Tag Archives: sums

A Nutty Problem

Mama Squirrel found a bag of nuts in the driveway of the house near her tree home. She decided to separate the nuts into equal portions for her children.When she tried putting 6 nuts on each plate, unhealthy one squirrel baby was left out and received no nuts at all. (So there was one sad little squirrel now.)

But when she tried putting 5 nuts per plate, cialis there was one nut left over. (At least nobody was unhappy this way.)

What is the sum of the number of the nuts and the number of squirrel children?

NOTE: Before writing out your answer, please check our Guide lines for Writing POW Answers.


The Professors’ Ages

If you were solving AlgPoW problems from the Math Forum in the year 2001, and in particular the one titled “The Professors’ Primes“, (May 7) you met three of my favorite university professors: Drs. Ken Travers, Peter Braunfeld, and Wilson Zaring. Before I left them to finish their lunch in the cafeteria that day, I inquired as to what their ages would be when students would be returning to school for next year (September, 2001). They were kind enough to oblige; hence I have some data for this problem you are about to work now.Upon careful examination of their three ages, I have formulated the four following facts. I am using the initial letter from each man’s first name to represent his age, respectively.


  1. 60 < K < P < W < 80
  2. The sum of the older two professors’ ages is a square number.
  3. The sum of the oldest and youngest professors’ ages is the first palindrome less than the square.
  4. The sum of the younger two professors’ ages is the 2nd prime less than the palindrome (of statement #3).

With that information, give me the “digit sum” of the product of the three professors’ ages.

NOTE: show enough steps of work so that any reader can follow your reasoning easily. The variables for your equations are already given, but restate them for clarity as part of your solution process.

BONUS: What is the “digital root” of the product of the three professors’ ages?

Before you try writing up your solution, it is highly recommended that you check out the Guidelines for Writing PoW Answers.

You may send your answers via email to me at: trottermath@gmail.com or ttrotter3@yahoo.com. Or click on the icon below.


A “Mean” Product

My mother once taught in a small country three-room schoolhouse, ampoule where students of different age levels often had to study the same course together. For the course in American History, salve the kids in the 6th, 7th, and 8th grades were combined into one class. This means that their ages were between 10 and 14.

If the product of their ages was 41,760,576, what was the average, or arithmetic mean, of their ages?

Two Differences to a Sum

Two positive numbers are such that their difference is 7 and the difference of their squares is 105. What is their sum?
Your task:

1. Show your kid brother (or sister) what this means by solving it in an arithmetical manner. Be patient.
2. Show your parents how much you’ve learned in Algebra 1 this year by solving it with basic algebra. This means, you just demonstrate what the sum is, without determining what the two numbers are.

A Tale of Two Problems

When I see a good problem somewhere, I like to investigate its properties further and deeper, with the intention of using it to develope problem solving skills in my students. Such was the case when I saw the two problems that you will see in this page of WTM. I hope you will agree with me that they can be useful to bring inter-esting math challenges to young students at the upper elementary levels (3rd-6th).

Problem #1

The first problem appeared in the 1994 MATHCOUNTS contest exams (School level, Sprint #5). It said:

Two positive numbers are such that their difference is 6 and the difference of their squares is 48. What is their sum?

The foundation concept of this problem is a perennial topic in all high school Algebra I courses: the difference of two squares pattern. It occurs in the chapters on multiplying and factoring binomials. Solving such a problem in an algebra course is, there-fore, a somewhat regular, if not trivial, matter.

A possible solution process might go as follows:

1. x2 – y2 = (x + y)(x – y)

2. x – y = 6 and x2 – y2 = 48

3. 48 = (x + y)(6)

4. x + y = 8

However, elementary students are not expected to work at such an abstract level of thinking. But if they have access to calculators and a little basic guidance in understanding what the problem is all about, they can enjoy a meaningful experience just the same. We proceed by setting up a t-chart to organize our work.

