitals and Fibonacci

Surely you are familiar with a famous category of numbers in the mathematical world, called the Fibonacci Numbers. If not, we recommend that you do a search of the web and you will be amazed with this fascinating topic. Suffice it to say here that they are those numbers that appear in this sequence

1, 1, 2, 3, 5, 8,13, 21, 34, 55, etc.

where each number after the first two is found by adding the 2 preceding numbers. Study that example to convince yourself that it’s true.

Now we plan to connect the idea of the Fibonacci addition procedure and pandigitals in a clever and interesting way. Here’s how:

We will start with a certain number – it may be special in its own way, like being a palindrome, a prime, etc. – then we will try to get a pandigital result sooner or later by applying the special method of addition. Watch this:

Let’s start with the palindrome 9530359. Then add 9530360, the next counting number after it. The result is 19060719. Now add 9530360 and 19060719, to obtain 28591079. Continuing like this, our next two sums are 47651798 and 76242877. When those two are finally added, — voila! – we have our pandigital of 123894675. This only took us 5 steps of hard adding. Not bad, eh?

To avoid unnecessary re-copying of these big numbers, we could condense our presentation like so:


+ 9530360











Looks rather easy now, doesn’t it? Well, we’re just getting started. Here are some more starting numbers for you to play with. Some take more steps to arrive at the “pandigital 9″, some require fewer. So be careful. One little slip and you’ll be going off in the wrong direction.

  1. 4187814
  2. 4870784
  3. 6097906
  4. 630036
  5. 6834386
  6. 4004
  7. 82466428
  8. 1993  (a prime year)
  9. 102013  (another prime)
  10. 30013  (a "lucky" number)


The activities above are based on a common theme – using the digits from 1 to 9, once and only once, to produce a special effect. It is a popular theme that has many variations, sometimes including the 0 as well. Some of the other variations are in other pages of WTM, while still others in other books and websites. For the individual who may be interested, we will give the web links here.

From the WTM:

  1. Double/Triple
  2. Nine-Digits-&-Equal-Products
  3. Ticket Number Trivia
  4. More Strange Multiplication
  5. Five Distinct Digits
  6. S.T.T.H.W.N. (Part II.2; Part IV.1)
  7. Special Numbers II (99066)
  8. FUN for Figurenuts

From the World!OfNumbers:


More About P = N2 + M3

This article originally appeared in the January 1971 issue of the Journal of Recreational Mathematics. For additional coverage of this topic, go to Trigg Numbers

In [1], C.W. Trigg discussed those positive integers less than 10,000, which are expressible in the form of N2 + M3, where N and M are also positive integers. Many of the special properties of the integers were listed such as being palindromes or squares, or having more than one representation, for example.

As I glanced over the square entries, I noted some extra relationships not covered by Mr. Trigg. When N = 6 and M = 4, not only does 62 + 43 = 100, but also (6 + 4)2 = 100. Likewise , when N = 10 and M = 5, then 102 + 53 = 225, and (10 + 5)2 = 225. While this pattern of N2 + M3 = (N + M)2 is true for only about one-third of the cases, I set out to discover what patterns, if any, existed in all of them.

To simplify some of the following, let P = A2, where P is one of the squares in Trigg’s list. Thus, in the two cases cited above, N + M = A. Of the 34 cases in the list, 12 possess this pattern. Eleven more can be expressed as the sum of N and some multiple of M, that is A = N + tM, where t = 2, 3, and 4. (See Table I for the breakdown of these 23 cases. The remaining eleven cases are not as simple, and will be treated separately.)

There are a number of patterns in Table I:

  1. For any given entry, the values for M and N are related by the general formula,
                               M(M - t2)
                        N = --------------- .
  2. The first M-value is always an integral multiple of t, therefore the values are always in arithmetic progression; and no multiple after the first one is omitted.
  3. For any given entry, the M-N values are related to the N-value in the next entry. For t = 1,the sum of the N and M values equals the following N-value. For all t > 1, the M-N sum must be diminished by an appropriate triangular number, according to the following recursion formula:
                     Nk+1 = Nk + Mk - ½ t(t - 1).
  4. For each value of t, there is always one entry where N = M. This occurs on the second entry of the section. (It is easily proved by substituting “t(t + 2)”, the formula for all second M-values, into the equation given above for pattern #1.) Also, it is interesting to note that the set of these equal values is the same as the sequence of N-values in the t = 2 section of the table!

Although Mr. Trigg restricted his discussion to positive integers, this need not be so. Other unique patterns emerge when negative integers are introduced. Table II shows the upward extension of the Table I columns so as to include negative integers and zero (the P-column being omitted).

