Tag Archives: numbers

UP-down Match

John has $160 in his bank account and saves $8 each week. Mary has $270 in her account and withdraws $3 each week. After how many weeks will they have the same amount? What will that amount be?


Here is an interesting “real-world” problem that young students can surely relate to, I would imagine. Two individuals have some money in a bank. One puts in a certain amount on a weekly basis, so the amount of money in the account goes “up”. The other person takes out a certain amount each week, so that amount goes “down”.

Once again, as has been discussed in recent entries in this website, the strategies for solving it will vary with the mathematical maturity and capacity of the solver. An algebra student would use variables and equations, and find the answers in rather short order (see Appendix below). But that approach is obviously inappropriate for the younger individual. Even more so I feel that it misses the fundamental “reality” of the problem. This problem treats of two persons whose bank balances are going in opposite directions; the lower one is rising and the higher one is decreasing. And our goal is to find when these balances match.

For this problem to be truly meaningful to the majority of students, it would be better if they could “see” the weekly balances as they are formed. Hence, there enters our old problem-solving standby: the “CHART“.

Here’s what it should probably look like in this situation:

Gradually, as the student begins to fill out the chart, it begins to take this shape:

Soon, the goal of equal balances is reached. And the two questions can be easily answered, without algebra, but with “real life” meaning. That’s all there is to it!

“After 10 weeks John and Mary each have $240.”


The procedure, while essentially quite easy, does prove to be somewhat of a challenge for many children. They are not accustomed to a “story” problem that has so many numbers in it, or so many computations either. And this situation can be made a bit more “tecky” if calculators and working partners are used.

For example, most calculators have a “constant operation” capacity, especially the non-scientific, four-function models. Once the “addition” mode (or “subtraction” mode) is operative, one only need to press the “[=]” key to produce the subsequent balances in the columns. And if this task is shared by two individuals — one taking the role of John, the other of Mary — you have “cooperative group” problem solving!

The secret to creating additional practice examples lies in some clever observations and thinking. To see what that entails, let’s return to the first chart and add on an extra column: the “Difference” of the various balances.

Now as John’s balance increases by $8 and Mary’s decreases by $3, this means there is an $11 net change each week; that is, the balances becomes closer together by $11 each week. Since they began $110 apart, this tells us it would take 10 weeks [$110 ÷ $11/wk = 10 wk] to bring the difference down to $0. (Actually filling out the chart for all those balances enhances the impact of this fundamental point.)

The procedure may now be summarized as:

1. Select two monetary values with a sufficiently large difference between them.

2. Find a divisor/factor of that difference, medium sized.

3. Separate that divisor/factor into two addends, denoting one as the “savings” value, the other as the “withdrawal” value.

4. Place those numbers in their respective places in the

“story”.


Appendix

The algebraic solution of our problem might go something like this:

Let w = the number of weeks to achieve equal balances.

Then John’s current balance takes on this equation:

j = 160 + 8w

Likewise for Mary we have:

m = 270 – 3w

Equal balances implies j = m. Therefore, we state:

160 + 8w = 270 – 3w

and the rest is textbook mechanics.

[To return to text above, click here.]

Ladder Math

This is a simple arithmetical activity, designed to give younger students an introduction to a topic not normally encountered until high school algebra, as least in a formal, abstract sense:

arithmetic sequences.

Here is how I present it to my students…


Below you see a ladder standing up. The ladder has 7 rung (or steps). On the ground below the first rung is a number: 16. Beside the ladder is a small box with the number 6 on it.

Your job is to put numbers on each of the seven rungs by adding the box number over and over to the previous sum that you obtain as you go up the ladder. This means: on the first rung you will write 22 (because 22 = 16 + 6); then on the 2nd rung you will write 28 (because 28 = 22 + 6); and so on. When you have reached the 7th rung, your ladder picture should look like this:

And that’s all there is to it! Well, not quite, actually… You see there are four basic parts to each exercise in this Ladder Math activity: 1. the ground number, 2. the box number, 3. the number of rungs of a ladder, and 4. the top, or final, number. And this gives us four levels of difficulty, which I categorize as follows: 1. Easy: when the top number is not known; 2. Medium: when the number of rungs is unknown; 3. Hard: when the ground number is missing; and 4. Hardest: when the box number is not given. An example of each type is given in the chart below:

Category Ground Box Rungs Top
Easy 16 6 7 ?
Medium 20 9 ? 110
Hard ? 8 8 94
Hardest 16 ? 7 93

The example given above is obviously the one in the Easy category. This type of exercise is therefore the basis on which this activity is built. But when one of the other parts of information is missing, things take on a different color. The Medium category is actually rather easy as well. In it one does not know just how many rungs the ladder should have prior to doing the calculations. Once that is determined (in a rough draft form), the drawing can then be sketched.


More problem solving skills come into play, however, when an exercise from the Hard category is attempted. (See the following figure.)

For this one it is best to let the student work on it in any way he/she wishes: (a) trial-and-check on the ground number; or (b) using subtraction and working from “top down”. Obviously, the latter approach is more efficient and sophisticated, as it’s a matter of algebraic thinking, using inverse operations, etc. But it should not be forced onto a student until he is ready for it.


The highest degree of problem solving strategies come into play when considering the Hardest category. Here the missing part of information is the box number, the quantity that is to be added each time. (See next figure.)

Undoubtedly, most students will experience some stress on this one. Though the advanced math student knows the “shortcut” method, or could soon figure it out, the novice would probably be obliged to attack on the trial-and-check level. That in itself is a sign of intelligent thinking.


Additional Comments and Extensions

With what has been presented here, you can clearly see how to create more exercises for classroom use. But I would like to suggest one particular idea. Since all the examples given above dealt with positive integers (i.e. the simple counting numbers), it would prove interesting to include one or two instances in which “it doesn’t come out, teacher!” Depending on the maturity of the students, they may (1) give up and just say “There is no answer.”, or (2) try such things as fractions or decimals. These days with calculators in general, and fraction calculators in particular, this is an appropriate avenue to explore. Later, you can say up-front, “Some of these exercises need to be solved using a box number that is fractional and/or decimal.” For example, here are some that use “nice” decimals and fractions: 1. ground = 15, box = (4.6), rungs = 5, top = 38 2. ground = 23, box = (2¾), rungs = 8, top = 45


To summarize: For a student who can discern patterns in work such as this, a “formula” of sorts might be developed like this:

top = bottom + rungs × box

For the student of advanced algebra, the textbook formula for this work might be given as follows:

an = a1 + (n – 1)d

where an = the nth term, a1 = the first term, n = the number of terms being considered, and d = the “common difference”, i.e. the value being added each time.


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”.

PROOF

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

etc.

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


Variations

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.

Answers:

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

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

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

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

tt(1/5/86)

How Old Are You — In Days?

The question “How old are you?” or “How old is he (or she)?” actually has more than one type of answer, depending on the individual concerned. For instance, you, the readers of this webpage, will probably reply with statements like “I’m 14 years old”, or “I’ll be 25 next month”. (Scroll to the bottom of this article for a humorous treatment of this concept.)

But if the parents of a newborn baby or a very young child are asked the same question, their answers are would probably be
something like “He’s just 5 days old.”, or “She just turned 3 weeks old today.”, or “Last week he was 10 months old”. You see, we tend to use units of time that are appropriate for the situation.

This brings up an interesting question, that I bet you never thought of before: “How old are you today IN DAYS?” In reality it is nothing more than a basic math problem to calculate. After all, it
involves only a little simple arithmetic, perhaps a calculator, knowledge of a calendar, and genuine problem solving — all with a personal touch.


As a schoolteacher, I feel this activity is guaranteed to capture the interest of most students. When I present it (usually to students at the 4th-5th grade levels, with their teacher accompanying them), I begin by asking a child selected at random: “How old are you?” Usually the answer is: “I’m 9 (or 10, etc.) years old.” To which I reply: “Oh, nice, today is your birthday.” There follows a lot of giggles. “No, it’s not!” I then say: “Well, you’re not really 9 after all. You are 9 years and some days more, right?” Of course, they can agree to that. So we then dive into the main aspect of the activity: to determine one’s age IN DAYS, instead of the usual years.

