Tag Archives: addition

Digit Add-On Problem

One day while browsing on the Internet, I came across the following problem:

A 7 is written at the end of a two-digit number, increasing its value by 700. Find the original number.

Well, as I’m always hunting for good problems about numbers, I feel that I have developed a sixth sense for such things. And this problem proved to be as good as any I’ve run across recently. It is the kind of problem that looks very easy, at least to understand what’s going on, while at the same time proves to be a distinct challenge to young students. Of course, being a septophile, I was intrigued by the use of the numbers 7 and 700. And my interest only but increased when I found the solution: 77!

I solved the problem rather quickly, because I used some basic algebra. Then I set myself to thinking: if I didn’t know how to do it by algebra (like the vast majority of the students I teach), how would I proceed? Naturally, by “trial and check”. I would choose some two-digit number, place the 7 at the end, thus forming a much larger three-digit number. To find out how much the resulting increase was, it was a simple matter: subtract the original two-digit number from the new three-digit number; if the difference was 700, my search was through. If not, I would adjust my original trial appropriately, either up or down as needed.

That’s just what my students did, once they understood what the problem was talking about. Most of them had never been presented with a problem of this sort, especially in a math class. They often had difficulty knowing just which way to go. So I decided to devote a little more time to it. This sort of problem is easy to create variations for, such as:

  1. If 6 is written at the end of the two-digit number 53, the new number formed
    in this way would represent an increase over the original. How much was the increase?
  2. If a digit is written at the end of the two-digit number 48, the original
    number (48) would be increased by 434. What is the single digit?

I hear you saying about now, “Those are certainly easy problems.” And you are correct. After all, all these problems follow this basic pattern:

XY + increase = XYZ

where the capital letters represent the various digits, and Z being the special single digit part of the problem. Let’s analyze how this structure helps us to attack the three forms of the problem.

In form #1 above, we are given the X, the Y, and the Z. So it’s a simple matter to find the increase.

XYZ – XY = increase

Form #2 is easier still. Here we have the XY number and the “increase”. So it’s a trivial matter to add the two. As long as the sum starts out with “XY-“, we should have achieved our goal. The Z-digit is there for the picking.

Therefore, the tougher nut to crack is the original problem. In that form we only know the Z-digit and the increase. But as any good algebra student should be able to show, the positive difference of the single digit and the increase is divisible by 9, and further, the quotient of that difference divided by 9 is our desired two-digit number!! The proof might go something as follows:

Let n = the original two-digit number, d = the single digit, and i = the increase.

If we write a digit behind a number, the place value of each of its digits is multiplied by 10. [In effect, each of its digits is “shifted” to the left.] This means our three-digit number has this structure: 10n + d. Using the fact explained above for form #1, we can state

(10n + d) – n = i

Solving this equation gives us

n = (i – d)/9

Now solving problems like the original 7’s problem becomes a mere two-step process:

  1. Subtract the given digit from the increase.
  2. Divide that result by 9.

In the case of the 7’s problem, this means 700 – 7 = 693, and 693 ÷ 9 = 77.

Additional Comments

While working on the various extra homework exercises that I gave my students, Cristina found an interesting fact that had escaped my attention. And she did it without using algebra! She told all of us in her class the next day that if you add the digits of the “increase” number, and continue to add the digits if any sum is greater than 9 until a single digit is obtained, then that digit is the single digit in the problem statement. Using an example from above (form #2) we see that

434 + 48 = 482, so 2 was the so-called Z-digit.

But notice this:

4 + 3 + 4 = 11 and 1 + 1 = 2.

This is no freak accident. This is true all the time. Just get out your algebra and prove it; I challenge you to do this.

Next the two-step solution procedure explained above works equally well if the original number has three places (or even more). Try this problem and see.

A 3 is written at the end of a three-place number, increasing it by 5367. Find the original number.

At first glance it looks a little scary. But using the steps on it produces the desired result rather quickly.

Finally we “up the ante” just a bit here with this extension.

The number 17 is written at the end of a three-place number, increasing it by 16847. Find the original number.

(Put that one in your algebra “pipe” and smoke/prove/solve it!!!)

Products with Number Sums

Here is an interesting problem that only involves basic arith-metic, but with an unusual twist.

Find 5 numbers such that when each number is multiplied by the sum of the remaining 4 numbers, one of the following values will result:

44    63   95   108   128

WOW! Did you get that? I hope you didn’t get too confused.

All it really says is: let’s take five numbers — say 1, 3, 4, 7, and 8 — add any four of them, then multiply the sum by the remaining number. In this sample, try adding 1, 3, 7, & 8, getting 19; now multiply 19 by 4, getting 76.

Now it seems easy enough, right? But that’s because we knew the 5 numbers in advance. This is another of those problems I call “JEOPARDY” problems. When you are given the “answer”, you must find the “question”.

You know? Perhaps in this case I’ll give you some pretty heavy hints, as it does seem to be a “tough nut to crack” this time. We begin by forming factor T-charts for each of the 5 products.

                 44     63     95     108     128 
               1 44   1 63   1 95   1 108   1 128
               2 22   3 21   5 19   2  54   2  64
               4 11   7  9          3  36   4  32
                                    4  27   8  16
                                    6  18
                                    9  12

Next locate those factor pairs that have the same sum; in this case it should be 24. The smaller members of the required pairs are indicated in red. They are the 5 numbers which solve the original problem.

Perhaps you’re wondering how I knew the common sum should be 24. It’s really quite simple if you look at the chart that has the fewest entries: 95. Since it only has two factor pairs, whose sums are 96 and 24, this shortens our search quite a bit. Then since the first two charts (for 44 and 63) only have factor sums less than 96, this leaves only 24 to be considered. The rest, as they say, is “easy as pie”.


The reason the procedure works out so well can be proved by simple algebra. Assume the five numbers are a, b, c, d, and e. There-fore, we have

a(b + c + d + e) = 44

b(a + c + d + e) = 63


Notice that the sum of the two factors in each equation is always the same, namely a + b + c + d + e, the sum of the desired numbers. Hence, the procedure of finding the factor pairs possessing equal sums becomes rather obvious.

Teacher Notes:

Here are some problems that can be presented to your math class after the procedure has been explained. Many more can be prepared as needed.

1. 245 297 152 320 360
2. 220 328 468 490 360
3. 111 319 375 204 175
4. 522 742 816 570 882


1. Of course, more (or fewer) numbers than five can serve as the basis. It only depends on student interest, level, time, etc.

2. A problem can be given with one of the products missing. The problem will be considered as solved when that product has been found. (This is easier than it sounds at first.)

3. A competition between 2 students can be set up. Each player chooses, say 6 numbers, and prepares the set of products. A limit on the size of the numbers should be agreed upon in advance. Then they trade number sets and the first to declare the other’s 6 numbers is the winner.

Final Notes

Since the method of solution is being directly presented to the student from the start, the pedagogical objectives of this activity are these:

a) Important practice in finding factor pairs of whole numbers—a skill much needed in algebra courses;

b) Experience in following directions in a non-tradi-tional problem setting; and

c) Drill in systematic-search techniques for problem solving.

d) NOTE: Calculators should be considered as an option, especially if large numbers are employed.

REFERENCE: Maurice Kraitchik, MATHEMATICAL RECREATIONS, Dover Pub., 1953. p. 46.


(1) 4, 7, 9, 10, 12

(2) 5, 8, 9, 13, 14

(3) 3, 5, 6, 11, 15

(4) 9, 10, 14, 16, 18


Back-to-Front Multiplication


While contemplating on the number curiosity below, I discovered that it possessed deeper and unexpected characteristics, structure, and patterns. I have not seen any discussion of these findings in the recreational literature I have read. Hence, it is being offered with the belief that I have uncovered a new recurrent operation pattern of a cyclic nature. It is dramatic proof that there often is a lot more structure than we realize behind most of the very innocent-appearing number puzzles.