The most obvious fact here is that the N and A columns are “inversely identical”, that is, rotate the N-column 180° about the point where all three numbers, N, M, and A, are zero, and the A-column is the result. Next, the number of negative integers in either column is always one less than the t-value for the group. When the absolute values of those negative integers are arranged in a triangular array, the result is

                            3       3
                        6       8       6
                    10      15      15      10
                15      24      27      24      15
            21      35      42      42      35      21
        28      48      60      64      60      48      28
     36      63      81      90      90      81      63      36

*Here, 0 signifies “no negative integers are present.”

It’s readily seen that the rows are symmetric, analogous to the rows in Pascal’s Triangle. The sequences of integers along the diagonals have their corresponding matches in the N and A columns of Tables I and II. The values directly under the vertex are cubic integers. Finally, when the rows are totaled and denoted St, the sequence of sums is 0, 1, 6, 20, 50, 105,…. The terms obey the 4th-degree formula

                            t2(t2 - 1)
                      St = ------------ .

One final observation before we leave these particular P-squares: The digital roots of all the P-squares, represented by P = N2 + M3 = (N + tM)2, are either 1 or 9. As all squares have digital roots of either 1, 4, 7, or 9, it seems remarkable to me that only two of the four possible values ever occur. Also, whenever t is a multiple of three, the digital roots are always 9; when t is not a multiple of three, the frequency of 1 and 9 occurs in a ratio of 1:2. (The proof of these statements is left to the reader.)

Now we return to the remaining eleven cases referred to earlier. They are represented by the formula A = N + M ± c. (See Table III for the breakdown of these cases.)

Note that c is always an even integer. In the “minus-c” group, there are three cases each for c = 2 and c = 4. While this seems to be the limit for c = 2, I have found at least two more entries for c = 4, though they yield P-squares larger than 10,000 (see Table IV). The “plus-c” group requires a different c-value for each case; and while there is only one entry for each A-formula stated here, three of them (c = 6, 10, and 12) will yield other entries from the t = 2 section of Table I (when M = 6, 10, and 12, respectively). The fact that there are many even integers missing in the c-column (with no readily discernable pattern among the given five) does not imply that solutions do not exist for them. For example, when c = 8, we obtain an entry in Table I — 576, 8, 8, 24.

The value for c can be extended to other even integers as well. For example, I’ve found six P-squares for c = 6, again all larger than 10,000 (see Table IV).

Some interesting questions, but whose answers are unknown to this writer, arose while considering these cases. Can c be odd? Does there exist a limit to the value for c? is there a limit to the number of entries a given c may have?

All these observations, based on the curious relationships cited at the beginning of this article, only serve to illustrate that often there are more patterns behind a set of data than first meets the eye. I have, on occasion, taken the liberty of extending the tabular information beyond what was given by Mr. Trigg. This was done, of course, so that late appearing patterns could manifest themselves.


1. Charles W. Trigg, “Integers Equal to N2 + M3“, JRM, Vol. 2, No. 1, January 1969, pp. 44-48.

Pi Words

Every good math student knows that pi is an irrational and transcendental real number and defined as the ratio of the circumference of a circle to its diameter. You can find all that information in any good math book.

Now, WTM is going to introduce a new idea: pi words!

Definition: a “pi word” (hereafter denoted “PW“) is simply any word that contains the letters “p” and “i“, appearing in that order. The letters may be consecutive or separated (see below). A PW may occur in any language that has those two letters.

PW’s come in various types or categories.

Type I: the letters “p” and “i” are the first two letters of the word. Any dictionary can quickly provide an ample list of such PW’s.

Here are just a few examples:

pig, pip, pint, pilot, piece, pinch, pierce, pinecone, etc.

Type II: the letter “p” is the first letter, but the “i” occurs later in the word.

Examples include:

pain, prime, panic, paint, point, pallid, paradise, promise, etc.

Type III: here the letters “p” and “i” are consecutive again, but they appear within the word.

Examples include:

opium, shaping, compile, aspire, conspire, empire, spindle, despite, etc.

Type IV: now the two letters are within the word, but they are separated by one or more other letters, somewhat as was done in Type II.

Examples include:

composite, explain, expression, assumption, hyperbolic, separating, spacing, operation, etc.

By now, you may be wondering why we did not include such examples as “pineapple” (Type I), or “probability” (Type II), or “opinion” (Type III), or “opportunity” (Type IV). But upon closer examination, the reader should observe that these latter four examples have a “flaw” of sorts contained in them. Do you see it?

We admit it’s a little hard, but notice: all the examples given in the four lists above contain only one letter “p” and one letter “i” in the
word. The four cases given just above contain more than one of either letter! So they are not to be considered as “pure PW’s”. Perhaps they
deserve their own subcategory designation, as in Type I-x (the “x” could signify “extra”).