Filling out the entries — by educated trial and check — now becomes an easy task. In fact, for this MATHCOUNTS problem it is a rather quick one: A = 7 and B = 1; thus the sum is 8. This is merely because it was part of a large set of problems to be solved under a time limit. Hence it was
not intended to be a hard, time-consuming item. Also it should be pointed out that calculators are not allowed on this portion of the contest.

However, if number size is increased (moderately at first) and time is removed as a factor, many exercises can now be formulated. Here are some examples:

1. Two positive numbers are such that their difference is 6 and the difference of their squares is 180. What is their sum?

2. Two positive numbers are such that their difference is 7 and the difference of their squares is 161. What is their sum?

3. Two positive numbers are such that their difference is 10 and the difference of their squares is 260. What is their sum?

4. Two positive numbers are such that their difference is 15 and the difference of their squares is 555. What is their sum?


1. A = 18, B = 12, & sum = 30.
2. A = 15, B = 8, & sum = 23.
3. A = 18, B = 8, & sum = 26.
4. A = 26, B = 11, & sum = 37.

Calculator Connection

The work on these problems can be made a lot easier and more efficient if one uses certain special features of calculators. First, if one is using an ordinary 4-function nonscientific model, here is an interesting shortcut method that takes advantage of the memory keys. Using the answer of #4 above, the method goes this way:

0. Make sure the memory register is clear.

1. Press: 26, [x], [M+]. (This puts A2 into the memory.)

2. Press: 11, [x], [M-]. (This computes the square of B and subtracts it from the value from A2.)

3. Press: [MR] (or [MRC]). (This shows the difference.)

Of course, if one is using a regular scientific model, the steps are even shorter and memory need not be utilized to obtain the same results. The key sequence would be as follows:

26 [x2] [-] 11 [x2] [=]

Problem #2

This problem likewise appeared as part of the same MATHCOUNTS contest; it was #23 on the Sprint round.

What is the smallest multiple of 5 the sum of whose digits is 18?

We should remind ourselves once again that this is a timed contest and no calculators permitted. Hence, one might be expected to solve this question analytically, perhaps as follows:

Since all multiples of 5 end in 5 or 0, and our desired multiple must contain at least 3 digits, we are looking for a value in one of these two forms:
aa0 or bc5. The only number in the first form to have a digital sum of 18 is 990. But it could not be the smallest one because the b-c digits of the
second form will certainly be smaller due to the help of the 5. Of course, the sum of the b-c digits will then be 13, the only possibilities being 4 & 9, 5 & 8, and 6 & 7. Therefore, the smallest multiple will be produced by the pair containing the smallest digit, which is 4 & 9. So the problem’s answer is 495.”

It might be noted here that a new question can be asked using the facts presented in the above solution. It would be:

How many multiples of 5, less than 1000, have a sum or their digits that is 18?The answer of seven is easily seen by making a list like this:

495 585 675 990 945 855 765

But now let’s bring this whole situation down to a basic, more elementary level, one that uses calculators, mental addition, and emphasizes more strongly the concept at the heart of the original problem, multiples of a number. We might proceed as follows:

1. Present the problem with as little initial explanation as possible and allow the student time to wrestle with it.

2. Then as necessary, direct the student to use the calculator to produce a series of multiples of 5 by utilizing its constant addition feature. [For many nonscientific models, this simply means pressing “5”, [+], [=], then continuing with the [=] key as often as needed.

3. It is here that the attack could take two different directions: (a) making a t-chart of the multiples and their digital sums, observing patterns along the way; or (b) simply adding the digits mentally, and as rapidly as possible, as one presses [=]. Each strategy has its positive and negative side; the learner should choose whichever way seems best.

It should be obvious that many more problems of this nature can be posed by changing either the multiples’ factor, the digit sum, or both. Here is an example of each style:

1. What is the smallest multiple of 6 the sum of whose digits is 18? [Ans. 198]

2. What is the smallest multiple of 5 the sum of whose digits is 15? [Ans. 195]

3. What is the smallest multiple of 7 the sum of whose digits is 16? [Ans. 196]


This is a prime example of how one simple problem can be turned into many, and in which important concepts are present and yet basic skills can be practiced. Additionally, it is a case where students could be encouraged to invent their own problems, thus becoming a more integral part of the learning process, a factor often overlooked in many math classrooms today.