The Problem

In the April 1962 issue of Recreational Mathematics Magazine, there appears the following multipication oddity:

421,052,631,578,947,368 can be doubled by shifting

the last digit to the front. (p. 34)

Desiring to use such fascinating problems in my school teaching at the junior high level, I wondered, “Are there more such problems, or is this just a freak case?”

The Solution

To my great surprise, once the simple secret of their construcion is understood, I found it very easy to find quite a few such examples. But, often one has to be rather patient for the required factors to reveal themselves. To facilitate the following discussion, I will demonstrate the construction method with a simple example. (This will also serve to motivate the rationale behind the recurrent operation to be described shortly.)

Say that we desire to have a problem where “4” is our single-digit multiplier, and “7” is to be the shifted unit’s digit. We begin by setting those digits in their normal places (see Figure 1a).

		          2           2  	3 2
        	7	  _ 7         8 7       _ 8 7
            ×   4       ×   4       ×   4      ×    4
		            8	        8	  4 8
              (a)        (b)         (c)	 (d)

				Figure 1

The indicated multiplication is carried out as shown in Figure 1b. Then as the “8” is the unit’s digit of the product, it follows that it must be the ten’s digit of the factor as well (see 1c). Figure 1d shows how the process is to be repeated.

But when does it stop? It’s quite simple: when you obtain a “7” to place in the product and there is no “carry” involved. In the example before us, we are fortunate to arrive at another “7” in only six steps:

			3   3   1   3   2
			1   7   9   4   8   7
			  		×   4
			7   1   7   9   4   8

In general then the process continues until such time that the shifted digit itself is the outcome of one of the steps.

By using this procedure for other original digit choices, one can produce multiplication problems that possess the back-to-front shift property for the multi-digit factor.

Table 1. Back-to-Front Multiplication Integers
Single-digit factor The Special Integers
2 105,263,157,894,736,842
3 1,034,482,758,620,689,655,172,413,793
4 102,564 128,205 153,846 179,487 230,769
5 102,040,816,326,530,612,244,897,959,183,673,469,387,755
6 1,016,949,152,542,372,881,355,932,203,389,830,508,474,-
7 1,014,492,753,623,188,405,797



8 0,253,164,556,962





9 03,370,786,516,853,932,584,269,662,921,348,314,606,741,573

Table 1 presents a summary of what I choose to call the “primary” solutions. These solutions were developed in the following systematic manner:

1) First I selected various digits (destined to be the shifted unit’s digits) that were equal to or greater than certain single-digit multipliers.
2) While computing all such cases, I observed that frequently a result contained the same sequence of digits; it just began at a different position.

Here is a simple example to illustrate:

		1  2  8  2  0  5	 2  0  5  1  2  8
		           ×   4	            ×   4
		5  1  2  8  2  0	 8  2  0  5  1  2

So, the “shifted-8″ case was considered redundant for my purposes, and was not included in the table. This should make it clear why there are fewer integers listed than are possible, i.e. the others are merely “embedded” in those given.

3) There were only a few cases where this embedded aspect did not account for instances in which the shifted-digit was less than the multiplier. This happened only when “8” and “9” were the multipliers, occurring three times and once, respectively. It is in such situations for all multipliers that a “leading zero” must be appended to the front of the large factor. The reason for this becomes obvious when it is pointed out that no positive integer times another can be less than the former one (whether or not there is a “carry”). In view of the way it makes things more uniform and regular, the use of the leading zero is not too great a liberty to take with standard notation., one thing is clear at this point: it preserves the equal length property of the integers, a fact that will become more apparent later.

Since further structure and pattern considerations for the integers in Table 1 will actually be covered in the discussion of the recurrent operation, we should now turn our attention to that topic.

The Recurrent Operation

While carrying out this iterative process, the tedium of writing the work in the customary manner made me aware that the digits of the product were of little use to me. So I ceased to write them at all. Then I noted that the same information could be conveyed by a vertical method. Figure 2 shows how this would be done for the “7-over-4″ example discussed earlier.

	  3  3  1  3  2        3  3  1  3  2
	  1  7  9  4  8  7     1  7  9  4  8  7         /×4/
	             ×   4                ×   4	          7
	  7  1  7  9  4  8				 28
                (a)                   (b)    	         37

				Figure 2

The recurrent operation algorithm then emerges in the 2c part, by observing these steps:

  1. Two digits are selected; one is the constant multiplier, the other is the first term of our cycle.
  2. Their product is the second term of the cycle.
  3. The next term, and all succeeding terms, is the product of the constant multiplier and the unit’s digit of the preceding term, increased by the ten’s digit (if any).

This can be summarized by the recursive formula:

Let k and a1 be the constant and first-term digits, respectively. Then,

			a2 = ka1 , and
			an = ku + t , where an-1 = 10t + u.

Table 2 was prepared by using this compact vertical format. Reading the unit’s digits from the bottom up produces the special integers given in Table 1. Table 3 (further below)is a summary of the patterns that can be found in Table 2.

Discussion of Table 3

Table 3.
Factor No. of primary cycles Cycle Period Max. Integer Omissions from
primary cycles
2 1 18 18 none
3 1 28 28 none
4 5 6 37 13, 14, 17, 23,
25, 29, 35
5 2 42 & 6 48 none
6 1 58 58 none
7 3 22 68 23 & 26
8 5 13 77 12, 14, 15, 17, 27, 33,

41, 57, 58, 61, 69, 71

9 2 44 88 none

This summary table reveals several intriguing properties for the cycles in Table 2. The first thing to note is that the multipliers 2, 3, and 6 each produce but a single primary cycle, whereas the remaining factors yield 2 or more cycles apiece. Turning to the multi-cycle cases, we see that all except the factor “5” had cycles of equal periods. The term “max. integer” simply refers to the largest integer that appeared in the normal production of the primary cycle for the factor. For the factors 2, 3, 6, and 9, it is relevant to note that the product of the cycle period and the number of cycles is equal to the “max. integer”. This shows that there are no “omissions” in the range of 1 to that largest integer. The factor of 5 also qualifies here by merely adding the two values for its cycle periods.

However, the factors 4, 7 and 8 present the most interesting facets so far. Number 7 has the fewest omissions, so let’s examine them first. Applying the R.O. algorithm. It so happens that each integer (23 and 46) yields itself immediately. In line with current terminology in such cases, I call such integers “self-generators”.

Testing the values in the omissions list for “8” shows that none of them are self-generators. However, all but one of them (i.e. 69) generate another entry in the list. Yet 69 itself generates an interesting value nonetheless: “78”, which is one greater than the max. integer. So, if we temporarily include 78, we can construct a “secondary” cycle, namely,

12 — 17 — 57 — 61 — 14 — 33 — 27 — 58 — 69 — (78) — 71 — 15 — 41; (and 41 returns to 12).

This augmented cycle now has 13 terms, just like the primary ones!

Finally, we look at the case for “4”. That set has two self-generators (13 and 26). (At this point we can note a common property shared by our two pairs of self-generators: namely, the smaller one is half the larger.) As we did for the 8-case above, another secondary cycle can be formed by using the remaining values. This time we must also temporarily include the integer that is one greater than the max. integer. This last cycle is

14 — 17 — 29 — (38) — 35 — 23 — 14.

And the equal-period property is preserved!

Now, upon glancing back to Table 3, we can appreciate the full significance of the max. integer column by noting that only the “4” and “8” cases had a largest entry ending in a “7”; and only those two yielded secondary cycles that necessitated the restoration of the “missing link”, so to speak. With the inclusion of those two missing links, everything “ends with an 8″.

That strange finding naturally raises the question: “What about all those values that immediately follow those 8-enders, namely, 19, 29, 39, … , 89?” Well, oddly enough they all turn out to be self-generators for their respective single-digit factors! And this fact extends beyond the table, and is easily proved as follows:

[10(n – 1) + 9]n = 9n + (n – 1) = 10n – 1.

(The final expression is the same as what is inside the brackets.)