I use an example to demonstrate, often my own son’s age. [Those are photographs of him, his mother and myself at the top of this page, if you haven’t figured that out. 😉 ] Since he was born on Dec. 22, 1983, I usually do the activity around March or April. [The reason will
soon be apparent.]

Let’s assume a presentation date of March 14, 1997. The following is written on the board and discussed/explained step-by-step. [The words inside the brackets are presented merely here for explanation, not written.]

Kevin Trotter: Dec. 22, 1983
1. 365 x 13 = 4745 [age at last birthday party (1996 – 1983) times the number of days in a basic year; this is where the calculator helps with younger students]
2. LYD(96, 92, 88, 84)      = 4 [LYD = leap year days; the years in which a Feb. 29 occurred; here counting backwards by 4s is fun, and instructive.]
3. Dec = 9

Jan = 31

Feb = 28

Mar = 14

—-

82

[no. of days left in the birth month, after the party, so to speak]

[no. of days lived in January]

[no. of days lived in February]

[no. of days lived SO FAR in March]

[no. of days lived SINCE his last birthday party]

4. 4745 + 4 + 82 = 4831 days

Then we finish things off with a “summary statement”: “Kevin
is 4831 days old today.”


Teaching Advice

I often then repeat the procedure for the class’s own teacher’s age. That really perks up the kids’ interest! Then they find their own ages, or mine as well. Later they are encouraged to do likewisefor their parents and other family members. A wealth of practice is thus generated that is quite meaningful to any student.

I’ve found that it’s better to use, as an initial example, a date of birth (i.e. month/day) 4-5 months prior to the lesson date. It’s so that in Step 3 there is a medium length list of post-birth-month entries — not too long, not too short. In practice, if it’s too long, it worries the kids (it’s a LOOONG problem!); if it’s too short, they get “the wrong idea” (they generalize that this is a quick list). However, they must learn to accept variations and surprises as they come up. Some Step 3 lists are long; some are short. “It all just depends.”

One of the most interesting aspects of this involves whether or not the kids know the number of days in the months of the year. This causes a lot of errors and discussion. It provides a good
opportunity to introduce the famous nursery rhyme: Thirty days hath
September…

Thirty days hath September,

April, June, and November;

February has twenty-eight alone,

All the rest have thirty-one,

Excepting leap-year, that’s the time

When February’s days are twenty-nine.

The fun thing about this activity is that one’s age sounds so BIG! And it keeps changing by the day.


Footnote:

Additional challenging questions can be posed, depending on the mathematical maturity of the students, that use this basic concept. For example: On what date were you exactly 1000 days old? (That number could be changed to 2000, 3000, etc. to quickly provide more problems at no great cost of preparation time.)

When I do this with older students, at the middle school level, I make it into a little quiz, giving two invented dates, one that would be appropriate for a parent’s birthdate and another for their child’s birthdate. After finding the respective ages, in days, of course, I ask them to solve this problem: “The child’s age is what percent of the parent’s age?”

After this had been done one year, a student presented me with this poem that she wrote:

Mr. LeeI told Mr. Lee

to please tell me

what percent of his life

had he spent with his wife

tt(12/28/96)

Kevin & Dad Kevin Kevin & Mom

The STAGES of AGES

Do you realize that the only time in our lives when we like to get old is when we’re kids? If you’re less than 10 years old, you’re so excited about aging that you think in fractions. How old are you?…. “I’m four and a half” …. You’re never 36 and a half …. you’re four and a half going on five!

That’s the key. You get into your teens, now they can’t hold you back. You jump to the next number. How old are you? “I’m gonna be 16.” You could be 12, but you’re gonna be 16.

And then the greatest day of your life happens …. you become 21. Even the words sound like a ceremony …. you BECOME 21 … YES!!!

But then you turn 30 …. ooohhh what happened there? Makes you sound like bad milk …. He TURNED, we had to throw him out. There’s no fun now.

What’s wrong?? What changed?? You BECOME 21, you TURN 30, then you’re PUSHING 40 ….. stay over there, it’s all slipping away ……..

You BECOME 21, you TURN 30, you’re PUSHING 40, you REACH 50 ….. and your dreams are gone.