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.

Amicable Numbers

[Note: In another article we presented the concept of Perfect Numbers, and the related two classes of numbers, Deficient and Abundant. If you have not read that one, it would be a good idea to do so now before proceeding here. The topic before us now is a natural extension of that article.]

For centuries men have felt that numbers had powers over the natural affairs of their lives. One small piece of evidence of this concept is that of Amicable, or Friendly, Numbers. These are merely

pairs of numbers that are connected in a special way. The math behind the relationship is the same that was used to classify numbers as either perfect, deficient, or abundant: the sum of the proper divisors of the number, or numbers, being considered. An example will make this clear:

The number pair of 220 and 284 are considered amicable, because the sum of the proper divisors of 220 is 284, while the sum of the proper divisors of 284 is 220. Got that? Look at these factor charts:

	          220		  284
		---|----	---|----
		 1 | 		 1 |
		 2 | 110	 2 | 142
		 4 |  55	 4 |  71
	         5 |  44
	        10 |  22
	        11 |  20

[NOTE: In this sort of work, divisor and factor are synonyms.]
The sum of the divisors in the larger chart {1, 2, 4, 5, 10, … , 110}
is equal to the other number (284), whereas the sum of the divisors in
the smaller chart {1, 2, 4, 71, 142} is the first number (220).

Nice, huh!

I have read that in times long ago when romantic love was, well, more cerebral than material, amicable numbers were used by a young man to express his love to his beloved. He would prepare two metal charms with the numbers 220 and 284 engraved on them. These were then put on chains to place around their necks. Maybe we should do it these days, too.

Just as mathematicians have searched long and hard to find perfect numbers, they have done the same for amicable ones. And as
you might expect, the sizes of the numbers soon become very large. (Some are so large that if a lover used them to make a love token, the charm would be a “back breaking” burden for the wearer.)

About 1200 pairs are known by the last count that I have access to.* Here are the smaller members of two different pairs: 2620 and 5020. I leave it as “homework” for you to find the partners that apply
to each of them.

But there is one pair that mathematicians missed in their search down through the centuries which was found by an Italian schoolboy in the last century. One number of the pair is 1184. We leave it to you, dear reader, to find its “loving” companion. [Hint: look in the cartoon above.]


* Joseph S. Madachy, Mathematics on Vacation, 1966, p. 155.

Postscript: 7/5/99

Recently David Einstein informed me that the figure given by Madachy’s
book is now very outdated. There are now more than 450,000 pairs of
amicable numbers. For more information about this, go to this website:

Postscript: 2/2/02

Here are some more sites with additional information:

Perfect Numbers et al.

How many perfect people do you know? Or, better, how many perfect people have lived in this world of human beings? Not many, I’m sure you all would agree. Well, the same is true, in a relative sense of the word, in the world of numbers. The ancient Greeks, who held a great reverence for the mysticism of numbers, had a certain category of numbers called “Perfect Numbers“. A perfect number is nothing more nor less than a positive integer whose proper divisors have a sum of the number itself. (It should be understood here that while any number is a divisor of itself — that is, 12 is a divisor of 12 — that’s considered to be an improper divisor. And so it is not used in this activity.) Now it just so happens, that like the scarcity of perfect people, there are not many perfect numbers, hence they become special and worthy of our attention.

The first, and therefore smallest, perfect number is 6. Its proper divisors are 1, 2, and 3. Their sum is, right, 6. The next perfect number is 28, as its proper divisors are 1, 2, 4, 7, and 14. Again, the sum of those integers is 28.

Simple as it now appears, why did I state that there are so few of this kind of number? Well, it’s because the 3rd one does not appear until nearly 500; the 4th one is over 8,000; and the 5th one is over 33 million!!! Does that put things into perspective now? You just don’t bump into a perfect number every day, at least those of the larger size.