I admit to now becoming more aware of just how many PW’s there are in English. They’re literally all over the place. Why I’ve even found a couple that start with “p” and end with “i”. (If you can find my pair, or others that are new to me, send them to me. I’ll post them on this page with your name as the discoverer.)

We have even found some interesting words that perhaps deserve their own special category. One example is:: peripatetic. Written in this form – PerIPatetIc – we see that it contains two of each letter, and in P-I-P-I order. We would like to call it a PI-PI Word. Okay?

I hear you now asking: “Where’s the math in all this?” We offer the following as our answer.

First, this activity increases student awareness of the number pi in general. And does it through the field of language arts, thus complying with NCTM Standards of making connections between math and other school subjects. Compiling long lists of PW’s of the four types increases a student’s vocabulary and spelling skills. Geography might even be included by looking for PW’s on a map. Then there are groups like Plants & Animals, Food, and so on. Let your creativity soar!

Next, once a nice list has been prepared, real math of a numerical style can begin by playing the game WORDSWORTH. Activities such as these can be pursued, depending on the mathematical level of the individual:

  1. What is the smallest (or largest) numerical value for PW’s of each type for words of the same number of letters?
  2. Can you find PW’s with specific numerical values, such as 100?
  3. Combine another important category of numbers, primes, by finding PW’s with prime values. (It is curious to note that the word “prime” has a prime value. What is it?) Or look for squares, cubes, and other powers.
  4. We can introduce a special item of advanced math notation: [n]. This is called the greatest integer function. It means the largest integer contained in the number n. In simple language, it merely means “truncate, or cut off” the decimal part of a number and keep the rest. Some examples will make this clear:
    [4.367] = 4	[72.61] = 72	[0.851] = 0 	[pi] = 3    etc.

    So now we can search for PW’s whose numerical values are

    [10pi] = 31	   [100pi] = 314          [20pi] = 62    etc.

    The first case (31) provides a distinct, perhaps impossible, challenge. The numerical value of every PW is greater than or equal to 25 (p = 16 and i = 9, so 16 + 9 = 25). Therefore, any PW with 3 letters must use the letter “f”, and “pif” is not a proper word in the dictionary. However, if we take the PW’s of pie and pig, and calculate the arithmetic mean of their values, we do get 31 after all.

    The second case (314) is quite the opposite. As 314 divided by 26 (the value for z) is in excess of 12, any PW with that value must have a lot of letters. At present, we do not know of any word that has that value. Perhaps you, dear reader, could find one and send it to us? We would then announce your great discovery on this page. On the other hand, we could possibly just change the problem, letting 2, 3, or more PW’s to be added together to make 314. Ideally, the words should have a common theme, but that’s not absolutely required.

    When we use other multiples of pi, the possibilities of success increase. And all the while, the student is becoming more familiar with an important symbol of math notation, not to mention a little more practice in multiplication.

Speaking of math symbols and notation, this would seem to be an appropriate moment to introduce some more. It will facilitate our discussion of results for a PW’s numerical value, and simultaneously teach “function” notation, a basic topic in all math courses from Algebra I and up. So we propose the following symbolism:

#(PW) = n

This simply means: the numerical value of a particular PW is equal to the integer n.

An example will make this clear.

#(pig) = 32

because P + I + G = 16 + 9 + 7 = 32.

And that’s all there is to it. Easy, huh?

[N.B. 32 = 25. If we pretend that the exponent “5” grows “fat as a pig” and then falls down to the line beside the “2“, forming the two-digit number “25“, we have the “square” of our original exponent.]

Here’s another idea that can add some variety to our topic, not to mention more results in which we can look for interesting results and patterns. Let’s merely reverse the value of our letters! This simply means:

A = 26, B = 25 C = 24, … , Y = 2, and Z = 1.

So now the numerical value of “pig” becomes P + I + G = 11 + 18 + 20 = 49.

Of course, we’ll need to adapt our function notation to indicate in which order we’re counting the letters. We’ve decided that the follow manner should suffice:

#'(PW) = n.

Notice the little apostrophe mark ” ” next to the symbol ” # “. That’s called a “prime” mark in mathematics. (I caution you, however, it has nothing to do with “prime numbers”.) Therefore, we can now write the following:

#'(pig) = 49

[N. B. 49 = 72. So while #(pig) equals a power of 2, #'(pig) equals the square of 7.]

This new notation now permits us to form algebraic statements such as

#(pig) < #'(pig)

Can you find a PW that satisfies this relationship?

#(PW) > #'(PW)

Special challenge:

While playing with this concept, I realized that once I had calculated #(PW) for a given word, there was a neat short-cut to find #'(PW) without resorting to the second alphabet counting system and doing more laborious addition work. I could develop a simple, time-saving formula to find either of those values once I had calculated the other. Can you find that formula, and send it to me?