This brings me to the end of my discussion into this puzzle, except for two closing comments and a challenge. The systematic manner in which the data was gathered proves that the contributor of the original puzzle could not have presented a shorter integer than one with 18 digits. The cycle periods given herein are all minimums.

Also, it is very easy to see that all the integers (no matter how large), under this R.O. procedure reduce to only the cycles or self-generators given in this paper. This is a factor that parallels the well-known R.O. procedure concerning the sums of the squares of the digits of all integers (see the Steinhaus or Trigg references). [Also click on Happy & Dizzy Numbers for more information in this website.]

Finally, a challenge to the reader. What patterns, and periods, and other phenomena occur when integers in other bases are put through this new recurrent operation? Try them, and see for yourself!


  1. Steinhaus, Hugo. One Hundred Problems in Elementary Mathematics. New York: Basic Books, 1964, pp. 11-12, 55-58.
  2. Trigg, Charles W. “A Close Look at 37,” Journal of Recreational Mathematics, 2(2), April 1969, pp.117-128.

[Published in The Oregon Mathematics Teacher, Feb. 1978, pp. 22-27.]

Palindrome Power

PALINDROMES: A Teacher’s Guide

A number that remains unchanged in value upon writing it in reverse is called a palidromic number, or simply a palindrome. The idea is borrowed directly from a popular form of word play. Certain words, like dad, noon, radar, etc. possess this reversal invariant property. Even many sentences can be formed that also exhibit it. Examples: Madam, I’m Adam. and Rats live on no evil star. Of course, here we will mainly be concerned with the numerical version, though the use of palindromic words should not be ignored as a means of explaining this concept to students.

[Note: The word palindrome is derived from the Greek palíndromos, meaning running back again (palín = AGAIN + drom-, drameîn = RUN). A palindrome is a word or phrase which reads the same in both directions. (Source: What are palindromes?)]

The teacher has two possible methods of introducing palindromes to the student: the direct and the indirect. The former consists of a straightforward definition (“A palidrome is ….”), followed by examples and appropriate exercises.

However, this writer believes that the indirect approach provides an excellent opportunity to provide students with a moderately easy situation to test their observation skills. Our success with students at a wide variety of levels has led us to this conclusion. However, you, as the teacher, may devise your own method of presentation. Here is our basic method.

We begin by presenting a few examples of numbers that are palindromes, and asking if anyone sees what makes these numbers special. We start with three-place numbers and gradually present larger ones. Depending on the responses received, we at times present numbers that are NOT palindromes, for contrast. Other times, we give a number that meets the criteria of the students’ attempts of definition, yet is not a palindrome. By being patient, the proper definition usually is deduced with little help from us. Then we proceed to the reverse-and-add activity to be described below.

Once the definition has been “discovered”, we then introduce “things to do” with it. The one that provides the most interest is the Reverse-and-Add procedure. It is quite simple:

  1. Pick a number, any number.
  2. Reverse that number, writing it beneath the other.
  3. Add the two numbers.
  4. If a palindrome results, “Voila! Eureka!”; if not, reverse that sum underneath itself, and continue until such time as one is obtained.

Simple as it is — and it REALLY is — it is impossible in a practical sense to just say when, if ever, a palindrome will be encountered. Sometimes it occurs at the first or second addition, other times it requires lots more. For example, 89 does not yield one until the 24th addition, consisting of 13 digits. [To see an actual account of what happened one day in my class, click here.]

Another enigma is the number 196. Here it is literally not known whether a palindrome will appear. A computer has performed over 4000 additions on it [see update below], and none has shown up yet. However, this, like Goldbach’s Conjecture, does not prove the case. It just says the problem is beyond the paper-and-pencil method of solution.

But to return to our classroom work, we present a variety of examples in the following manner (with variations, of course):

                      38      85        475
                    + 83    + 58      + 574
                     121     143       1049
                           + 341     + 9401
                             484      10450
                                    + 05401

From this point on, various lines are possible. One is to give the three numbers 195, 196, and 197 and ask the students to find palindromes. Usually 195 is done first, with a palindrome appearing at the 4th addition step. This is followed by 197; its palindrome appears on the 7th step. But as stated above, 196 is a mystery. So to spark interest and enthusiasm, we offer a “reward” to the first person who finds a palindrome correctly computed from 196. This never fails to get them working frantically. Of course, while this may seem to be an unfair problem, there are many lessons than can be learned in this sort of exercise. Invariably, students start announcing that “I’ve found one!” You know in advance that an error of some kind has been committed. With care, you will locate a basic fact error, or one involving improper carrying, or improper reversing, etc., etc. A bit of advice is in order here: prepare yourself with a list of the first 25 or so sums to make it more efficient in locating the point at which an error was made (it may not be the only one, or even the first). Also, take extreme care in making your list, lest you commit an error yourself. (It CAN and does happen, even to the best of us.)

Another approach we have taken, with equally interesting results, is to offer 25 cents to the first person who finds a palindrome based on 89, correctly computed. [Note: this article and work was originally done in the late ’70s, when a quarter was a bit more attractive. But, even then, 25 was an essential part of the activity.] This of course, is possible; and the payoff is “fair” — virtually 1 cent per addition. Yet, to discourage sloppy or careless work, we offer a counter-bet of 5 cents that the student must pay us IF an error can be found. Of course, we don’t take the students, as somebody either wins from the class, or the money is returned to them. This makes some students be a little more careful and aware that even in an apparently simple problem, one can make mistakes. As time passes and a few (2 or 3) losers have had to pay up, we give the hint that they had better do more than 20 addition steps, or refuse to look at their work (knowing they haven’t spent enough time on it and should look for their own mistakes). After this problem is conquered, we present the 196 case, with an appriately greater reward (e.g. $10). Other times we give them the choice of problems.

There are other types of palindrome processes. The reverse-and-add procedure can be modified to ask the question: If you keep on going with the R&A work, how many palindromes appear in 10 (or 15) addition steps? Other questions include: Does a palindrome yield a palindrome? How many palindromes can come up “in a row”? Can a palindrome only occur if there has not occurred a “carry”? (Can you think of your own?)

Palindromes can arise in other settings. For example, squaring, cubing, or finding other powers of certain numbers will yield palindromes. How many can you find?

Encouraging students to be on the lookout for strange or unique numbers ca have interesting ramifications. Ask them to be alert and see who can find a license plate number that is a palindrome. Or a phone number, a social security number, a house address, a zipcode. Since numbers are all around us in this modern world, we ought to make them our friends. In other words, “Let’s don’t fight it. Learn to recognize interesting numbers with strange properties. Then it is more like hunting for diamonds or nuggets of gold. We appreciate them more for their rarity or unique characteristics.”

[NOTE: (July 2001) I have recently been introduced to Patrick De Geest of Belgium. He has a marvelous website devoted largely to alindromes, called WORLD!Of Numbers. Go there soon.]

[UPDATE: (February 2002) The number of reversal steps has been increased to many millions. Click here to learn more.

Below is a sample of a worksheet that could be developed to give elementary students some nontrivial investigation experience. Whether calculators should be allowed on this work is an open question. But sometimes, as in the “196” and “89” problems, the display window of the calculator is too small to contain the sums; so there it is “back to basics”, I guess: good ol’ paper-and-pencil.

1. JUST FOR PRACTICE: make sure that you understand the “reverse- and-add” procedure by completing this table.

Seed No. No. of additions Palindrome
283 3 .
185 . 4774
3947 2 .
275 . 44444

2. in each pair below produce the same palindrome, but they don’t need the same number of additions. So, for each pair, (a) find the palindrome for each pair; and (b) state the number of additions for each number.

(a) 6228 & 2673 (b) 197 & 5873

3. The seed numbers in this group all produce palindromes that have a strange common property. Find the palindrome for each, then state what the strange thing is. (Can you find a seed number of your own that will do the same thing?)