It is, in part, for this reason that I’m always pleased to point out to my friends, collegues, and students a curious fact about my personal life. My wife’s name is Gloria, and mine is Terrel. Both names contain 6 letters. But wait, the best is yet to come. We were married on June 28. Note that June is the 6th month, so the digit form of that date is (ta-dah!) 6/28. Two perfect numbers.

In fact, the only two perfect numbers small enough to form a date in the calendar. In other words, 6/28 might be considered as the “perfect date” for a mathematician to get married!

Interesting facts:

  1. There are only 37 known perfect numbers! Most of them are so large that it would take many, many papes of paper just to write one such number. The number of digits for the 37th one, just recently found by the way (1998) is 1,819,050. Here is the way we can represent it via exponents: 23021376 × (23021377 – 1)[Note: See the formula in the heading of this article.]
  2. All known perfect numbers are even. No odd ones have been found as yet. But this is no proof that none exist.

You should be asking yourself the question: what about all those other numbers? How do they fit into the framework of the Greek mind? The answer is quite straightforward, actually, if you think about it. There are 3 things that can happen with the sums of the proper divisors of numbers: they can be equal to the number itself, as just shown above, or they can be lesser, or they can be greater, than the number itself. Some examples will illustrate what we mean by this:

  1. Take 10. Its proper divisors are 1, 2, and 5. Sum: 8.
  2. Take 12. Its proper divisors are 1, 2, 3, 4, and 6. Sum: 16.

There you have it. The Greeks called numbers that fall into category 1 as deficient, and those that fall into category 2 as abundant. The names are really quite self-descriptive, don’t you think?

Now armed with this new knowledge, you are ready to classify any number as deficient or abundant. Or if you are lucky, perfect.

Have fun!


In case you’re still wondering about the 3rd, 4th, and 5th perfect numbers alluded to above, I’ll give hints for the 3rd and 4th:

— one can be found between 490 and 500.

— the other can be found between 8,120 and 8,130.

The 5th one, however, is quite big, so I’ll just state it:


A reasonable math class challenge is to find all its divisors. With a

calculator it’s a snap!

Update: 2/2/02

For a listing of the 39 (it’s not 37 anymore!) Perfect Numbers, click HERE.

Charlene Numbers

The basic concept needed to solve this problem is that of the digit sum of a number. The digit sum of a number is nothing more nor less than the “sum of the digits of a given number.” For example, the digit sum of 283 is 2 + 8 + 3, or 13.

To make an interesting and reasonably challenging activity for elementary students, we can proceed as follows:

1. Pick any number. 283
2. Find the sum of its digits. 13
3. Add these two numbers. 296

That’s it. But that in itself is not much, you say. Of course, you’re right. The challenge lies in reversing the whole situation. Ask the students to find a number that will produce a specific result or sum at Step 3. That is, given something like 296 for the “sum” in Step 3, the students are expected to come up with 283 as the “answer.”

Let’s illustrate what we mean. “Okay, kids, now that you know how to do the 3-step procedure, let’s reverse things just a bit. Can you find a number that after you do the process on it you will have a sum of say, 152?”

[Draw a rectangular box to represent the top addend. Put a smaller box under it to represent the digit sum. Underline, and place the “+” sign. Write the number 152 under the line to form the vertical format that most students are accustomed to.]

Now say: “What could we put in the large box?” What do you think would be a reasonable guess for the number we’re looking for?” Try to elicit something smaller than 152, of course, but not too small. (This is where good number sense comes in handy.) A possible line of reasoning might go as follows:

1) Try 145 145 + 10 = 155 [Too high]
2) Try 144 144 + 9 = 153 [Still too high but closer]
3) Try 143 143 + 8 = 151 [Too low]
4) Try 139 139 + 13 = 152 [That’s it!]

Continue the investigation with other numbers, especially larger numbers like the one used in the MATHCOUNTS problem. Warning! Some “answer/sum numbers” can not be produced by the digit-sum addition process. Can you find some of them? Also, some “answer/sum numbers” can be produced by two different numbers. Can you find some of this type as well?


