Tag Archives: Division

Guidelines for Writing Math Solutions

Writing a math answer for a Problem of the Week is very different from writing an essay in English class or a term paper in History class, so we would like to give you some guidelines. You write only one document, but we receive sometimes as many as 300-400 (or more) answers per week to read and analyze; when your presentation style is at its best, much time can be saved, a more efficient service can be provided, and everybody will be happier.


The first thing that would speed up the evaluation process can be called readability. Sometimes an individual sends us an answer in one long continuous paragraph, with equations embedded in it. Such paragraphs are very hard to read.

The solution is simple: just break the long paragraph up into several short ones, each one with its own concept, and leave a blank line between paragraphs.

Another matter regarding readability concerns polynomial expressions and equations. Notice the difference between these items:

    EASY TO READ                      HARDER TO READ
    x^2 + 2x + 1                      x^2+2x+1
    x^3 + 4x^2 - 6x + 10              x^3+4x^2—6x+10
    (3a + 4b)(3a - 4b)                (3a+4b)(3a-4b)

See how a space on either side of a plus or a minus sign makes the reading easier? (This is what good textbooks do.)

Similarly, when you show the steps in solving equations, add spaces and align the equals signs, like this (when you align text, never use tabs!):

    EASY TO READ                      HARDER TO READ
    2x + 48 = 58                      2x+48=58
         2x = 10                      x=5
          x =  5

Getting Off To a Good Start

After carefully reading a problem, it is essential to determine just what you need to find to answer the question posed. You should now select a letter, or letters, that will represent your unknown quantity, or quantities. This is the famous “Let” statement. Then, and only then, are you ready to begin forming your expressions or equations.

Be careful here, however. Many times “Let” statements aren’t clear. Examples:

GOOD ==> Let x = the number of apples in the basket

BAD ==> Let x = apples

In the latter case it’s not clear if we’re counting apples, or weighing pounds of apples. So be as precise as possible. It will save troubles later in your solving process.

Use of Guess-and-Check Procedure

In general, the method of “guess-and-check” is not allowed in AlgPoW as your primary strategy to solve the problem. This is not saying guess-and-check is not a good way to solve problems. In fact, it is often a good way to start to understand a problem, and therefore recommended for that. But for most of our problems, you must define variables or unknowns, then form equations to solve by logical steps.

Historically, it was the main way that problems were solved. But as advances were made in symbolic notation, mathematicians moved away from it and toward the more efficient and time-saving methods of step-by-step manipulations on equations.

One of our mentors advises students in the following way:

Guess and check is a valid problem-solving approach. However, it also one of the most difficult to explain. If you are going to use guess and check, you must list every guess, along with the reason that you know the answer is incorrect. You also must explain why you know your final answer is the only possible answer. In all, a pretty long process; however, since this is the Algebra Problem of the Week, you might want to try algebra. Please read the “Guidelines for Writing POW Answers.” The link is at the bottom of the problem.
So, unless otherwise indicated, please do not use guess-and-check as your principal solving procedure.

Writing a Complete Answer

The Problem of the Week (PoW) project here at the Math Forum, as you probably know, is a very unique one. Unlike other math tests in school or competitions (such as SAT), here we are not only interested in the right answer, but also how you arrived at it. This means, you must show your procedure and steps and thinking along with the final answer.

Even more so, your presentation must be explained well as you go from start to finish. Just imagine if you were to show your solution to a friend who was unfamiliar with the problem. Would that friend be able to read it and understand what you were saying?

ElemPoW has its own Guidelines document such as this one. Here is what is said there:
One good way to make sure you include enough information in your solution is to pretend you are explaining the problem to a friend who does not know anything about it. Imagine yourself leading your friend on a tour of your thinking as you solved the problem. How did you start? Where did you find the information you used? What were your calculations? How did you check your solution?

Math steps without a math explanation in words is much like watching a talented magician on stage. You see all the moves go by rapidly and you are “amazed”, but still you are left with the question, “How’d he do that?” Problem solving in PoW is not magic. Our goal is for everybody to understand as much as possible, according to his/her capacity.

Again, a thought from the ElemPoW service:

Our focus here at the Math Forum is not only on getting the correct answer, but also on communicating the steps involved in finding the correct answer.

To see the entire ElemPoW document about writing good answers, consult this page: How do you write a good math solution? There is much good advice to be found there.

E-Mail Notation

Sometimes we cannot write certain symbols (like exponents or square roots) in e-mail as we do using paper and pencil. Here are some examples:


    It is standard now in e-mail to use the ^ (caret sign) found above the 6 on the keyboard for exponents. If we wish to say ‘four squared’, we write 4^2. For higher powers we do the same: ‘The volume of a cube is e-cubed’ would be V = e^3.

Square roots

    •                __                 _____
                    V64       or      \/a + b
  • Some people use the notation popular in spreadsheet applications, e.g. sqrt(16), to mean ‘the square root of 16′. This even applies in formulas; for the Pythagorean theorem, we can write:


    c = sqrt(a^2 + b^2).

    Other students ‘draw’ a square root symbol this way:

    [A few people try decimal or fractional exponents: 64^0.5 or 64^(1/2),
    but depending on the font this method can be difficult to read, so it is not recommended. However, there are occasions in which such exponents are better.]