(a) 338 (b) 676 (c) 726 (d) 7482 (e) 53877

4. Here the “problem numbers” tell you how many additions are required to find a palindrome.

(1) 47 (2) 238 (3) 86 (4) 78

(5) 6358 (6) 62354 (7) 118471 (8) 8779

Can you find a seed number that requires 9, 10 or more additions? There are many.

5. The seed numbers 428, 527, and 626 all yield the same palindrome with the SAME number of additions. Look at them closely to see the pattern; what is it? Then give another number (less than 428) that belongs to this family BEFORE trying it. Finally, prove your choice is correct. (Using that idea, can you give several seed numbers that produce 15851? HINT: See exercise #1 above.)

*6. Prepare a short report by investigating the patterns you find when you use the ten numbers in this sequence:

606, 616, 626, 636, … , 696.


Happy & Dizzy Numbers


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

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

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

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

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

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

82 + 32 = 64 + 9 = 73

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


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

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

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

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

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

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

Part II: Dizzy Numbers

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

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

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

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

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

Have fun!

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

Update: (6/24/02)

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

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.)

Calculator Poker

Mathematical games are always a popular adjunct in any teacher’s repertoire of motivational activities. This game promotes learning and fun at the same time by combining two unlikely allies: a deck of ordinary playing cards and a calculator. The material presented here utilizes a simple, four-function calculator with a square root key. With minor adjustments, the more powerful scientific models may be used.

There are many ideas that a student can pick up while playing this game, not the least of which is the added familiarity with the calculator’s utility as a tool. Reading and writing the displayed digits carefully and precisely are skills that need to be developed in certain students. Of particular relevance to the game is the careful identification of which digits produce one’s best hand.

The basic game involves 3 players using 3-digit numbers. Remove the king, queen, jack and ten of any particular suit from the deck of cards. The nine remaining cards (Ace to 9) constitute the playing deck.

The value of the Ace will be 1. Each player uses his/her own calcuator, a recording sheet with headings: Number, Root, Combination, Points, a copy of the scoring chart (Figure 1), and a pencil.


  1. The dealer shuffles the deck and deals 3 cards face down to each player.
  2. Each player arranges his/her cards left-to-right, without turning them over, in any way desired.
  3. When all players have arranged their cards, the dealer says, “Turn over.” All cards are then turned face up without changing their left-to-right order. The 3-digit number thus produced is entered into the first column of the recording sheet.
  4. All players enter their numbers into their calculators and press the square root key. The displayed value, which represents one’s playing hand, is written in the next column of the recording sheet.
  5. Each player then carefully studies the digits of the square root to determine the highest scoring combination (see Score Chart) which is then written in the third column. For example, the number 142 yields the square root of 11.916375, and the three ones give the highest scoring combination of”three of a kind”.
  6. The points are determined by adding the base score given in the Score Chart to the value of the largest digit involved in the combination. The largest digit in the combination constitutes the Bonus Points. So, in the illustration above, 142 has a square root of 11.916375, for which the base score is 30 points (three of a kind: 1, 1, 1) plus 1 Bonus Point for the largest digit in the combination. After adding, the total (i.e. 31) would bewritten in the Points column of the recording sheet.

    Examples: If one has two pairs (6,6 & 2,2), the score would

    be 20 + 6 = 26 points.

    If one has a full house (4,4,4 & 7,7), the score is 50 + 7

    = 57 points.

    For a run (1,2,3,4,5), the player scores 100 + 5 = 105 points.

  7. For the 2nd and 3rd hands of a game, Rules 1-6 are followed as before. But for the 2nd hand, the square root key is pressed twice; for the 3rd hand, the square root key is pressed three times.
  8. After 3 hands are completed, the points are totaled and the winner is declared.
A Sample Game
Number Root Combination Points
795 28.195744 1 pair: 4, 4 14
641 5.0316972 all different 200
436 2.1376461 2 pairs: 6,6,1,1 26

This player’s score is 240, a good game.


First, from time to time a perfect square is formed by the cards. Since here a square root (i.e. one press of the key) only has 2 digits and that isn’t enough to form any hand in the first part of the score chart, an award of 100 points is given. And once in a while, all the eight digits that appear in the display are different, or distinct. We feel that such an event is unique enough to merit the award of 200 points! Experience has revealed that occasionally only 7 digits (or even 6) appear, all of which are distinct. So a point value is assigned to each of these cases. However, no bonus points are given in any of the above cases.

When using scientific calculators with 10-digit display capacity, just instruct the players to copy the first 8 digits that appear, that is, “truncate to 8 digits”. While rounding the decimal part to achieve 8 digits is possible, our experience has shown that truncation is simpler to explain to the majority of students.


There are many possibilities to vary the basic game: use more number cards at once (i.e. 2 sets of Ace through 9), deal more cards for each hand (i.e. 4 cards to form 4-place numbers), include one or

more jokers (they equal zero, or become a “wild card”), allow more hands to constitute a game, etc. Presented below are several more variations that have proved popular and instructive in our classroom. Of course, the reader is encouraged to devise his/her own rules.

  • Decimal Numbers. Once the three cards in the basic game are turned face up and the digits are noted on one’s paper, a decimal point can be inserted between the second and third digits, or prior to the first one. Example: for 382 we would have “38.2” or “0.382”. It is of interest to note that placing the point between the first and second digits does not producea different set of digits for one press of the square root key, as can be demonstrated by the example shown here.The reason behind this is well known to anyone who has studied the properties of radicals in an algebra class. However it is not necessary to go into the details of why in order to play the game. But it could be explained if students express an interest. So turn it into another of those special “teachable moments” we all look for. However, for 2 (or 3) presses of the square root key, different digits do appear.
  • Fractions Game. Form a fraction using the cards dealt to you.For instance, when using 4 cards per deal, let the first two form the numerator and the latter two form the denominator. In most calculators, this only amounts to converting the fraction to its decimal equivalent by dividing before pressing the square root key. For those who have calculators with fraction keys(as the TI Explorer and others), this is made even easier. This game variant, therefore, reinforces the connection between decimals and fractions.
  • Add-to-your-neighbor Game. Your playing number is found by adding your 3-digit “card number” to that of the player to your immediate left (or right) and putting the sum into the first column of the recording sheet. For instance, if the cards gave 329 for A, 416 for B, and 785 for C, then each player’s new number would be 745 for A, 1201 for B, and 1114 for C.
  • Third-Round-Sum Game. You form your playing number for the thirdround by adding the numbers used for the first and second rounds. With 624 for the first round and 179 for the second round, you use 803 for the third round. (This has the advantage of speeding up the game as there’s no need to deal the cards again.)
  • Constant-Multiples Game. Throughout your game you multiply yourcard number by some pre-set number, say your age, the day’s date, a favorite prime, etc. If Player A is 14 years old and his card value is 291, the playing number would be 4074. For students who can appreciate it, this is a good time to introduce the constant operation feature of the calculator and show that pressing the root key does not inactivate the constant multiplier as one goesfrom round to round.


An important spinoff can involve a statistical analysis of the data that can be collected after many games have been played by the whole class or merely a small group. For example, have the students

compute the arithmetic mean of various final scores: per hand, per game, per group of players, etc. An additional connection to probability can also be explored by constructing frequency histograms of the types of the hands that occurred throughout. The raw data is easily obtained from the students’ recording sheets.

One of our students showed us an important example of that “what-if” curiosity that needs to be promoted more in our classes. She thought: “What if I had arranged my cards differently? What might

have been my score?” So she took the card numbers from several of her games that day and proceeded to form all the various possibilities, then obtained the square roots and points. She was so proud of herself, as we were of her. What makes this more outstanding is that she was a below average performer, yet did this quite independently.

This game, in its own special way, provides a connection between the world of regular mathematics and that of simple gaming pastimes. It’s fun and one learns all the while.