Then you MAKE IT to 60 ….. you didn’t think you’d make it!!!!

So you BECOME 21, you TURN 30, you’re PUSHING 40, you REACH 50, you MAKE IT to 60 …… then you build up so much speed you HIT 70!

After that, it’s a day by day thing. After that, you HIT Wednesday ….

You get into your 80’s, you HIT lunch. My grandmother won’t even buy green bananas …. it’s an investment you know, and maybe a bad one.

And it doesn’t end there …. into the 90’s you start going backwards …. I was JUST 92 …

Then a strange thing happens. If you make it over 100, you become a little kid again …. “I’m 100 and a half!!!!”

–found on the internet.

Apple Bobby

Once upon a time there was a young man named Robert Appleson. He lived in a place called Appleland. It was famous for growing delicious red apples. Robert, who was known as Bobby to all his friends, had the largest apple orchard in all of Appleland. Everyone loved his apples, including the family of forty monkeys who lived in the forest nearby.

Now when it came time every year to pick his apples, Bobby did so in a very peculiar way. He would take his basket to the entrance of the orchard, go to a tree to get some apples, taking just seven apples at a time back to his basket. When he finished his day’s work, he would take the basket to his house and count his apples there.

All this was fine, except for one thing. Those monkeys! When Bobby would return to the trees to get more apples, those mischievous rascals would run up to the basket and steal some of the apples. But they did so in a very peculiar way also. Each monkey would only take
four apples at a time.

Strange though it may seem to you and me, that’s just the way things were in Appleland.

One Monday Bobby went to pick apples. It was at the beginning of the season and not many apples were ripe for picking. So he did not go to many trees. But when he came back to his home, he was surprised to find only 10 apples in his basket. He couldn’t figure out what had happened. He knew he had gone to at least 2 trees. Can we help Bobby figure out just what happened?

SOLUTION:

Bobby had gone to 2 trees, bringing 14 apples (2 × 7) in all. But while he was gone to the second tree, a little monkey came up and took 4 apples. (7 – 4 + 7 = 10) See? It’s easy.

On Tuesday Bobby went to his orchard again. He visited some trees to get apples. But you know what happened. Of course, some monkeys stole some apples while he was away. This day he counted 15 apples when he got home. How many trees did he go to, and how many monkeys stole apples from the basket?

[Use your calculator and paper-and-pencil to solve this problem carefully. It’s a little harder than the last one.]

When you finish that, find solutions for the other days of the week.

The numbers of apples in Bobby’s basket each day were these:

Wednesday 20
Thursday 25
Friday 40
Saturday 50

tt(9/7/96)

APPLE BOBBY

and the

Forty Monkeys
(Part 2)

When we last saw Bobby, he was having many problems picking his apples in Appleland. Those little monkeys were stealing so many apples, that he couldn’t make any money selling them in the village market. So the following year he thought he would try a new idea: he would pick more apples from each tree. He decided to bring eight apples from each tree to put in his basket. “Now,” he said to himself, “I’ll surely have more in my basket each day.” When he arrived home on the first Monday of the next harvest season, he still found only 10 apples in the basket. He remembered that he had picked apples from exactly 5 trees. He felt he should have had 40 apples, that is, if no monkeys had come around that day.

But this time he had done something else different. He had set up a video camera to watch his basket while he returned to the trees. He wanted to see just how many monkeys came to the basket. When he played the video, he counted the monkeys. “One, 2, 3, 4, 5, 6! Six monkeys!” he exclaimed.

“That’s strange,” he thought. “Six monkeys and 4 apples for each monkey should mean 24 apples taken. [6 × 4 = 24] Then by subtracting, I should have 16 apples left. Hmmm.” [40 – 24 = 16]

Can you help Bobby figure out just what happened?

SOLUTION:

The monkeys now are stealing more apples each time they run up to the basket. Now they take five apples. Six monkeys taking 5 apples means 30 apples stolen. [6 × 5 = 30 and 40 – 30 = 10]

Well, on Tuesday Bobby went to his orchard again. As always, he visited some trees to get apples. And, of course, some monkeys stole some apples while he was away. This day he counted 14 apples when he got home. How many trees did he go to, and how many monkeys stole apples from the basket?