    •  1               15               3a  +  4b
      ---             -----            ------------
       2               25               5c  -  6d
      Two-fourths  2/4     five-sixths  5/6     etc.
      (3a + 4b)/(5c - 6d)
      Vertical: --- x      Horizontal: (2/3)x
      Vertical: ----       Horizontal: 2/(3x)
  • Writing fractions is more complicated. There are two basic styles: vertical (sometimes called ‘stacked’) fractions, and horizontal fractions. Vertical fractions are what we are used to writing with pencil and paper, and are what you see in books. We can make them in e-mail as well; it just takes more effort and more keystrokes. But they are more readable when we need to write algebraic fractions.

    Horizontal fractions consisting only of numerals are easy to write, as these examples show:

    Even fractions that contain binomials, as shown above, can be written horizontally, if you employ parentheses. Observe:

    The difficulty arises when you need to express something like ‘two-thirds of x’. If you write this as 2/3x, it could be misinterpreted as 2 over 3x. Luckily we have ways of clarifying our meaning:

    Now if your intention really was 2 over 3x, you still have two options:


    • a1, a2, a3, a4, …
           y1 - y2
      m = ---------   instead of  m = (y_1 - y_2)/(x_1 - x_2)
           x1 - x2
      m = (y1 - y2)/(x1 - x2)
  • Unlike exponents, which go above the line (that’s why they’re sometimes called ‘superscripts’), subscripts go below the line. Unfortunately, the standard keyboard doesn’t have a true subscript key. Some people write a_1 for ‘a-sub-one’, but since many cases that need subscripts occur in sequences, we could write the following:

    to stand for a sequence of terms (a-sub-one, a-sub-two, …). In this context there is no real confusion with multiplying ‘a’ by 4. We universally write that as 4a.

    Notice how nice the slope formula can look using vertical fractions with this subscript style:

    The vertical equation looks almost like a line from a textbook, but even a horizontal equation like this one would be preferable:

Quadratic Formula

The quadratic formula is often needed in algebra problems. Here are two good ways to write it in email answers:

x = (-b +/- sqrt(b^2 - 4ac))/(2a)
                            -b +/- sqrt(b^2 - 4ac)
                       x = -----------------------


    When you are using determinants to solve a system of equation by Cramer’s
    Rule, they may be nicely formed as shown here:
        |   3      5  |
    D = |             | = (3)(6) - (-1)(5) = 18 - (-5) = 18 + 5 = 23
        |  -1      6  |
    For a 3-by-3 case, the same idea applies:
        | a    b    c |
        |             |
    D = | d    e    f | = aei + dhc + gbf - gec - dbi - ahf
        |             |
        | g    h    i |


The method you use will often depend on the needs of the specific problem you are working; these comments should be understood as suggestions and general guidelines only.



Calculating Expressions

One evening Elisa was doing her math homework when Dina came by to visit. “The exercise I have to do,” said Elisa, “is this one.”

Convert this expression to its calculator key-sequence form, then use your calculator to find its value.

“So far I have this much done,” she added.

12 [×] 5 [-] 8 [÷] 4 [+] 7 [×] 2 [=]

“That looks fine to me,” said Dina. “Let’s evaluate it now.”

Each girl took her own calculator and confidently entered the numbers and symbols as Elisa had given them. But their smiles quickly turned to a puzzled look on their faces when they realized that different results came up.

“How can that be?” they asked almost simultaneously. “Let’s do it again. Maybe we pressed a key incorrectly.”

Once again they entered the expression, only to have the first results to be confirmed. “Ah, I think I know the trouble,” said Dina. “Let me see your calculator.”

Elisa showed her this:

Dina then said, “Here’s mine.”

You see, it’s obvious now. They were using different kinds of instruments. So your task for this POW is to state what value each girl obtained and explain why it happened, based on the type of calculators being used.

Calculating Expressions

Number Sequences

[Preface NOTE: The material presented below was lifted from an article I saw some time ago by Dan Brutlag, “Making Your Own Rules”. THE MATHEMATICS TEACHER, November 1990. pp. 608-611. What is being given below is how I prepared a handout to give to my classes, mostly as anactivity for use by a substitute.]


Students: I found the information below in a math magazine. I think you will find it interesting, as I did.


  1. Study/investigate the results of applying the rules below to many different numbers. Keep good records to see if something “special” always happens. Describe what you find.
  2. Then make up a set of rules of your own to investigate, similar to or completely different from the samples given. Use the table of ideas that you find at the end of this handout to help you with this. Write up a little report about what you discover.

A. Lori’s Rule

  1. Start with any three-digit number.
  2. To get the hundreds digit of the next number in the sequence, take the starting number’s hundreds digit and double it. If the double is more than 9, then add the double’s digits together to get a one-digit number.
  3. Do the same thing to the tens and units digits of the starting number to obtain the tens and units digits of the new number.
  4. Repeat Steps 2 and 3 as often as necessary to find the special happening.

Example: 567, 135, 261, 432, ???, ???, …

B. Barbette’s Rule

  1. Start with any three-digit number.
  2. Add all the digits together and multiply by 2 to get the hundreds
    and tens digits of the next number in the sequence.
  3. To get the units digit of the next number, take the starting
    number’s tens and units digits and add them. If the sum is more than 9, add the digits of that sum to get a one-digit number for the units place.
  4. Repeat Steps 2 and 3 as often as necessary to find the special happening.