Score Chart

           Combination                          Base Score 

	One Pair                                  10 points

	Two Pairs                                 20 points

	Three of a Kind                           30 points

	Three Pairs                               40 points

	Full House (3 of one kind, 2 of another)  50 points

	Two Trios (2 sets of 3 of one kind)       60 points

	Four of a Kind                            70 points

	Five of a Kind                            80 points

	Run (5 consecutive digits)               100 points

	Special Cases:

	*Perfect Square                          100 points

	*All 8 digits different                  200 points

	*Only 7 digits showing, all different    150 points

	*Only 6 digits showing, all different    125 points

	[*No bonus points in these cases.]

				Figure 1

Sample Hands & Scores
Number Root Hand Points
158 12.569805 5, 5 15
145 12.041594 4, 4 & 1, 1 24
142 11.916375 1, 1, 1 31
138 11.74734 1, 1, 4, 4 & 7, 7 47
149 12.206555 5, 5, 5 & 2, 2 55
152 12.328828 2, 2, 2 & 8, 8, 8 68
485 22.022715 2, 2, 2, 2 72
147 12.124355 1, 2, 3, 4, 5 105
169 13 square 100
786 28.035691 all different 200

[Note: The square roots in this table were obtained from a simple, 4-function, 8-digit calculator.]

This article of mine is from MATHEMATICS TEACHING in the MIDDLE SCHOOL. NCTM. Feb. 1998.
pp. 366-8. Reprinted with permission. (See photos following.)

Egyptian Fractions Target Game

Recently I had the opportunity to play around with a novel middle school level calculator that “does fractions”. [For those who are curious, it is the Sharp EL-E300 model.] In fact, not only does it “do” them in the normal vertical format, but also it shows more than one fraction at a time in the display window. “Neat-o torpedo”, I thought. “Now what could I do with this clever instrument that would capture the interest and attention of my students while at the same time teach them some good stuff about this time-honored bug-a-boo of math at this school level?”

After spending some time reflecting on the matter, I decided I wanted to create a game that would involve various concepts: addition of 2 (or 3) common fractions, perhaps their subtraction as well, comparing relative sizes of 2 or more fractions, and decimal equivalents of fractions would be nice, too. Since estimation and number sense are hot topics these days, I wanted to include them also. And there was one more thing: how about a little historical connection thrown in for good measure!

The latter goal was easily met: Egyptian Fractions. (See my page on Egyptian Math for more on this topic.) From there, things just took a natural course, and I developed the little game that you will see described below. By the way, that’s an Egyptian fraction up there on the left side of the title bar; its modern form is given on the right.

Rather than presenting a lot of complicated rule descriptions of how to play, I think I’ll just describe how two students might go about a sample game. Our players are the ubiquitous John and Mary.

Play begins when their teacher announces that the “target fraction” will be “7/19“. The timer is set for 2 minutes. At the word “GO!“, John and Mary must each try to find two Egyptian fractions whose sum is as close to 7/19 as possible, without exceeding it, before time runs out.

If you recall your math history, you will know that John and Mary are looking for two fractions of the form


Another name for such fractions is unit fractions, perhaps because the numerator is the unit “1”. However you choose to call them, the addition
procedure for such a pair is really quite easy. Observe:

                  1     1      b      a     b + a
                 --- + --- = ---- + ---- = -------
                  a     b      ab     ab     ab

In other words, the sum of two unit fractions is equal to the sum of the two denominators over the the product of the denominators. For more about all this, go to this WTM article: Fraction Addition.

[Recall that in Egyptian fraction work, all denominators must be distinct values, no repeats. So a is NOT equal to b.]

The two minutes are up. Let’s see how John and Mary are doing now. John chose 3 and 30 as his denominators, whereas Mary chose 4 and 11. So their respective sums are these:

          1      1     33     11             1      1     15
         --- + ---- = ---- = ----    and    --- + ---- = ----
          3     30     90     30             4     11     44

The next step is to determine which sum is closer to the target fraction (7/19) without going over. This can be done in two ways: fractionally or decimally. If speed of play is more important to you, then the decimal approach is probably better; otherwise, doing it by fractions could provide a different view of the situation. First, let’s look at the decimal way.

We convert the 3 fractions (target and 2 sums) to their decimal forms by simple division in our calculators, obtaining

target: 7/19 = 0.368421053

John: 11/30 = 0.366666667

Mary: 15/44 = 0.340909091

It’s clear that both players have obtained fractions whose values are less than the target’s value. It is considerably more helpful if we round our decimal expressions to some convenient level. Here I would recommend “to the nearest 1000th“, giving us these numbers. target: 7/19 = 0.368

John: 11/30 = 0.367

Mary: 15/44 = 0.341

Mere inspection, or simple subtraction, allows us to declare a winner here: John. His error was approximately “0.001” (“Very good, John.”) while hers was approximately “0.027” (“Not bad at all, Mary.”) “How about another game, kids? Try 13/47 as your target fraction.”

While they are busy on that, let’s return to the first game and discuss it some more, particularly the other, fractional, way to compare the results. This time we need to subtract the fraction sums from the target fraction. That is not so difficult as you might believe if we use the formula that was presented in the page, Fraction Addition, or rather a minor alteration of what was given there, namely:

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

Using this formula on our target and sums fractions, these results are produced:

 7     11     210 - 209      1
  ---- - ---- = ----------- = -----
  19     30        570        570
  7     15     308 - 285      23
  ---- - ---- = ----------- = -----
  19     44        836        836

Again the results appear rather lopsided, even in fraction form, that 1/570 is smaller than 23/836. Checking it in general can be done in 2 ways: fractional and decimally. And again, decimally wins for ease of computation. The 2 differences have these decimal values:

1/570 = 0.001754385964912 or about 0.002

23/836 = 0.02751196172249 or about 0.028

[These figures agree, allowing for rounding of course, with what we found earlier.]

But to do it by fractions requires a bit more time; here’s how I would proceed. I like the “cross products” test.

                          836          13110
                    1         23
                  -----  <  -----
                   570       836

And since 836 is definitely less than 13,110, the left fraction is smaller than the right fraction. [I leave it up to you to ferret out the why’s and wherefore’s. :>) ]

Well, it’s about time to check in on our game players and see who is the winner this time. How’d you do, kids?

John says he used 4 and 50 as his denominators, while Mary chose 5 and 14. So the sums this time are

1      1      54      27             1      1     19
--- + ---- = ----- = -----    and    --- + ---- = ----
4     50     200     100             5     14     70

Getting right down to the nitty-gritty (i.e. decimal comparison strategy), we have these figures:

target: 13/47 = 0.276595745

John: 27/100 = 0.27

Mary: 19/70 = 0.271428571

While John’s sum produced a nice fraction with a terminating decimal form, and pretty close to the target fraction we might add, it’s easy to see that Mary’s choice is just a little bit better. (Girls get the point here!)

Extensions & other comments

Throughout the whole discussion of this game use of the calculator has been assumed at all moments. If one has access to a model which handles fractions, things are a bit more interesting. But such models are not required; the game can be played with even simple four-function types. The players just have to “know” a bit more mathematics. (And that’s not a bad thing either, is it?)

The time element of the game is enhanced if one’s calculator has a “replay” feature (as does the Sharp EL-E300 refferred to at the start of this page). Changing the numbers to investigate a “better choice” is more effecient with such calculators. And after one has gained a lot of experience there is a “trick” that is useful to recognize to make one’s search more efficient. Namely, after all is said and done, the error value (and not the sum of the fractions) is the most important information to be obtained. So one could combine and streamline the computation by using parentheses, or even not using them. For John’s work in the first game it would look like this:

7/19 – (1/3 + 1/30)
or 7/19 – 1/3 – 1/30

Now, by using the cursor keys and the replay feature, the denominators can be adjusted or changed rapidly, then the difference is more quickly
computed, even in decimal form.

Finally, one way to make this game a bit more challenging would be to require three fractions to be added. Now, things really start to get interesting.


While playing around with the target fraction in the first game, I noted this:

1/5 + 1/6 + 1/570 = 7/19, exactly

Trotter’s Own Prime Oddities

For the present, this page is only meant to be a storage place for curious and odd facts that WTM has discovered while researching prime numbers and the prime factorizations of various composite numbers. The dates indicate when the item was submitted for posting in the website Prime Curios.