When you finish that, find solutions for the other days of the week.

The numbers of apples in Bobby’s basket were these:

Wednesday 26
Thursday 33
Friday 48
Saturday 59

[Be sure to keep a record in an “ACTION – RESULT” chart like last time. Also write a sentence that tells how many trees were used and how many monkeys came.]

tt(10/13/96)


TEACHER’S GUIDE

for

APPLE BOBBY and the FORTY MONKEYS

This activity is an excellent experience in true problem solving, because it can be dealt with on various levels: easy, medium, and hard. It can also be done with manipulatives, then symbolically, or just symbolically. It involves the recording of data in a little chart. One can
allow calculators, or forbid them, depending on your desires and/or objectives. And for the students, it’s a lot of fun!

It is based on an advanced idea from number theory, but presented in a way that 4th graders can manage it. Number theory says that the expression “ax + by” can be made to equal any integer if and only if a and b are relatively prime numbers, i.e. their GCF is 1. The mathematics behind this is subtle and complicated, but need not concern us here. In our little story, it simply means that there’s always a way to solve Bobby’s dilemma no matter how many apples are in his basket when he returns home to count them.

For Part 1 of our story, the algebraic form is merely 7t – 4m = A, where t = the number of trees visited by Bobby, m = the number of monkeys who come to steal the apples, A = the number of apples counted by Bobby at his home. So now we say: we can always find values for t and m for any A we wish to dream up. And there’s a bonus here: there is more than one pair of t-m values that can form the A. (In fact, there is an unlimited number of pairs! But that’s another story.)

Now none of the foregoing need be mentioned to the average student, of course. Our presentation manner goes something like this:

  1. The story page is put on a transparency and projected on the board with an overhead projector. The students sit on the floor in front while I read the story as dramatically as possible. (This sets a unique mood for a math class.)
  2. We discuss the first problem (Monday’s result) orally to see if the students are beginning to “get the general idea”, following the suggested solution in the text.
  3. We then suggest that to “organize our thoughts and work”, we should make a T-chart like the one shown (below left).
                    Monday                       Monday
               action  | result             action  | result
               --------|--------            --------|--------
                 + 7   |    7                 + 7   |    7
                 - 4   |    3                 + 7   |   14
                 + 7   |   10                 - 4   |   10

    The 2nd one is then presented to illustrate that there is another way to do it! We add another scenario to justify the 2nd method by saying: “Bobby went to 2 trees. But because his hands were dirty from the picking (or he was thirsty after his work), he went to a well to wash his hands (or get a drink of water). While he was away from his basket, a monkey slipped up and took 4 apples.” This aspect to the Apple Bobby activity is essential: there is more than one way to arrive at a solution. And that both are correct and equal.

  4. The Tuesday part comes next. One possible solution is demonstrated on the board, using suggestions from the group to find it. This is done because for most students this idea is still too new, and they need to see a longer solution. (See below.)
  5. Now the students are “turned loose” to solve the Wednesday through Saturday problems. (And the fun really begins!) And they can solve them.

One more comment. The students are expected to finalize their work with a “summary statement”, that is, a simple sentence that answers the given question in the story. This is so important, now and in the future of one’s math experience. The solver needs to reflect back after all the computation has been done, to see if the problem really has been, in fact, solved.

                Tuesday
	action	   |	result
	  ---------|-----------
	   +7	   |	  7
	   +7	   |	 14
	   +7	   |	 21    Bobby went to 5 trees and
	   -4	   |	 17    5 monkeys came to steal apples.
	   -4	   |	 13
	   +7	   |	 20
	   -4	   |	 16
	   -4	   |	 12
	   +7	   |	 19
	   -4	   |	 15

Finally, Part 2 of the story allows the teacher a natural follow-up extention for assessment: to see if the students can apply the concepts learned to a new, yet similar, situation. Again, the number theory part of the math is at work. The expression now becomes 8t – 5m = A, of course. But everything else remains pretty much as it was before: T-chart presentation, multiple solutions, summary statements, etc. And a lot of good
math is learned by all!

tt(11/02/96)

The 100th Letter

INTRODUCTION

A popular category of puzzle-problems often involves something like the following:

What is the 1998th digit in the decimal expansion of the fraction 1/7?