Example: 563, 289, 388, 387, ???, ???, …

C. Ramona’s Rule

  1. Start with any three-digit number.
  2. Obtain the next number in the sequence by moving the hundreds digit of the starting number ito the tens place of the next number, the tens digit of the original number into the units place of the next number, and the units digit into the hundreds place.
  3. Then add 2 to the units digit of the new number; however, if the sum is greater than 9, use only the units digit of the sum
  4. Repeat Steps 2 and 3 as often as necessary to find the special happening.

Example: 324, 434, 445, 546, ???, ???, …

D. Lisa’s Rule

  1. Start with any three-digit number.
  2. If the number is a multiple of 3, the divide it by 3 to get a new number.
  3. If it is not a multiple of 3 then get a new number by squaring the sum of the digits of the number.
  4. Repeat Steps 2 and 3 as often as necessary to find the special happening.Example: 315, 105, 35, 64, ???, ???, …

    Example: 723, 241, 49, 169, 256, ???, ???, …,


    Add ___ to ___ Divisible by ___
    Move to another position Hundreds, tens, or units place
    Take positive difference Prime
    Double Composite
    Square Even, Odd
    Halve (if even) Multiple of ___
    Multiply by ___ Greater than
    Replace by ___ Less than
    Exchange Equal to

    Postscript (7/29/99)

    Here is a personal sidelight to this activity. On April 23, 1991, I designed my own sequence rule-set. Here it is.


    1. Start with a four-digit number.
    2. To get the thousand’s digit and hundred’s digit of the next number, double the sum of the first three digits of the original number, that is, the thousand’s, hundred’s, and ten’s digits. (If this result is only a one-digit number, append a 0 to the left of that number, so a “6” would become “06”.)
    3. To get the ten’s and unit’s digit of the next number, double the sum of the last three digits of the original number, that is, the hundred’s, ten’s, and unit’s digits. (If this result is only a one-digit number, do as was done in Step #2.)

Niven Numbers

The topic of Niven Numbers is briefly mentioned on another page in this website. But since writing that, I have been working on the concept a little bit more with some students and I now feel that it deserves its own separate page. So what follows here is a discussion about how it can be utilized in the regular school math class.

We begin by defining once again the concept of what a Niven Number is:

A Niven Number is any whole number that is divisible by the sum of its digits.

The only real trouble with this is the meaning of the word divisible. In number theory it simply means “division with no remainder“. Another way to express that is to say that the divisor in the division process is also a factor of the dividend.

An example or two should make all that quite clear. Let’s take the number 126. First, we must find the sum of the digits.

1 + 2 + 6 = 9

Next, we try to divide 126 by 9. And we find…

126 ÷ 9 = 14

There it is. 14 and no remainder; or put another way: 9 is a factor of 126 because

126 = 9 × 14

So 126 is a NIVEN NUMBER.

On the other side of the coin 121 is not a Niven Number. Its digit sum of 4 is not a divisor (or factor) of the number itself. 121 divided by 4 yields a quotient of 30 and a remainder of 1.

Doesn’t sound too hard now, does it? Well, it all depends. For young students this has proved to be sufficiently challenging, mostly because it is all so different from the usual math fare that is served up to them by their textbooks and teachers.

For example, I have posed the following simple question to students:

How many Niven Numbers can you find from 100 to 200?

It has been interesting to observe their initial reactions. Mainly they don’t quite know where to begin, what numbers to pick, and so on. They seem to be overly concerned about being unable to state the number of Niven Numbers in that range in a short amount of time, rather than realizing that it will take some time and patience to arrive at it. Next, they start out in an unorganized way, choosing numbers at random to test, rather than “beginning small and working up”.

So, after allowing them to struggle with the whole matter for a while, I make the suggestion, “Why not make a list of all the numbers from 100 to 200, inclusive; then start at the 100 and work your way up? We can circle the Niven Numbers, and cross off those that are not.” This shows a good problem solving strategy that many had never thought of. And at the same time, it allows us to review the divisibility tests that the students had studied earlier, but for lack of use in any meaningful context, had forgotten.

This organized strategy also has a unique advantage. One begins the find some easy Niven Numbers rather early: 100, 102,
108, 110, 111, and 112. Nothing breeds confidence and understanding better than some early success.

Furthermore, in this range of 13 numbers (100-112) there are several good opportunities for discussing another number property that aids our search. You see, there is no real need to carry out the long division and find the quotient IF the main objective is to determine if a number is Niven or not. If the number in question is odd and the digit sum is even, it is clear that the number can be crossed off the list. For example, 101. It is odd, but its digit sum is even. No odd number can be divided by an even number without leaving a remainder of some sort. Or to put it another

No odd number is the product of an even number and something else.

So numbers like 101, 103, 105 and 107 can be eliminated rather easily. Another simple case arises when numbers have a digit sum of 10 and the number itself does not end with zero, as is the case for 109. Finally, in this limited range that we are considering, 104 is of interest. It has a digit sum of 5, yet the number does not end with a zero or 5. Basic and obvious as these points may seem to some, they still need to be brought into the discussion.

So with all these powerful ideas and strategies working for us now, just how many number from 100 to 200 are indeed Niven? We intend to leave that up to you to find. It’s not nice to spoil your “thrill of victory and agony of defeat”, right?