  • 11: Begin with 11, and continually [i.e. recursively] add the first five powers of 2, but in reverse order (32, 16, …, 2). All sums are primes (43, 59, 67, 71, and 73). Sent 8/8/01
  • 41: The sums of the squares of the first digits with the cubes of the second digits of the primes in the first prime triplet (41, 43, 47) — i.e. ab gives a2 + b3 — are primes as well (17, 43, 359). [Note: 43 produces itself.] sent 8/7/01
  • 164: Its prime factorization is 2 x 2 x 41. Then 12 + 62 + 4(4-1) = 101, a pal-prime. Then changing (4-1) to (4×1) and (4+1) produces two more primes, 293 and 1061, respectively. Sent 8/6/01
  • 168: A factorization of 168 is 3 x 7 x 8. So, 13 + 67 + 88 = 17057153, which is prime. [Note: 17, 05, 71, and 53 are primes as well.] sent 8/7/01
  • 263:Cloning the digits of this prime as exponents in this way — 22 + 66 + 33 — yields another prime: 56687. sent 8/4/01
  • 323 Patrick De Geest has said: “323 doubled up (i.e., 323323) has five consecutive prime factors which when squared and summed, yield 989, another palindrome!” WTM adds: And when those prime factors (of 323323) are merely summed, or cubed before summing, a prime number is the total each time (67 and 15643, respectively). [Note: 15643 happens to be part of a twin prime pair and of a prime quadruple.]
  • 463: Cloning the digits of this prime as exponents in this way — 44 + 66 + 33 — yields a composite (46939), which upon deleting the 9’s, leaves 463. sent 8/7/01
  • 643: Cloning the digits of this prime as exponents in this way — 66 + 44 + 33 — yields a multiple of itself: 46939 (= 73 x 643). Sent 8/4/01
  • 881: Cloning the digits of this prime as exponents in this way – 88 + 88 + 11 — yields a rather interesting result: 33554433, whose prime factorization is 3 x 11 x 251 x 4051. (Note the lengths of the 4 primes). Sent 8/4
  • 881: [2nd var.] Using the clones of the digits of this prime, in reverse manner – 81 + 81 + 18 – yields the prime 17. 8/7/01
  • 997: Cloning the digits of this prime (the largest 3-digit prime) as exponents in this way – 99 + 99 + 77 — yields another prime: 775664521. sent 8/4
  • 997: The largest 3-digit prime AND the sum of the squares of its digits is also a prime (211). sent 8/4

Palindromes and Prime Factorizations

Observe these interesting patterns. Let’s begin with this basic palindrome: 98789. It’s main claim to fame is that it’s the largest 5-digit palindrome that is the sum of three consecutive primes. (Can you find those primes?)
The palindrome is not a prime because its prime factors are 223 x 443. But now notice this: 223 + 443 = 666, the number of the Beast!
We’re not through with this. Watch! Let’s “squeeze” the digits of 98789 from the sides until the “7” disappears and the “8’s” merge into one. This gives us another, smaller palindrome: 989. Still not a prime, but check out its prime factorization and the sum of those factors:

23 x 43 and 23 + 43 = 66.
(Might we not consider 66 as a “baby” beast?)

It looks like we’re on to something here. Let’s continue with a larger palindrome: 9876789. Its prime factorization is

9876789 = 33 x 13 x 19 x 1481

No “beastly” number here, you say. Ah, but look closely as we re-arrange those prime factors a little…

(3 x 3 x 13 x 19) x (3 x 1481)

which yields the following…

2223 x 4443 and 2223 + 4443 =6666

…and it looks as though our beast is growing up!

There are certainly more palindromes to investigate. Try these. Your task is to re-arrange the primes to produce a pair of numbers that has a sum of “all 6’s”. Sometimes it’s easier than others.

987656789 = 72 x 71 x 313 x 907

98765456789 = 61 x 3643 x 444443

9876543456789 = 34 x 17 x 97 x 1697 x 43573

987654323456789 = 172 x 29 x 5303 x 22222223

98765432123456789 = 449 x 494927 x 444444443

9876543210123456789 = 32 x 13 x 6353 x 8969 x 1481481481


Postscript (9/3/01): Another presentation of the concept above can be found in Patrick De Geest’s The World of Numbers, as WON plate 112.

A Closer Look at Another Pattern

In the work above, we highlighted 2 numbers in blue: 1481 and 1481481481. The reason, of course, is that there is something special about them in addition to being primes. The second number shows why: it is composed of a block of digits “148”, repeated three times, then it ends with a “1”.
That should make you wonder about 1481481. It is easily seen that it is not prime — the sum of its digits is a multiple of 3 — so it must have a prime factorization. If you divide it by 3, then 3 again, then by the largest 2-digit prime, you will see a nice result.
Now we have three numbers that form a “family”: 1481, 1481481, and 1481481481. And two of those were prime.
I’ll bet you know what the next question will be, right? Naturally, what happens if we use more blocks of “148”? It should be obvious that our numbers become rather large; so we feel it’s time for a little new notation. We will illustrate our method with the 3rd number: 1481481481.
It has three blocks of “148”. We will show this as (148)3. So with the final “1”, our number looks like this:

(148)3 … 1.

In general, we denote our numbers in this way:

(148)k … 1, where k = 1, 2, 3, 4, …

It just so happens that we have checked the values of k up to 14. Here is what we found:

(148)k … 1 is prime for k = 1, 3, and 4.

Can anybody go further?

Mirror, Mirror, On the Wall

Let’s now turn our attention to the “mirror images” of our numbers. Reversing 1481 gives us 1841. But while 1481 is prime, its reversal is not. Proof: 1841 = 7 x 263. You see, changing the positions of the “8” and “4” made a big difference.

Does reversing digits in 1481481 make any difference? That is to say, could its reverse (1841841) now be prime? Unfortunately, the answer is NO. The reason is that changing digit-order does not change the sum of the digits of the number. It is still a multiple of 3. (Can you find its prime factorization?)

However, for (148)3 … 1, change does have a big effect. Now (184)3 … 1 is composite. Here is a partial factorization. Can you finish it?

1841841841 = a2 x b x 196799

Are you ready for a big surprise now? Here ’tis…

(148)4 … 1 and (184)4 … 1

are both PRIME!

Continuing with this theme, we can now state: (184)k … 1 is composite for k = 5 to 11. Beyond that is unexplored territory.

Sandwich Primes

After further thought, WTM has decided to call any prime that starts with the digit “1”, and ends with the digit “1”, as a sandwich prime.

Our first such prime occurred in the palindrome investigation above: 1481. The extreme digits, the 1’s, serve as the “slices of bread”, and any other digits represent the fillings.

And if we continue repeating the block of digits as shown in other numbers above, we have a refinement in our new name: Dagwood primes! (Recall the famous character in the comics, Dagwood Bumstead, who often made multi-layer sandwiches with extra slices of bread separating his fillings.)

So our first Dagwood prime to be offered is this: 1481481481.

Our investigation of sandwich primes has turned up some interesting results, which we will share with you now.

We begin by noting that our research of 1481 was inspired by the factorization of the palindrome 9876789, and then the factor 4443. The factors of 4443 are 3 and 1481. So it seemed a natural extension to examine the number 5553.

Step 1: 5553 = 3 x 1851. But 1851 is not prime; it is 3 x 617.

Step 2: Let’s repeat it in this manner: 1851851. Bingo! A sandwich prime, of the Dagwood variety!

Step 3: And repeating again — 1851851851 — yields an even bigger prime!

Step 4: Unfortunately, further repetitions, up to k = 12, yield no more primes.

We may summarize the foregoing this way: (185)k … 1 is prime for k = 2, 3.


Reversing the digits in this manner gives this:

(158)k … 1 is prime for k = 2, 12.