Here’s a tip; it starts out like so:

#'(PW) = …

Now, the rest is up to you.

Another alphabetical numbering system

Here is yet another way to assign numerical values to the letters of the alphabet. Observe the following chart, and I think it will be virtually self-explanatory to you.

1 2 3 4 5 6 7 8 9

Since this is is rather a “cyclic” arrangement of the alphabet in relation to the numbers 1 to 9, we’ve decided to use the following function notation for this case:

@(PW) = n

Returning to our practice word of “pig”, we have:

@(pig) = 23

As before, we can easily prove it: P + I + G = 7 + 9 + 7 = 23.

[N.B. This time our result is the unique prime “23“, a concatenation of our first two primes, “2” and “3”. And “23” is the reversed form of our first result of “32” above.]

And putting all three systems together, we have the following mathematical statement:

@(pig) < #(pig) < #'(pig)

What this will do for the betterment of the world of mankind, I have no idea. But it’s certainly a little piece of Truth.


Finally, we may be coming to the end of the material presented on this page until we hear from you, the reading public, who will hopefully start sending me your own reactions. So we present our last item regarding Pi Words.

Two very popular parts of pi recreations are: (1) memorizing large numbers of digits of the decimal form of pi (to hundreds, even thousands of digits) or (2) writing prose passages or poetry in which the number of letters in each word correspond to successive digits in pi. The former activity is usually rather straightforward, but the latter style is, in my opinion, much more difficult to do. It even has a special name; it’s called constrained writing. When done well, it produces some very interesting results, ranging all the way from the humorous and ridiculous to the sublime and elegant. There are many websites that contain more information about these two types of activities.

One of the famous humorous phrases says:

How I want a drink, alcoholic of course, after the heavy lectures involving quantum mechanics. All of thy geometry, Herr Plank, is fairly hard…

For those who want to check each word-length, here are the necessary digits of pi to do so:


But to see a very elegant poem, go to Liz’s Pi Poems.

So what do I have to offer in this fascinating field of recreational mathematics about pi ? Let me show you.

As you have undoubtedly read all, or most, of the material above, you know that we’re talking about pi words here, right? Then, look at my first attempt at constrained writing:

3 -->  I eat pizza
1 -->  pie
4 -->  when I'm in Pittsburgh,
1 -->  Pennsylvania
5 -->  because when I'm there partaking
9 -->  of my favorite form of baked food, I promise
2 -->  to purify
6 -->  my dietary habits before I expire.

Do you see the pattern? Each line contains the number of words corresponding to the digits of pi , AND each line ends with a PW. Notice that in this example all pi words are pure PW’s.

Well, maybe it’s not the greatest gimmick for pi lore, but I like it. Can you make up one of your own, using more digits of pi ? Be careful! When a zero (0) appears, that means a break must occur. That means that the PW just before the “0-line” must be the finish of a thought, and the line that follows must be the beginning of a new sentence.

I really hope that someone with more talent than I possess at constrained writing will compose something like this and send it to me. I’d love to post it on my page.

(okay, I’m waiting…)

Email at or

Distinct Digit Fraction Sums: Revisited

On the page Distinct Digit Fraction Sums, we presented an activity wherein fractions formed by using various combinations of distinct digits had sums that were always 1. To refresh your memory, here is the introductory example again:

                      1     6
                     --- + ---  =  1
                      4     8

On this page, we intend to expand on that concept in a simple and natural way. Now we are allowing the sum of our fractions to be any number from 2 to 9.


  2     5             2     8           4     7
 --- + ---  =  2     --- + --- = 3     --- + --- = 4
  6     3             6     3           8     2

See? It’s not so difficult now, or is it?

We challenge you to find more such examples and submit them to WTM. We’ll post your creations in the charts below.

Please note: that in order for your solution to even be considered for posting, you must write “DDFS” in the subject line of your email; otherwise we will merely ignore it and delete it. Thank you. or

# Solution Name Date
1 2/6 + 5/3 = 2 WTM 9/27/03
2 2/6 + 8/3 = 3 WTM 9/27/03
3 4/8 + 7/2 = 4 WTM 9/27/03
4 4/8 + 3/2 = 2 Nicholas Kruckenberg 10/1/03
5 6/8 + 5/4 = 2 Nicholas Kruckenberg 10/1/03
6 5/2 + 6/4 = 4 Nicholas Kruckenberg 10/1/03

Distinct Digit Fraction Sums

Observe the following fraction addition carefully:

                  1     6
                 --- + ---  =  1
                  4     8

On the left side of the equation there are four distinct digits — 1, 4, 6, and 8. While that may not look like earth-shattering news to some people, I think it looks nice.
Can you make up a similar example?
This means, can you find another equation of the form

                  a     c
                 --- + ---  =  1
                  b     d

where a, b, c, and d are distinct digits?
You know, it may not be as easy as it looks. This shall be called a Type I expression.