Where next?, you might be asking yourself now. Well, several ideas come to my mind; perhaps you could dream up some of your own. How about these two?

  1. What is the largest 3-digit Niven Number? (Also, the 2nd largest and 3rd largest ones?)
  2. What is the smallest Niven Number formed by the digits “1”, “6”, and as many zeros as needed?

Of course, we should mention the natural extension of the problem presented above: How many Niven Numbers are there in each “CENTURY” from 0 to 999? That is to say: 0-99, 100-199, 200-299, …, 900-999. The point here is to investigate these groups of one hundred numbers to see if there are any interesting patterns. Also to determine which century has the most (or the fewest) Niven Numbers. [Perhaps a bar graph could be the final presentation format?]

This would not be as tough a job as it might first appear if calculators, or even computers, were utilized. After all, it is the thinking, the mathematical investigation that is important here, not one’s ability at looonngg division. And don’t worry if your calculator doesn’t give division answers in “quotient-remainder” style. [Nowadays some do have that feature.] Any ordinary model will suffice, because of this basic fact:

If the division would produce a remainder in the traditional sense, then the calculator will give a “decimal” result instead.

Using our example earlier of 121, we would have this:

121 ÷ 4 = 30.25

And that is sufficient to tell us that there had to be a remainder of some sort, so 121 is therefore not Niven. (For the teacher and student willing to delve deeper into the matter, a discussion could be held about the relationships between ordinary remainders and the decimal part of the result. Thus making this project into a worthwhile learning experience.)

It’s now time to “get Niven”, and find those NN’s out there.

Good luck, and let me me know what you find.


After writing the above article, I received an e-mail from the individual who first introduced me to this topic, a math professor in Colorado named Monte Zerger. Here I present selected portions of what he told me:

“You got me thinking about Niven numbers again, so I
did a little quick research. According to an article in the
Journal of Recreational Mathematics the origin of the name
is as follows. In 1977, Ivan Niven, a famous number theorist
presented a talk at a conference in which he mentioned
integers which are twice the sum of their digits. Then in
an article by Kennedy appearing in 1982, and in honor of
Niven, he christened numbers which are divisible by their
digital sum “Niven numbers.”

In most all the literature I have seen this is the accepted
name. However, I note that in his “Dictionary of Curious and
Interesting Numbers” Wells calls them “Harshad numbers.” …

In an article appearing in Journal of Recreational Mathematics in 1997, Sandro Boscaro defines and discusses Niven-morphic numbers. These are Niven numbers which terminate in their digital sum. For example 912 is such a number since it has a digital sum of 12, is divisible by 12, and ends in 12. In this article he proves a startling result. There is no Niven-morphic number which has a digital sum of 11, is divisible by 11 and ends in 11. 11 is the *only* such exception.

Of course 1, 2, 3, …,9 are all trivially Nivenmorphic. I played around this morning with a simple BASIC program and found the smallest (I think!) Nivenmorphics for 10, 12, 13, …, 19. They are [answers withheld to give you the chance to find them. Sorry, that’s the way things are in the World of Trotter Math. :=) ]”


Later our correspondence grew a bit more, and the following curious trivia resulted:

  1. In my investigation for NNs under 1000 in order to understand what would be involved for the “century” question above, I found a string of 4 consecutive ones: 510, 511, 512, 513. And that was the longest string in that range.
  2. To which Monte replied, “My computer program found three other strings of four terms: 1014-1017, 2022-2025, and 3030-3033. I still like the string starting at 510 the best since 510 is the Dewey Decimal classification for mathematics and “cloning” it we have the product of the first 7 prime numbers as well as the product of the 7th through 10th Fibonacci numbers.”
  3. And Monte’s BASIC program uncovered the following item: “The smallest string of five consecutive Nivens begins with 131,052. I checked all the way up to 10,000,000 and could not find a string of six.”

    (Well, I think that’s going high enough for most purposes with young students in school. Perhaps a worthwhile activity would be to ask them to confirm that the numbers mentioned above are indeed Niven Numbers.)

    Then later Monte came through with a “really big” piece of information: “Ran across something today that will interest you. It has been proven there cannot be a string of more than 20 consecutive Nivens. A string of 20 has been found. Each member has 44,363,342,786 digits.”

  4. I also thought of the reverse side of things… How about the idea of strings of non-NNs? Between what two NNs can we find great numbers of non-NNs? In my limited, yet accessible-to-young-students list, I found these 17-term “droughts”: 558-576, 666-684, 736-756, 846-864, & 972-990. No Nivens here!
  5. Here’s more from the mind of Monte: “I suppose you noticed that whereas the current decade had only one Niven year (1998), the next decade has four! 2000, 2001, 2004 and 2007.”

[Stay tuned, dear reader. There’s no telling where this is going to end!]

Well, it took awhile, but there is some new action about Niven Numbers. On May 17, 2001, I received the following email:

I just noted your page on “Niven Numbers” and enjoyed it…………………I’m still very much interested in such numbers. Ivan Niven died last year but at least his numbers will help people to remember him. By the way, have you investigated Niven numbers in differenct bases? Have you considered which powers of two are Niven?


Robert E. Kennedy

Ok! Different bases and powers of two… hmm… Looks like we have
more work to do.