After the above material was originally written, I came across a nice idea in math: naming categories of numbers according to some particular characteristic they have. (See Keith Numbers.) So I have decided to call those numbers that do not have a solution in the procedure described above as CHARLENE NUMBERS.

tt(May 1998)

Update (7/24/98)After the above material was written, I was rumaging through some old notes and clipped magazine articles that I had gathered over past years (and were frankly gathering dust!). And lo and behold! The concept of the MATHCOUNTS problem goes back a lot further than I had remembered. It seems it has its genesis with that mathematician from India, named D. R. Kaprekar. (Clickhere to see another of his activities that I wrote about in this website and that is often mentioned in the literature.)

Now the situation is this: what I had called Charlene Numbers earlier already had a name, Kaprekar’s self-numbers. Kaprekar called them by that name because they were “self-born”, that is to say, not the result of adding the digit sum of a smaller number (which he called the generator of N) to produce that number. All numbers that are not self-numbers he called generated numbers because they were the results of the digit sum addition process, that is to say, “generated” by a smaller number.

My source of this and what will be explained below are two articles from Martin Gardner’s Mathematical Games columns in the March and April 1975 issues of SCIENTIFIC AMERICAN magazine.

First, Kaprekar called this unusual algorithm “digitaddition”. Sequences of numbers using this process can easily be constructed by any 3rd-grade schoolchild. For example, beginning with 47, one would have

47, 58, 71, 79, 95, … .Of course, such a sequence has no end, so there could be no sum of the terms of an infinite sequence of this type, as any schoolboy is supposed to learn in advanced mathematics. But, unlike that advanced math class, there is no “easy” (formula) way to find the sum of a finite series either. You just have to “add ’em up”. For the partial sequence given above — 47 to 95 — the sum is 350.

But, Kaprekar did find an interesting short cut process for finding the sum of all the digits involved in such partial sequences. He said you merely subtract the first and last terms and add to that result the sum of the digits of the last term. For our example, we have

95 – 47 + (9 + 5)
= 48 + 14
= 62.

As he himself said, “Is this not a wonderful new result? The Proof of all this rule is very easy and I have completely written it with me. But as soon as the proof is seen the charm of the whole process is lost, and so I do not wish to give it just now.” (Gardner, March 1975, pp. 113-4)

As I discovered and mentioned earlier, some numbers can even be produced by more than one number or generator. Of course, Kaprekar knew all about this, too, back in 1949, in fact. He had found that many numbers could have more than one generator, calling such numbers by the name junction numbers. The smallest junction number, having 2 generators, is 101; its generators are 91 and 100. The smallest junction number having 3 generators is

10,000,000,000,001; its generators are 10,000,000,000,0009,999,999,999,901, and 9,999,999,999,892. That’s getting pretty large, I’d say.

But in 1961, Kaprekar said he found the smallest junction number with four generators, a 25-digit number. It is
The generators, however, were not given in the article. Can you find them?

Gardner mentioned that numbers that were primes and self-numbers could be called self primes. (Sounds logical, I’d say.) 3, 5, 7, 31, and 53 are some small examples of this. And likewise for year numbers, mentioning that the years in this century prior to the time he wrote these articles that were “self-years” are the following:

1908, 1919, 1930, 1941, 1952, 1963, and 1974.[I found this interesting because my birth year is one of those self-years! Which one? Ah, I’ll never tell.]

Gardner then comments that 11,111,111,111,111,111 (or repunit-17) and 3,333,333,333 are self-numbers.

Next, what about special sets of numbers. Gardner says that the powers of 10 produce some interesting results.

Power of 10 Generator
10 5
100 86
1000 977
10,000 9968
100,000 99,959

But one million is not there. Why? You guessed it; it’s a self number! It has no generator. (Could we say that like many millionaire people are “self made”, likewise the number 1,000,000 is “self born”?) The next self-power-of-ten is 1016.

The March article closes with this comment: “No one has yet discovered a nonrecursive formula that generates all self-numbers, but Kaprekar has a simple algorithm by which any number can be tested to determine whether it is self-born or generated.” He challenges the reader to find it and find the next self-year after the one given in the list above. In the April issue he gave the answers, which I give below verbatim.