Now, how about another variation on that theme? Observe this structure…

                  a     c       e
                 --- + ---  =  ---
                  b     d       f

where a, b, c, d, e, and f are distinct digits, and e/f < 1.

Can you find a solution to that?

This shall be called Type II.

Want to go for more? Well, then look at this.

                  a      d
                 ---- + ---  =  1
                  bc     e

where “bc” represents a “two-digit number” (like 27 or 83), and not the algebraic multiplication of 2 values.

This shall be called Type III.

Hey, I’m not done yet. Try your luck, er skill, on this one.

                  ab     e
                 ---- + ---  =  1
                  cd     f

where again “ab” and “cd” represent “two-digit numbers” (like 14 or 65), and not the algebraic multiplication of 2 values.

This shall be called Type IV.

If you can show me an answer to any of these questions, send it to me by email and I will present it here on this page in the charts below.

Please note: that in order for your solution to even be considered for posting, you must write “DDFS” in the subject line of your email; otherwise I will merely ignore it and delete it. Thank you. or

For an important UPDATE, see below Chart IV…

Type I

# Solution Name Date
1 1/4 + 6/8 Daniel Lu 4/30/01
2 1/2 + 3/6 Konstantin Knop 8/9/01
3 1/3 + 4/6 Jacqueline Hu 10/24/01
4 3/4 + 2/8 Jacqueline Hu 10/24/01
5 1/2 + 4/8 Leonard Lee 11/4/01
6 2/4 + 3/6 Leonard Lee 11/4/01

Type II

# Solution Name Date
1 1/4 + 2/8 = 3/6 Konstantin Knop 8/9/01
2 3/9 + 1/6 = 2/4 Leonard Lee 11/4/01

Type III

# Solution Name Date
1 2/10 + 4/5 Konstantin Knop 8/9/01
2 2/16 + 7/8 Jacqueline Hu 10/24/01
3 2/14 + 6/7 Jacqueline Hu 10/24/01
4 8/14 + 3/7 Jacqueline Hu 10/24/01
5 5/10 + 4/8 Leonard Lee 11/4/01
6 4/12 + 6/9 Leonard Lee 11/4/01
7 5/20 + 6/8 Leonard Lee 11/4/01
8 7/21 + 6/9 Leonard Lee 11/4/01

Type IV

# Solution Name Date
1 13/26 + 4/8 Konstantin Knop 8/9/01
2 15/30 + 2/4 Leonard Lee 11/4/01
3 15/30 + 4/8 Leonard Lee 11/4/01
4 15/60 + 3/4 Leonard Lee 11/4/01
5 19/38 + 2/4 Leonard Lee 11/4/01
6 16/48 + 2/3 Leonard Lee 11/4/01

On August 9 and 10, 2001, Konstantin Knop, from St. Petersburg, Russia, sent in some solutions to our problems posed above. But he extended the concept to include more types. And he provided solutions as well.
So we now present his extension ideas with two samples of solutions for each one. Wouldn’t you like to join him and send in a solution or two of your own?

Here is Type V.

                  ab      e
                 ----- + ---  =  1
                  cde     f

Type V

# Solution Name Date
1 34/102 + 6/9 Konstantin Knop 8/9/01
2 26/130 + 4/5 Konstantin Knop 8/9/01
3 35/140 + 6/8 Leonard Lee 11/4/01
4 72/108 + 3/9 Leonard Lee 11/4/01
5 53/106 + 2/4 Leonard Lee 11/4/01
6 78/156 + 2/4 Leonard Lee 11/4/01

Next is Type VI.

                  ab      f
                 ----- + ----  =  1
                  cde     gh

Type VI

# Solution Name Date
1 64/208 + 9/13 Konstantin Knop 8/9/01
2 85/136 + 9/24 Konstantin Knop 8/9/01

And now Type VII.

                  ab      fg
                 ----- + ----  =  1
                  cde     hi

Type VII

# Solution Name Date
1 24/136 + 70/85 Konstantin Knop 8/9/01
2 96/324 + 57/81 Konstantin Knop 8/9/01
3 45/180 + 27/36 Leonard Lee 11/4/01

This is Type VIII.

                  abc     gh
                 ----- + -----  =  1
                  def     ij


# Solution Name Date
1 148/296 + 35/70 Konstantin Knop 8/9/01
2 204/867 + 39/51 Konstantin Knop 8/9/01

This is Type IX.

                  ab      fg
                 ----- + -----  =  1
                  cde     hij

Type IX

# Solution Name Date
1 57/204 + 98/136 Konstantin Knop 8/9/01
2 59/236 + 78/104 Konstantin Knop 8/9/01

We like Type X.