Egyptian Math

In this page, WTM is going to travel back in time and do some really “old-fashioned” mathematics. Specifically we will look at how the ancient Egyptians multiplied quantities and dealt with fractions. To make things more convenient for our discussion, we will use our own numbering system and symbols. This is a piece of good news/bad news. First, it facilitates our presentation: you will understand the numbers right away and I can write things with my computer keyboard. However, such does not how “what it really looked like to the ancient Egyptian’s eyes and mind” and we can’t truly appreciate why he did things that way (which seems more difficult to us with our modern notation). But it’s the best I can do for now.

With that preface, step into our time travel machine, strap on your seat belts, and let’s go!

We will begin with the idea of multiplication, as it is the easier of the two topics. Let’s say we wish to multiply 13 by 28. We write things this way:

			1	 28
			2	 56
			4	112
			8	224

Notice that in the two columns each number is the double of the number above it (that is, after the first row). That’s the first step — just multiply by 2! Everybody should know their “2 times table”, right?

Next, we make use of a very important fact from number theory: Every positive integer (number) can be uniquely expressed as a sum of powers of 2. Our first column consists of just that, the powers of 2. So from among those numbers we should be able to find some that add up
to 13. That would be:

			1	 28
			2	 56
			4	112
		        8	224

Now we will highlight the numbers in the second column that correspond to 1, 4, and 8, then add them:

			1	 28
			2	 56
			4	112
		        8	224
		       13       364

Therefore, we can say

13 × 28 = 364

That this is not a cute parlor trick that only works in special situations can be shown by using the famous Distributive Property of Multiplication over Addition. Note:

	13 × 28  =  (1 + 4 + 8) × 28
		 =  (1 × 28) + (4 × 28) + (8 × 28)
		 =     28    +    112   +    224
		 =  364

By way of reviewing another famous number property, the Commutative Property of Multiplication, we will do it all over again, reversing the factors to 28 × 13.

			 1	 13
			 2	 26
			 4	 52
			 8	104
			16	208
			28	364

You see? It works, just like it’s supposed to. Now practice this idea with some problems of your own choosing. Some working advice: instead of using color to highlight the numbers you need to add, you could mark then with a star(*), an “x”, or some other convenient mark; or you could cross out the ones you DO NOT need. Do whatever seems helpful to you.

How about some fractions now, Egyptian style. (Get ready for something really different here.)

One thing you must know right off: the Egyptians didn’t write their fractions like we do, even allowing for the different way that they symbolized their numbers. This means they didn’t write things like

		 1     5     4     8         12
		---,  ---,  ---,  ---,  or  ----.
		 2     6     7     9         17

No sir, no numerator-over-denominator for them. In fact, they had no numerator at all. Well, that’s not quite true; they did have a numerator of sorts: 1. That’s it, just 1. So all their fractions are what we nowadays call by the name unit fractions, because 1 means “unit”. Some “unit fractions”, therefore, are:

		 1     1      1      1          1
		---,  ---,  ----,  ----,  or  -----.
		 3     8     14     26         150

Now, as I said, they didn’t write any numerator-over-denominator, you should be wondering now just how did they express their fractions.
It’s quite simple, actually. They wrote the denominator, then placed an oval or “mouth” above it. If I recall my Egyptian numbers, “4” was
made with four sticks, like “||||”. So, one fourth (1/4) would be

Sure, it looks strange to us to call that a fraction. But it worked for their time. (Actually, I kinda like it.) And remember, they didn’t know about things like calculus or computers, or sending spaceships to the moon. As their day-to-day needs were somewhat simpler than ours, their mathematics could certainly be written differently as well.

That still doesn’t account for non-unit fractions, though, does it? True enough, and here’s where the real fun begins. If they wanted to use a fraction such as 3/4, they would express it as the sum of two unit fractions. In our modern notation, we would write it like this:

			 3       1       1
			---  =  ---  +  ---
			 4       2       4

Now that certainly does the job, doesn’t it? This presents us with an interesting problem: given any particular non-unit fraction, how might an Egyption student of that time express it as the sum of two (or more) unit fractions? Let’s take 2/7 as a example.

You might think that should be rather easy. Just do this:

			 1	 1	 2
			---  +  ---  =  ---
			 7	 7	 7

To which I must say, “Sorry, Charlie!” Those Egyptians had another strange rule to follow: there could be no repeated fractions in the expression. To be a properly written Egyptian fraction, each denominator must be distinct (different). So in our example, we need to write

                         1        1       2
			---  +  ----  =  ---
			 4       28	  7

See? We have two different unit fractions and still have the sum of 2/7. That’s the Egyptian way. (Nowhere did I say it was going to be easier, just interesting.)

But how does one get from the 2/7 fraction to the unit fraction form? The Egyptians had their way, perhaps rather complicated to our way of thinking, and they also used prepared tables with many simple cases already worked out in advance. What I will show here are two ways that can be done more easily with our modern notation. They will give you some excellent practice in the concept of fractions and factors, while at the same time learning something new.

Method I:

	Step 1:  Divide the denominator by the numerator, old-
		 fashioned style.

		7 ÷ 2 = 3 r 1  or 3½  or 3.5

	Step 2:  Round UP the answer to Step 1.  It is 4.
		 That will be the denominator of our first
		 unit fraction: 1/4.