Wow! Look at that last value for k. That’s special. Here it is, in full glory:


Here is a table, summarizing all the data gathered to date (k < 13):

Number Form Primes when k =
(148)k … 1 1, 3, 4
(184)k … 1 4
(158)k … 1 2, 12
(185)k … 1 2, 3
(123)k … 1 1, 2
(132)k … 1 1, 6, 10
(147)k … 1 1, 7
(174)k … 1 1, 2
(138)k … 1 1, 2
(183)k … 1 1, 2, 3, 4, 6, 11
(115)k … 1 1,
(151)k … 1 1, 3, 4, 6
(102)k … 1 1, 4, 5
(120)k … 1 1, 2, 3, 7, 12
(103)k … 1 1,
(130)k … 1 1, 3
(106)k … 1 1,
(160)k … 1 1,
(109)k … 1 1, 4, 12
(190)k … 1 1, 7

Trotter in Prime Curios


In the latter part of May of this year (2001) we discovered a very interesting website all about prime numbers, titled appropriately The Prime Pages. There is a companion page connected with it, called Prime Curios, a collection of clever and interesting trivia, moderated by G. L. Honaker, Jr. It is to this 2nd site that this WTM page is concerned.

Naturally, we began submitting our own contributions right away. First, we sent the one about 1992 that appears at the beginning. Then others began to follow in rapid succession. The list soon grew rather lengthy, so we decided to place ours in one location. So what you will see and read below is the results of our number play since that time. Nothing is in any particular order, unless otherwise indicated.

Each entry in the list is preceded by a link (the highlighted number) that will take you to the specific page in the website of Prime Curios, so that you may read all the other interesting facts that other people found about that particular number. We think you will be quite surprised and well rewarded.

So go forward now, and most of all, have fun with numbers!

  1. 1992 1992 = 8 x 3 x 83. The only other two years in the period 1000 to 1999 to share the structure of “a x b x ‘ab'” (where ‘ab’ is prime and the concatenation of the factors a and b) are 1316 = 4 x 7 x 47 and 1533 = 7 x 3 x 73.
  2. 1992 If you separate the digits of 1992 like 199 2, you have two primes. Note that 199 is the largest prime less than 2 hundred. Separation in the middle gives us 19 92, and 19 turned upside down along with 92 reversed are prime.
  3. 2310 2310 = 2 x 3 x 5 x 7 x 11 = 112 + 132 + 172 + 192 + 232 + 292. Note the consecutive digits: 0, 1, 2 and 3.
  4. 197 197 is the only three-digit prime Keith number.
  5. 26 The prime factorization of 26 uses the first three counting numbers.
  6. 17 17 is the smallest Trotter prime, i.e., a prime of form 10 x (n2) + 7, where n = 1, 2, 3 …
  7. 1234567 The prime factors of 1234567 (127 x 9721) form a peak-palindromic arrangement of digits (1279721). It is curious that the prime factors of 1279721 are all three emirps (79 x 97 x 167) which upon concatenation form yet another prime (7997167). [Note: see this reference…World of Numbers for more information.]
  8. 36 The smallest square that is the sum of a twin prime pair: 17 and 19.
  9. 8 8 is the smallest cube which is the sum of a twin prime pair {3 + 5}.
  10. 23 2n + 3n is prime for n = 0, 1 and 2.
  11. 23 23 is the smallest prime for which the sum of the squares of its digits is also an odd prime.
  12. 31 The number of letters (in English) required to write the word names of the first six primes is the reverse of the sixth prime (13), namely 31.
  13. 73 73 is the smallest prime whose square (5329) is the concatenation of two multi-digit primes.
  14. 73 389 and 17 are primes, as is their concatenation (38917). Inserting the lowly 0 between them transforms the prime into a power, i.e. 389017, the cube of 73, a prime itself.
  15. 83 The cube of 83 (571787) is the smallest case of the concatenation of a pair of 3-digit primes.
  16. 211151 The smallest Xmas Tree Prime with 3 rows, i.e. 6 digits.
  17. 2111511013 The smallest Xmas Tree Prime of 4 rows, i.e. of 10 digits.
  18. 211151101310867 The smallest Xmas Tree Prime of 5 rows, i.e. of 15 digits.
  19. 3883 This palindrome is transformed into a pal-prime upon the insertion of a single 0 in the middle.
  20. 121661 121661 is the smallest OP-PO Prime.
  21. 72727 72727 is a pal-prime, and separating the digits so — 72 and 727 — we can now say the sum of the digits of the first 72 primes (2, …, 359) is 727, another pal-prime. (P.S. I’m submitting this on 7/27 of this year!)
  22. 277 76729 is the square of the prime 277 and the smallest square with 5 or more digits that is the concatenation of three primes (7, 67, and 29). (Note: the square contains the digits of the root embedded in reverse order.)
  23. 159 159 = 3 x 53, and upon concatinating the prime factors, we have a peak palindrome, 353, which is itself a prime.
  24. 159 Its square (25281) is the concatenation of 2 primes: 2 and 5281.
  25. 77 The square of 77 is 5929, the concatenation of two primes, 59 and 29.
  26. 181 181, a pal-prime, is the sum of the digits of the first 23 primes (2, …, 83).
  27. 217 47 and 89 are primes, as is their concatenation (4789). Inserting the lowly 0 between them transforms the prime into a power, i.e. 47089, the square of 217.
  28. 311 311 is the 11th three-digit prime for which the sum of the squares of its digits is also a prime; and the sum here is 11 as well.
  29. 2357 21 + 33 + 55 + 77 and 22 + 33 + 55 + 77 are twin primes. [Trotter, Kulsha]
  30. 2357 Letting A=1, B=2, …, Z=26, then 2357 is the sum of all the values of the U.S. Presidents’ last names from Washington to Coolidge. [Ref. Wordsworth.]
  31. 2357 2357 is also the sum of consecutive primes in at least two ways: (773 + 787 + 797) and (461 + 463 + 467 + 479 + 487).
  32. 17 Using A = 1, B = 2,…, Z = 26, 17 is the smallest non-negative number whose numerical value of its word form is also a prime (109).
  33. 7 Using A = 1, B = 2, …, Z = 26, the sum of all the letter values in the word names of the numbers from 1 to 7 is a prime (367). If 0 is included, the sum is yet another prime (431).
  34. 64 Using A = 1, B = 2, …, Z = 26, the sum of all the letter values in the word names of the numbers from 1 to 64 is a prime (7369). If 0 (whose alpha-numeric value is 64) is included, the sum is yet another prime (7433).
  35. 5 Only 5 U.S. Presidents have 5 letters in their last names (Both Adams, Grant, Hayes, and Nixon).
  36. 23 23 = 14 + 23 + 32 + 41 + 50.
  37. 13 Using the first three primes we have: 23 + 5 = 13.
  38. 111 111 equals the sum of 2 + 3 + 4 + … + 17 minus the sum of the primes less than 17.
  39. 131 131, a palindromic prime, equals the sum of 2 + 3 + 4 + … + 19 minus the sum of the primes less than 19.
  40. 414 The exponential factored form of 414 (2 x 32 x 23) consists of three 2’s and two 3’s; whereas its expanded form (2 x 3 x 3 x 23) has two 2’s and three 3’s.
  41. 434 434 is also the sum of the cubes of the digits of the emirp 347.
  42. 821 The smallest prime of the first prime quadruple for which the sums of the cubes of the digits of the 4 primes (821, 823, 827, 829) are primes themselves (521, 547, 863, 1249).
  43. 440 The sum of the first 17 primes (2 to 59) and also the number of yards in the quarter-mile race in track-and-field competitions.
  44. 997 The largest 3-digit prime AND the sum of the cubes of its digits is also a prime (1801).
  45. 137 The sum of the squares of the digits of 137 is 59, another prime, and all five odd digits are used (Ref. Father Primes).
  46. 317 The sum of the squares of the digits of the prime 317 is 59, another prime. Note that all odd digits are present.
  47. 165 165 is a multiple of (16 – 5), which is its largest prime factor.
  48. 132 The concatenation of its three distinct prime factors (2, 3, and 11) forms primes in three ways: 2311, 2113, and 1123.
  49. 131143 This prime is composed of three 2-digit primes — 13, 11, and 43.
  50. 123456789 Replacing each of the digits, one-by-one with a 0, yields primes in three cases: 1, 2, and 7 (023456789, 103456789, and 123456089). Note that 127 is a Mersenne prime.
  51. 37 37 + 4n yields primes for n = 1, 2, 3, 4, 5, 6, 7.
  52. 45 If the first five powers of 2 (2, 4, 8, 16 & 32) are each subtracted from 45 all results are primes (43, 41, 37, 29 & 13).
  53. 170 The 170th Trotter number (289007) has an all-emirp prime factorization: 37 x 73 x 107. Note: the 3rd prime is a permutation of the digits of the original number.
  54. 304589267 A prime containing 9 distinct digits, where upon inserting symbols [30/45 + 89/267], we discover the missing digit “1”. The only other prime for which this is possible is 536948207. [Trotter and Knop]
  55. 13831 13831 is the smallest multi-digit palindromic prime such that the sum of it with the next prime (13841) is a palindrome (27672).
  56. 10501 A palindromic prime that is the sum of 3 consecutive primes (3491, 3499, and 3511), while at the same time serving as the middle prime of a set of three consecutive primes whose sum is another palindromic prime (31513).
  57. 97679 96769 is the largest 5-digit palindromic prime that is the sum of 3 consecutive primes (32251 + 32257 + 32261).
  58. 94949 The only 5-digit palindromic prime that is undulating and the sum of 3 consecutive primes (31643 + 31649 + 31657). Note that by adding 3 consecutive primes we only get one other undulating palindrome (16161), which is a non-prime.
  59. 98789 The largest 5-digit palindrome that is the sum of 3 consecutive primes (32917 + 32933 + 32939). Its prime factorization is 223 x 443 and 223 + 443 = 666!
  60. 13124…97909 (24-digits) A prime composed of eight 3-digit palindromes of a “consecutive” style, and one-nineth of (118)8 … 1.
  61. 11111117 11111117 and 71111111 are both primes, thus emirps.
  62. 742950290870000078092059247 (27-digits) The first prime in an arithmetic sequence of 10 palindromic primes. It was found by Dubner and his assistants and has common difference of 10101 x 1011.
  63. 144169 It is also the concatenation of three squares (144, 16, and 9). Note that: sqrt (144) = sqrt (16) x sqrt (9). [Note: this is an extension to another person’s “curio”.]
  64. 8609The largest distinct-digit pime. Pimes (pronounced with a long i) are primes whose digits contain circles, i.e., using only the digits 0, 6, 8, 9. Note: 6089 and 8069 are also distinct-digit pimes.
  65. 174 174 = 72 + 53 (using the first four primes).
  66. 199 199 is also a Permutable prime, meaning that 919 and 991 are primes as well.
  67. 2213 2213 is a “sum of cubes” as follows: 23 + 23 + 133.
  68. 2222 The smallest number divisible by a 1-digit prime, a 2-digit prime, and a 3-digit prime.
  69. 1429 The prime sum of two famous baseball records: 714, number of homeruns hit by Babe Ruth, and 715, number of the homerun hit by Hank Aaron to break the Babe’s record (on 4/8/1974).
  70. 202 A semiprime palindrome equal to (2 + 3 + 5 + 7)2 – (22 + 32 + 52 + 72). It’s the only such case for all primes < 2,000,000,000. [Trotter and De Geest]
  71. 576 A square equal to (2 + 3 + 5 + 7 + 11)2 – (22 + 32 + 52 + 72 + 112). It’s the only such case for all primes < 2,000,000,000. [Trotter and De Geest]
  72. 223 The sums of the nth powers of its digits are prime for all n between 1 and 6 inclusive: sum of digits = 7, sum of squares of digits = 17, sum of cubes of digits = 43, sum of fourth powers = 113, sum of fifth powers = 307 and sum of sixth powers = 857.
  73. 607565706 A palindrome resulting from this “prime based” expression: (257 + … + 607)2 + (2572 + … + 6072). There are 57 consecutive primes inside each parentheses. Note that this palindrome starts with the last prime added: {607}565706. “57” appears also as a substring, 60756{57}06. The number of the beast {6}075{6}570{6} is included. [De Geest and Trotter]
  74. 107 Rudy Giuliani was the 107th mayor of New York City.
  75. 2003 There is only one way to use consecutive integers to produce a sum of 2003.
  76. 20022002 The prime factorization of 20022002 is 2 x 7 x 11 x 13 x 73 x 137, which when grouped thus, 2 x 11, 13 x 73, and 7 x 137, yield 3 palindromic semi-primes: 22, 949, and 959.
  77. 1951 1951 is prime, and appears as the 9th term of the sequence 1+9+5 = 15, 9+5+15 = 29, 5+15+29 = 49, etc.
  78. 98689 The first centered triangular number (i.e. of the form the form of (3n2 – 3n + 2)/2) that is a palindromic prime.