                  abcd     i
                 ------ + ---  =  1
                  efgh     j

Type X

# Solution Name Date
1 1278/6390 + 4/5 Konstantin Knop 8/10/01
2 1485/2970 + 3/6 Konstantin Knop 8/10/01

August 12, 2001…

Let’s continue our patterns. Here’s another variation on Type II.

                  a     c     e
                 --- + --- + ---  =  1
                  b     d     f

Type XI

# Solution Name Date
1 1/4 + 2/8 + 3/6 Leonard Lee 11/4/01
2 3/9 + 1/6 + 2/4 Leonard Lee 11/4/01

Russian Peasant Multiplication

Egyptian method of multiplication on another page in this website, you might assume that here again we will select certain numbers from the second column to add, the result of which will be our desired product. That’s right, but which ones are they? The answer: the ones that are opposite odd numbers on the left!!

22	 35
11	 70
 5	140
 2	280
 1	560

So 70140560770, which is the correct product. (Now that wasn’t so hard, was it?)

If we use the Commutative Property of Multiplication, we can work it “the other way around”, to compare and check results.

35	 22
17	 44
 8	 88
 4	176
 2	352
 1	704

And again 770 is our product, by adding 2244, and 704.


Which way was better with this problem: 22 × 35 or 35 × 22? The first time, with the smaller number on the left (“halving”) side, definitely required fewer lines of writing. But both ways left us with only three numbers to add. So perhaps in this case, there wasn’t too much of a difference either way we go. But it would seem to be good advice in general to use the smaller factor as the one to cut in half. There would be fewer lines of halving-&-doubling in the long run.

Perhaps you are wondering why this strange process works. Here is part of an explanation. Note:

 22	  35	 (1 group of 35)
 11	  70	 (2 groups of 35)
  5	 140	 (4 groups of 35)
  2	 280	 (8 groups of 35)
  1	 560	 (16 groups of 35)

The parentheses that contain a blue number correspond to the numbers that were added to give the product, which in their turn were chosen because they were opposite an odd number in the first column. And they just happen to add up to 22, the other factor in our problem!

So now we can say: (2 × 35) + (4 × 35) + (16 × 35) = (2 + 4 + 16) × 35 = 22 × 35 = 770

Try some more instances of this marvelous procedure and you will soon be convinced of an interesting and historical aspect of mathematics.

Feedback (8/3/02)

We received the following email from Poul Wulff Pedersen, from Denmark, who wishes to add some relevant information to this topic. He wrote:

I saw it many years ago in a Schaum Outline book about Numerical Analysis – and have wondered since if any russian peasant has ever used it. But it seems to have been known in other cultures as well.Unfortunately I don’t understand your “part of an explanation” – but actually it is quite simple: If you compute 22 (in your example) in binary form you get

22 = 10110 = 24 + 22 + 21


22*35 = (24 + 22 + 21)*35 = 560 + 140 + 70

which are just the multiplies [sic.] picked out by the algorithm.

Another algorithm for multiplication is

a*b = (1/4)*((a + b)2 – (a – b)2)

where the multiplication is replaced by two squares. Squares are easy in binary and – more important – if you use table-look-up for speed you need only a one-dimensional table for squares instead of the two-dimensional multiplication of our school days (a long time ago !). The 1/4 can (with a little skill) be absorbed in the two squares so that you only need the same number of bits for a+b as for a and b. [WTM note: see Multiplication with Squares in this website.]

A third trick in binary is the Booth algorithm which replaces a string of ones by one 1 and a (-1),

11111 = 10000[-1]

so that the additions and shifts corresponding to 11111 is replaced by a shift and one subtraction. The “anti-Booth” is useful for binary subtraction by hand.

To learn more about this topic, click HERE.

Adding Fractions

A Case Approach

Everyone, or nearly everyone it seems, hates fractions, especially their addition and the other operations. Of particular note is the fact that the rules for addition and multiplication are often confused and interchanged, and the rule for division is just plain forgotten.

This article attempts to present WTM’s own approach to one aspect of this situation, the addition of just two fractions. The thesis is: that if this is thoroughly understood, then the other operations will cause less misunderstanding and frustration; and later the student will operate with “algebraic” fractions in an Algebra I or II course with
greater comprehension of the basic structures involved.

This method, here called a “case approach”, takes the a priori assumption that the student already knows (or can acquire soon) the simple fundamentals of elementary number theory, e.g. such terms as least common multiple (LCM), greatest comon factor (GCF), and relatively prime numbers. Also it is assumed that the basic concepts and terminology of fractions (numerator, denominator, reduce, equivalent fractions) are likewise understood.