	Step 3:  Our situation looks like this now:

			 2	 1	 1
			---  =  ---  +  ---
			 7	 4	 ?

		 So to find the fraction "1/?", we need to

			 2	 1	 1
			---  -  ---  =  ---
			 7	 4	 ?

			  8	 7	 1
			---- - ----  =  ---
			 28	28	 ?

				 1	 1
			       ----  =  ---
				28	 ?

It appears we were lucky this time, as the result of Step 3 gave us a unit fraction right away: 1/28. So we have done it! 2/7 = 1/4 + 1/28. Case closed. If our difference had not been a unit fraction after Step 3, we would have merely repeated the entire procedure with the difference(s) obtained until we had all unit fractions. [You see, it is another fact from number theory that: Every proper fraction has an Egyptian representation as a sum of distinct unit fractions. (Hurd, p. 593)] This is a simplification of a method due to Fibonacci, a famous mathematician of the 13th century.

To see if you have caught on to this idea, try Fibonacci’s process on the fraction 3/7. If done correctly, you should obtain 1/3 + 1/11 + 1/231 as your result.

Now, to extend this idea just a bit, look again at the result for 2/7. If we multiply all the fractions in the representation by 2, we have this interesting result:

			 4	 2	  2
			---  =  ---  +  ----
			 7	 4	 28

			 4	 1	  1
			---  =  ---  +  ----
			 7	 2	 14

See? We have an Egyptian fraction form for 4/7. With very little extra effort on our part. Sort of like something “new” from something “old”. This doesn’t work all the time, but sometimes it does.

Method II

Many proper fractions have more than one way to be represented. For our example above of 3/7, there is another way:

			 3	 1	 1	  1
			---  =  ---  +  ---  +  ----
			 7	 4	 6	 84

This is somewhat nicer that the other style, as the denominator values are smaller. To get this result, we must first utilize the concept of equivalent fractions. For 3/7 some equivalent fractions are

		  6	   9	   12	    15	     18
		----  =  ----  =  ----  =  ----  =  ----  ...
		 14	  21	   28	    35       42

As you might guess from looking at the solution, we need the fraction whose denominator is 84, obtained by multiplying both parts of the fraction by 12.


We now “decompose” the number 36 into three numbers, all of which are factors of 84. So next we need the factors of 84. I like the T-chart listing; it helps me to “keep things straight”.

			       1  |  84
			       2  |  42
			       3  |  28
			       4  |  21
			       6  |  14
			       7  |  12

The highlighted numbers in the chart have a sum of 36, so…

			 36	  21 + 14 + 1
			----  =  -------------
			 84	      84

			 36	  21       14	     1
			----  =  ----  +  ----  +  ----
			 84	  84	   84	    84

			  3	  1	  1	   1
			 ---  =  ---  +  ---  +  ----
			  7	  4	  6	  84

[A little review problem is in order here since all our denominators are even numbers; can you show me an Egyptian expression for 6/7?] Return to our T-chart for 84 once again. The highlighted numbers show another way to make 36.

			       1  |  84
			       2  |  42
			       3  |  28
			       4  |  21
			       6  |  14
			       7  |  12

So that should provide us with yet another way to express 3/7:

			 36	  28 +  7 + 1
			----  =  -------------
			 84	      84

			 36	  28        7	     1
			----  =  ----  +  ----  +  ----
			 84	  84	   84	    84

			  3	  1	   1	    1
			 ---  =  ---  +  ----  +  ----
			  7	  3	  12	   84

Now how about a “really big show”? Observe: the red numbers make a six term way to make 36, right?

			       1  |  84
			       2  |  42
			       3  |  28
			       4  |  21
			       6  |  14
			       7  |  12

This allows us to make a really big expression for our fraction.

	 36	  14 + 12 + 4 + 3 + 2 + 1
	----  =  -------------------------
	 84	            84

	 36	  14       12	     4	      3	       2	1
	----  =  ----  +  ----  +  ----  +  ----  +  ----  +  ----
	 84	  84	   84	    84	     84	      84       84

	 3	  1	  1	  1        1        1        1
	---   =  ---  +  ---  +  ----  +  ----  +  ----  +  ----
	 7	  6	  7	  21       28       42       84

Here’s a homework problem for you. Can you find a way to form 3/7 with 4 unit fractions? With 5 unit fractions? With 7 of them?

[NOTE: Hurd called this method the “convenient denominator” way. This means that a multiple of the original denominator was sought that would “conveniently” possess some factors that could be used to form a sum equalling the numerator. Actually, it’s pretty easy to do, don’t you agree now?]

Closing Notes

While writing this page, I was reminded of a fact of modern technology: the hand-held scientific calculators which so many students have contain a “reciprocal” key. It looks like this on most models:

Now, doesn’t that look a little familiar to you? Like a unit fraction used in our work above. This suggests that we could also verify our solutions using that key. First, find the decimal form of the original fraction, such as 3/7; it is 0.428571428 (in a 10-digit display model). Then taking our first solution for this fraction (1/3 + 1/11 + 1/231), we can use this key sequence to check it out:

3 [1/x] [+] 11 [1/x] [+] 231 [1/x] [=]

Since the display shows the same result, we can be reasonably confident that our work was correct. Wouldn’t an ancient Egyptian scribe be impressed by that! How fast it is these days!