D. R. Kaprekar’s method of testing a number, N, to see if it is a self-number is as follows. Obtain N’s digital root by adding its digits, then adding the digits of the result and so on until only one digit remains. If the digital root is odd, add 9 to it and divide by 2. If it is even, simply divide by 2. In either case call the result C. Subtract C from N. Check the remainder to see if it generates N. If it does not, subtract 9 from the last result and check again. Continue subtracting 9’s, each time checking the result to see if it generates N. If this fails to produce a generator of N in k steps, where k is the number of digits in N, then N is a self-number. For example, we want to test the year 1975. Its digital root, 4, is even, so that we divide 4 by 2 to obtain C = 2. 1975 minus 2 is 1973, which fails to generate 1975. 1973 minus 9 is 1964. This also fails. But 1964 minus 9 is 1955, and 1955 plus the sum of its digits, 20, is 1975; therefore 1975 is a generated number. Since 1975 has four digits, we would have had only one more step to go to settle the matter. With this simple procedure it does not take long to determine that the next self-year is 1985. There will be only one more self-year in this century: 1996.

Closing Comments

In light of the information gleaned about Kaprekar’s work in Gardner’s articles, I feel I must change my original intention about the name “Charlene numbers”, because they are in reality already named as self-numbers. But I now propose that all the non-self-numbers with only one generator carry that name; I don’t see any charm in calling them generated numbers, beyond the fact that such is rather more descriptive of their true character. It’s just that I like the sound of Charlene Numbers better. Also the year number used in the MATHCOUNTS problem is a generated number anyway, not a self-number; nor was the year of her birth, 1974. [By the way, did you figure out that Charlene was 21?] Then the remaining non-self-numbers can be called junction numbers, as Kaprekar himself did.As a classroom math activity, I would hope that the teacher would not reveal all the secrets covered in the articles by Gardner, especially the shortcut for finding a given number’s generator, if it indeed exists. Much math exploration can be provided if such concepts are revealed after the benefits of the activity are no longer important. Then the shortcut can be introduced to infuse new life into the problem.

And finally, if you really want to see some high-class math formulas related to self numbers, I recommend that you check out this page: Self Numbers.

Equal-Sum Groups

Problem Set I

Separate each group of numbers into TWO smaller groups so that the sum of the numbers in each of the small groups is the same. NOTE: there may be more than one way to do a problem. If so, can you find the other ways?

a)  {3, 7, 9, 13}        f)  {5, 9, 10, 13, 19}

b)  {2, 3, 4, 7, 10}        g)  {1, 2, 7, 12, 23, 31}

c)  {2, 4, 7, 9, 10}        h)  {1, 3, 4, 6, 8, 12, 13, 15, 16}

d)  {1, 4, 5, 6, 9, 13}    i)  {3, 9, 13, 16, 28, 37}

e)  {5, 8, 9, 10, 12}

Problem Set II

Separate each group of numbers into THREE smaller groups so that the sum of the numbers in each of the small groups is the same. NOTE: again, there may be more than one way to do a problem. If so, can you find the other ways?

a)  {2, 3, 4, 8, 9, 10, 14, 17, 26}

b)  {4, 5, 6, 7, 8, 19, 21, 22, 28}

c)  {3, 4, 8, 9, 11, 15, 19}

d)  {2, 5, 7, 8, 11, 14, 15, 16}

e)  {1, 3, 4, 10, 12, 14, 15, 16, 17, 18, 23, 26}

f)  {4, 6, 7, 9, 12, 14, 17, 23 25}

g)  {1, 2, 4, 8, 11, 12, 15, 19, 36}

Problem III

Make up a problem or two like these to share with your classmates. Just be careful that it does have at least one solution.

Teacher’s Note: In problem set I there are two solution methods for two groups [d and i] and six solutions for group “h”.

In problem set II groups a, b, f, and g have two solutions each, and group e has at least 13 (!) different ways to be solved.