The key idea in this approach is to focus our attention first on the denominators, according to the following outline:

Case 1. Same denominators
Case 2. Different denominators
a) Relatively prime denominators
b) Not relatively prime denominators
c) Special case

Case 1 is practically self-explanatory. Both denominators are the same number, hence the addition comes naturally. Few individuals have much trouble here. Examples:

                                    2       5       7
                without reduction: ---  +  ---  =  ---
                                    9       9       9

                                  1        7        8       2
                with reduction: ----  +  ----  =  ----  =  ---
                                 12       12       12       3

However, Case 2 is the situation where the most difficulties arise. Let’s examine it more carefully now.

It is a fact in number theory that when two numbers are relatively prime (i.e. their GCF is 1), that their LCM is the product of the given numbers. Example:

The LCM of 5 and 8 is 5 × 8, or 40.

Therefore, when the two numbers are then used as denominators of fractions, the lowest common denominator (LCD) is, in effect, the aforementioned LCM. So, for Case 2a, the LCD is fairly automatic. Note:

                         2       3
                        ---  +  ---  =  ----  +  ----
                         5       8       40       40

In this problem the numerators are 2×8, or 16, and 3×5, or 15, respectively, yielding

                         2       3       16       15
                        ---  +  ---  =  ----  +  ----
                         5       8       40       40

The problem has now been converted to Case-1 style, hence the answer, 31/40 is now obvious. It is interesting to point out that if in a Case-2a problem the original fractions are given in “reduced” form to start with, then the sum itself will also be in “reduced” form, or as is often said, “in lowest terms”.

The foregoing example could be written in a condensed form as follows:

                         2       3       16 + 15      31
                        ---  +  ---  =  --------- =  ----
                         5       8          40        40

This mode of presentation suggests the formula for the addition of two fractions in Case-2a:

                         a       c       ad + bc
                        ---  +  ---  =  --------- 
                         b       d         bd

[As will soon be pointed out, this formula can technically be used for ANY two fractions, though this may not necessarily be the most desirable or practical thing to do in specific instances.]

Now let’s examine the most intriguing case: Case 2b. Here the denominators are NOT relatively prime, that is, their GCF is greater than 1. This means that their LCD (alias LCM) is some value less than their product. An example will clarify this:

			 3	 5
			---  +  ---  =   ?
			 4	 6

Here the LCD is not 24, but rather 12. So, we have (step by step):

		   3	   5
	Step 1:   ---  +  ---  =  ----  +  ----
		   4	   6	   12	    12

		   3	   5	    9	    10
	Step 2:   ---  +  ---  =  ----  +  ----
		   4	   6	   12	    12

		   3	   5	    9	    10	     19
	Step 3:   ---  +  ---  =  ----  +  ----  =  ----
		   4	   6	   12	    12	     12

Although the example above usually causes no great difficulty for most individuals (due to the smallness of the denominators), in other problems the search for the appropriate LCD can produce a certain degree of anxiety. [For example, try this problem and see how much time you spend finding the LCD, then using it:

			  7        8
			----  +  ----.]
			 26	  39

This then raises the question: What about the formula? Can it still be used? As was suggested earlier, the answer is “yes.” Note:

		   3	   5	   18 + 20	 38	  19
	          ---  +  ---  =  ---------  =  ----  =  ----
		   4	   6	    4 × 6	 24	  12

The only difference now is that a reduction step will always be necessary in order to give the answer in lowest terms, as is normally required.

Now it can be stated: Case 2b can be done in either of two ways: 1) using the LCD, or 2) using the formula. There are advantages and disadvantages for both methods.

A few words are in order for the formula’s disadvantages. In this modern era, the easy access to electronic calculators make the large number aspect no problem at all. The cross products are quickly done and (if you are using the simple 4-function models with memory keys) stored in the memory, where they are even added for you as well.
Note our big-number example earlier:

		  7        8      273 + 208	   481
		----  +  ----  = -----------  =  ------
		 26	  39       26 × 39        1014

The reduction step is also not as difficult as it often used to be before calculators either. This is because the GCF of the two original denominators (13 in the example above) is usually easier to find, or at least easier (and smaller) than the LCD. Dividing both parts of the fraction 481/1014 by 13 gives its reduced form: 37/78.

So when should one use a certain method? I suggest that it’s largely a matter of personal preference. We need to show students that there are times when a problem can be done in more than one way. This should tend to relieve a lot of the math anxiety prevalent in people’s minds.

Lastly, we turn our attention to Case 2c, the Special Case. This arises when one of the given denominators is already a multiple of the other, or equivalently the former is a factor of the latter.

		 3	 7	 6	 7	 13
		---  +  ---  =  ---  +  ---  =  ----
		 4	 8	 8	 8	  8

Note that here only one fraction had to be changed to an equivalent one, thereby keeping the numbers as small as possible. This case
should be stressed more than it normally is because it’s the one that probably occurs more frequently in “real life” situations.