Here is a little item, not strictly related to the topic of this page, but found in one of the websites below. To see a pretty result, use the reciprocal key of your calculator to find the sum of the reciprocals of these numbers:

6 7 8 9 10 14 15 18 20 24 28 30

I found many websites that discuss Egyptian fractions; here are some of them:

  1. http://home.clara.net/beaumont/egypt/egmath.htm
  2. http://www.math.buffalo.edu/mad/mad_ancient_egypt_arith.html#Egyptian
  3. http://www.ics.uci.edu/~eppstein/numth/egypt/why.html
  4. http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Babylonian
    and Egyptian.html

While looking in the first site given above, I found this information about Egyptian numerals, presented here for you to see.

Here is another nice website about Egyptian numeration in general: Egyptian Mathematics.

Finally, the magazine reference used was: Spencer P. Hurd, “Egyptian Fractions: Ahmes to Fibonacci to Today”. MATHEMATICS TEACHER. October 1991. pp. 561-568

Repeating Decimal Patterns

Many years ago I read some interesting articles in a British school mathematics journal about the various patterns that can be found in the decimal expansions of repeating decimal forms of certain rational numbers [aka fractions]. So I began my own investigation of this topic to see what I could discover on my own. Below I will try to explain what I found. I hope you find it interesting, too.


We begin with the first truly interesting case: 1/7. When we divide 1 by 7, we obtain the following repeating decimal pattern:

	---  = 0.142857142857142857...

The repeating group of six digits 142857 is called the period or repetend of the decimal form of the faction. It is a nontrivial fact — as will be seen later — that of the ten digits, only the multiples of 3 are missing: 3, 6, 9, and 0. [For convenience, 0 will be considered a multiple of 3 in this work.]

As was pointed out in the page of Special Numbers, this 6-digit period manifests some unique sums if it is separated into appropriate halves and thirds.

Separate it into two halves, and then add.

		142857		142  +  857  =  999

Next split it into thirds before adding.

		142857		14  +  28  +  57  =  99

Now we will do a “skip counting” sort of separation, or better said “look at the sum of the digits in the odd positions vs. the sum of the digits in the even positions.” Note:

		142857:		1 + 2 + 5 =  8
				4 + 8 + 7 = 19

Nothing there, you say? Perhaps so, but in this case we need to wait until we do the 13ths to see the significance of it.

Here I show something I call the “N-sequence”. N stands for numerator. Recall that when you compute the decimal respresentations
for the sequence of fractions — 2/7, 3/7, …, 6/7 — you merely obtain the same six digit period, just beginning in a different position. I will show it for 2/7, and you can do the others yourself.

	---  = 0.285714285714285714...

Again we see a 6-digit period (285714) which is our original case just starting two digits later in the expansion. If we compute the other four fractions the same phenomenon appears.

Now we have a sequence of six fractions and their period numbers. Each fraction’s numerator is a number from 1 to 6, and each period has a different “initial” digit. If we replace the digits of the basic
period for 1/7 with the corresponding numerator digit, we can form a sequence of “numerators”, hence the “N-sequence” in this fashion:

		    Period:  1  4  2  8  5  7
		Numerators:  1  3  2  6  4  5

If we perform the “split-half addition” operation on our new sequence, we get this:

		132645		132  +  645  =  777

Wow! All 7’s, just what our denominators are.

Let’s do the odd-even “skip counting” operation again, and observe what happens this time:

		132645:		1 + 2 + 4 =  7
				3 + 6 + 5 = 14

Hmmm, the sums are 7 and the double of 7. That’s sorta nice, if you think about it.


Now let’s proceed to the next fraction with an interesting decimal form, namely 1/13.

	----  = 0.076923076923076923...

Right away there are some big differences. One is that all the missing multiples of 3 from the 1/7 case are now present here. Another difference is that while the period for 7ths had 6 digits (7 – 1 = 6),
one could have expected that 13ths would have a 12-digit period (13 – 1
= 12); but it didn’t turn out that way. [As you will see later, it does
indeed happen for other cases.] As the song says, “Qué será, será“.

What about the split-half and split-thirds patterns, I hope you are beginning to ask. You’ll be pleased to know it still happens in the expected way:

Two halves:

		076923		076  +  923  =  999

Three thirds:

		076923		07  +  69  +  23  =  99

Interestingly, thirteenths behave differently than the 7ths as regards the sequence of the next eleven fractions: 2/13, 3/13, 4/13, …, 12/13. The first difference occurs when we find the decimal form for 2/13. Unlike the case of 2/7, this one produces a new set of six digits (rather than a rearrangement of the first period).

	----  = 0.153846153846153846...

But, there is still some good news here anyway. First, the split-half and split-thirds properties are retained. (I’ll let you
write out the steps this time.) And if we realize that we now have another six digits, this gives us 12 digits for our discussion; and the idea of 13 – 1 is encountered here afterall. In fact, if we do a “head count” of the digits present in these two periods taken as a pair, we find that every digit appears, AND 3 and 6 appear twice! Note that 3 + 6 = 9.

Remember that seemingly useless oddity of the oddeven skip-
counting that resulted in the sums of 8 and 19 for the 7ths period? Well, guess what? IT’S BACK!! See?

		076923:		0 + 6 + 2 =  8
				7 + 9 + 3 = 19

		153846:		1 + 3 + 4 =  8
				5 + 8 + 6 = 19

(Perhaps now it would be fitting to observe that the sum of 8 and 19 is 27, the triple of 9.)