This next group of 7 items is the result of some email correspondence we had with Carlos Rivera (6/27/01). [See Potpourri for that email.] We posed the basic idea, and Carlos provided us with the numbers. (See above on 36 and 8 for the cases of square and cube.)

  1. 253124999 The smaller of the smallest twin prime pair for which the sum is a 4th power (sum = 1504).
  2. 4076863487 The smaller of the smallest twin prime pair for which the sum is a 5th power (sum = 965).
  3. 578415690713087 The smaller of the smallest twin prime pair for which the sum is a 6th power (sum = 3246).
  4. 139967 The smaller of the smallest twin prime pair for which the sum is a 7th power (sum = 67). (Note: this prime ends with a 6 and 7.)
  5. 14097…72287 (26-digits) The smaller of the smallest twin prime pair for which the sum is an 8th power (sum = 15188). [The entire number is 140975 6730907423 9886172287.]
  6. 73099303486215558911 The smaller of the smallest twin prime pair for which the sum is a 9th power (sum = 1749).
  7. 8954942912818222989311 The smaller of the smallest twin prime pair for which the sum is a 10th power (sum = 16810).

The next items are the results of looking at someone else’s “curio”, then expanding a little on it. We suggest that you go to the Prime Curio page to see the full number and the name of the original submitter.

  1. 15555…55551 (33-digits) The digital sum of this prime is 157, another prime (whose digit sum in turn is yet another prime: 13).
  2. 10220…02201 (55-digits) The digit sum of this prime is 110, which is the double of its number of digits.
  3. 14444…44441 (67-digits) The digit sum of this prime is 262, a peak palindrome.
  4. 18181…18181 (77-digits) The digit sum of this 77-digit prime is 343, the cube of 7.
  5. 31313…31313 (83-digits) The digit sum of this prime is the prime 167.
  6. 19999…99991 (87-digits) The digit sum of this prime is a palindrome, 767.
  7. 37777…77773 (87-digits) The digit sum of this prime is 601, another prime.

Finally, we present some items that are harder to categorize. The first one was created by the moderator of Prime Curios, G. L. Honaker, Jr., after we wrote our webpage on Trotter Numbers and Trotter Primes. The second one is a two-person contribution, involving Monte Zerger (whose name and creations can be found in other WTM pages) and ourselves (WTM).

  • 735 There are exactly 735 Trotter primes less than 100,000,000. Note the first three odd primes in 735. [Honaker]
  • 510 The concatenation of 510 with itself (510510) is the product of the first 7 primes and also the product of the 7th through 10th Fibonacci numbers (13, 21, 34, and 55). [Zerger](Continuing the previous curio) The difference between the next prime (19) and the next Fibonacci number (89 – also a prime) is 70, which is the product of the Fibonacci subscripts above. [Trotter]
Note: Some of the above items have been removed from the Prime Curios page, though they at one time were indeed posted. Still it doesn’t alter the basic facts about any given entry. It was merely a decision taken later by the site moderator.