In this brief article, I have not pretended to be exhaustive in my presentation of all the ramifications normally encountered while doing the variety of textbook problems that students are expected to master. Nor has mention been made of a) improper-vs-mixed-number answers, or b) that with the calculator* all the fractions could be converted to decimal form and added in the memory even more easily still. Rather it has been to offer a basic structure than can be readily used to organize one’s thoughts along the lines of number theory and that the types of denominators govern what strategy must or might be taken in a particular situation.

*Footnote: The original text of this article I wrote at a time when most young students might have access only to simple 4-function models with memory keys. Now scientific models are very common. And lately there have even appeared models that “do fractions” in pure fraction form! My, how the world is changing!


An interesting fact from number theory not well known by the average student regarding LCM’s and GCF’s greatly aids the search for the LCD (LCM) in fraction work. It is that the product of the LCM and GCF of two numbers is equal to the product of the two numbers themselves. Symbolically this means:

LCM(a, b) × GCF(a, b) = a × b

This permits us to state a formula for LCD’s (LCM’s):

			                  a × b
			LCM (a, b)  =  -----------
			                GCF(a, b)

For our “big number” problem above, we can write:

                                   26  ×  39       26 × 39
                LCM(26, 39)  =  --------------  =  -------
                                 GCF (26, 39)         13

                             =  2 × 39  (or  26 × 3)

                             =  78

Super Squares

Here are four squares full of numbers. We will use them to perform a little math magic. Just follow the rules as given for each square.

  1. Choose any number in the square that you like. Draw a circle around it.
  2. Cross out all the other numbers that are in the same row as the number you chose. Also do this for all the other numbers that are in the same column. These
    numbers are now “gone”.
  3. Repeat steps 1 & 2 until all numbers are either circled or crossed out.
  4. Add the circled numbers from the square.
  5. Find the letter in the alphabet that corresponds with the sum in step 4. Let A = 1, B = 2, etc.
5 7 4 6
3 5 2 4
1 3 0 2
2 4 1 3
5 4 11 7
2 1 8 4
7 6 13 9
6 5 12 8
0 1 6 3
4 5 10 7
2 3 8 5
4 5 10 7
9 15 13 10
4 10 8 5
3 9 7 4
7 13 11 8

When this letter-finding process is done for each square, you will have spelled out a very important word in your life!!!

One note about counting out the sum to obtain the letter: when it exceeds 26, just return to the beginning of the alphabet with A = 27, B = 28, etc.

P.S. If any one wishes to learn how to construct your own Super Squares messages, contact me and I’ll explain how. It’s easy.

Almost Magic Squares

A New Twist to a Familiar Problem

The concept is deceptively simple: fix one of the entries in the square so that it is “incorrect”. The object of the exercise then is to find the “culprit”; determine by how much it should be altered to bring things back into balance; and make the correction.

Here is a simple example that can be used to demonstrate the idea to your class. Note that the second row and third column each have sums that are one greater than the other rows and columns. (This is why I call such squares almost magic.)

11 4 9 24
6 8 11 25
7 12 5 24
24 24 25 11

Now, if the entry that is common to both sets (11)is decreased by one, the magic property of the square is easily restored.

To construct a variety of almost-magic squares, follow these steps:

  1. Select a magic square of the size you desire.
  2. Add one value to all the entries, except one; add a different value to that final entry.

Before performing Step 2, it is often helpful to multiply all the square’s entries by a particular value. This increases the variety of possible problems, and more importantly, allows you, as the teacher, to adjust the difficulty level of the addends.

The following are some examples that I have used in my teaching of elementary school students. There is sufficient drill work in any one square and each provides a needed experience in problem solving. These exercises would make good worksheets to leave for a day when a substitute is needed.

4 14 12
19 10 2
8 6 16
8 18 5
6 10 14
16 2 12
6 21 18
27 15 3
12 10 24
7 22 19
28 17 4
13 10 25
9 29 25
37 21 6
17 13 33
8 28 24
36 20 4
16 12 33
11 38 31
46 26 6
21 16 41
35 7 27
15 23 31
21 39 11
16 2 3 13
5 13 10 8
9 7 6 12
4 14 15 1
32 4 6 26
10 22 20 16
18 14 12 24
8 28 32 2
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 21 2 9
19 5 6 16
8 14 13 11
12 10 9 15
7 20 18 4
17 3 4 14
6 12 11 9
10 8 7 19
5 15 16 2
48 6 9 39
15 33 20 24
27 21 18 36
12 42 45 3
52 32 48 4
12 40 24 60
8 44 21 56
64 20 36 16

This article appeared in The Oregon Mathematics Teacher, October 1978.