Are you now “getting the hang of it”? So, what about that thing called the “N-sequence”? We need to do a little dividing first; but I’ll let you do that this time. It’s not hard at all if you use
a calculator. We do have two periods this time, so you do have to be a little careful. But in the end you should agree with what is given next.

		    Period:  0   7  6   9  2  3
		Numerators:  1  10  9  12  3  4 

		    Period:  1  5  3   8  4  6
		Numerators:  2  7  5  11  6  8

Doing the oddeven skip-count operation here produces these
interesting sums:

		1 + 9 + 3 = 13   and  10 + 12 + 4 = 26

		2 + 5 + 6 = 13   and   7 + 11 + 8 = 26

(Do I need to direct your attention to the fact that 13 was our denominator this time, and 26 is the double of 13?)


We’ll do one more fraction, 1/17, in this page. Because after all this work, you should be getting the general idea. And it would be good for you to try it out yourself on other larger fractions.

	----  =  0.05882352941176470588235294117647...

Ah, now there is a period worthy of the name PERIOD! It has — count ’em — 16 digits! And 16 is 17 – 1. Here we have another case like 7ths, where the period length is one less than the denominator
being used. And also, like 7ths, we only have one period to concern us. So now the periods for 2/17, 3/17, …, 16/17 have the same digits, just starting at different positions.

If we “call the roll” of the digits present, the six digits we had in the 7ths period (1, 4, 2, 8, 5, 7) all answer twice, whereas the multiples of 3 (0, 3, 6, 9) each occur exactly once. [Keep this sort of thing in mind as you later investigate larger periods.]

While the period seems to be a mixed-up, mumble-jumble of digits in no particular order, there are structure and patterns here as well. For instance our old friend, the split-half property again produces a
sum consisting of sixteen 9’s. However, as 16 is not divisible by 3, the split-thirds property isn’t applicable here. But what about split-fourths? Let’s look at it.

	0588235294117647	0588 + 2352 + 9411 + 7647 = 19998

Not quite all 9’s there, but there are three and the digits at the beginning and end do add up to 9. It’s something anyway.

Let’s move along to our other pattern structures, like the oddeven skip-count one.

	0 8 2 5 9 1 7 4   <--- sum is 36
         5 8 3 2 4 1 6 7  <--- sum is 36

(And 36 is the quadruple of 9.)

Because we have 16 digits in our period, we might wonder what happens if we count every fourth digit before finding sums. Here’s
what happens:

	0   2   9   7     <--- sum is 18
	 5   3   4   6    <--- sum is 18    (And 18 is the
	  8   5   1   4   <--- sum is 18     double of 9.)
	   8   2   1   7  <--- sum is 18

Recalling the famous topic of Magic Squares and that a popular size is the the 4-by-4 square inspired me to wonder what would show up if I placed the digits of this period in a 4-by-4 array, then added up
the rows, columns and diagonals. I was pleasantly surprised, as I hope you will be, too.

. 24
0 5 8 8 21
2 3 5 2 12
9 4 1 1 15
7 6 4 7 24
18 18 18 18 11

The four column totals of 18 are not unexpected as we just saw that in the previous comment. But look at those row totals. They seem to be nothing more than some multiples of 3. But a closer examination
will prove otherwise.

	21 = 3 x 7	12 = 3 x 4	15 = 3 x 5	24 = 3 x 8

And the sequence of the 3’s co-factors — 7, 4, 5, 8 — can be seen in the chart in the lower-left-to-upper-right diagonal, in that same order!

The main 4-term diagonals don’t seem to provide anything of interest, but the “1-3 broken diagonals” do. Observe:

		0 5 8 8	   8		0 5 8 8     5
		2 3 5 2    2  		2 3 5 2     5
		9 4 1 1	   4		9 4 1 1     1
		7 6 4 7	   4		7 6 4 7     7
			  18			   18

		0 5 8 8	   0		0 5 8 8     8
		2 3 5 2    2  		2 3 5 2     3
		9 4 1 1	   1		9 4 1 1     9
		7 6 4 7	   6		7 6 4 7     7
			   9			   27

Finally, let’s examine the N-sequence for our 16-digit period.

	    Period: 0  5  8  8  2  3  5  2  9  4  1  1  7  6  4  7
	Numerators: 1 10 15 14  4  6  9  5 16  7  2  3 13 11  8 12

Using the oddeven skip-count pattern brings these sums:

	1    15     4     9    16     2    13     8     <-- S = 68
	  10    14     6     5     7     3    11    12  <-- S = 68

Guess what? 68 is the quadruple of 17.

The every-fourth term pattern provides this nice result:

	1           4          16          13           <-- S = 34
	  10           6           7          11        <-- S = 34
	     15           9           2           8     <-- S = 34
	        14           5           3          12  <-- S = 34

Perhaps by now you are gaining a deeper appreciation for the beautiful numerical patterns that can be discovered in the decimal forms of some fractions. And this is just a “drop in the bucket” of
what is possible to find. I hope you try investigating this topic for yourself. I suggest you begin with the 19ths, then move on to the 21sts, 23rds, and 29ths, to mention but a few. What you find depends mainly on your own creativity and sense of number.

Good luck.

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