Or here’s another one that involves a more basic idea:

What is the 1000th digit of this sequence:

1234567891011121314151617… ?

In another context, I have seen a problem like this one that uses letters instead of numbers:

What is the 100th letter in this sequence:

ABBCCCDDDD… ?

MY PROBLEMS

I began thinking about this idea the other day and came up with a problem that I had never seen before in any book, magazine, or contest exam. It goes like this:

If the word names of the counting numbers are written out as “one, two, three, … “, what letter in which number name would be the 100th letter?

Of course, a natural companion of that problem would be this:

If the word names of the first 100 counting numbers were written out as “one, two, three,…, one hundred“, how many letters would be written?

Now, dear reader, don’t be deceived by the apparent simplicity of these two questions. To paraphrase a popular saying: ease of solution lies in the mind of the solver. These problems were given to my 7th grade students recently [1998]. There was plenty of challenge here for these individuals, especially for the second problem. The results were interesting; some were even correct. Many different strategies were employed. And the important thing was that all would have been successful, except for a little error “here and there”.

And if you want a greater challenge, just increase the number “100th” to “1000th” or “1,000,000th” in the first problem; and likewise “one hundred” to “one thousand” or “one million” in the second problem.

I leave it to you to solve these letter problems.

E-mail me with
your solutions and explanation of how you arrived at your answers.

An R.O. for Fractions

I have always believed, contrary to general public opinion, that “FRACTIONS ARE FANTASTIC”. I have no great trouble with adding, subtracting, multiplying, or dividing these marvelous little creatures. In fact, I even take a certain amount of pleasure in doing problems with them. Of course, nowadays with the calculators that have the ability to do these computations, well, some of the fear should be removed from those of you out there who don’t feel the same as I.

Ah, I guess that’s progress; I don’t know.

But here is a little Recurrent Operation activity that involves fractions in a most unique and unusual way. Follow the steps that are stated below and you will see for yourself just how fantastic fractions can be at times.


1. Choose any two numbers to start the sequence.

2. Determine every number after the second by increasing the last existing number by one and dividing the sum by the number two places before the term being created.

Here is an example to explain the process. Let’s start with 7 and 4. The third term will be 5/7 because you add 1 to the “4” and divide the sum by the “7”. The fourth term can now be found by adding 1 to 5/7, then dividing by 4; this yields 3/7.

So far our sequence looks like this:

7,  4,  5/7,  3/7

Now, what would you do next? Of course, add 1 to the fraction 3/7, then divide by the fraction 5/7.

Then continue in like manner until you arrive at a surprising result. (Do you remmember what happened in the activities of Happy & Dizzy Numbers and Ulam?)


Of course, you’re curious if this will always happen every time, right? Well, try some more numbers. Try beginning with fractions for your first two selections; even mix in some negative numbers, if you feel adventuresome. That’s the way you really learn something in math: try it and see.

One warning here: just looking at many, many examples is not a mathematical proof that the surprising event that you have noticed by now always will occur. That, my friend, is where algebra comes to our aid! Find an algebraic proof that this activity produces a certain interesting outcome and email it to me. (See link below.)

By the way, can you find two starting numbers, whole numbers that is, that never bring about fractions as you build the sequence? It can be done.

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
                                      15851

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.

tt(1/29/77)

Happy & Dizzy Numbers

INTRODUCTION

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.


HAPPY NUMBERS

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.

Beyond Ulam

On the 14th annual American Junior High School Mathematics Examination (1998) there appeared the following problem, which I share with you here:


Problem #22:

Terri produces a sequence of positive integers by following three rules. She starts with a positive integer, then applies the appropriate rule to the result, and continues in this fashion.

Rule 1: If the integer is less than 10, multiply it by 9.

Rule 2: If the integer is even and greater than 9, divide it by 2.

Rule 3: If the integer is odd and greater than 9, subtract 5 from it.

A sample sequence: 23, 18, 9, 81, 76, …

Find the 98th term of the sequence that begins with 98, 49, …

(A) 6     (B)  11     (C)  22     (D)  27     (E)  54

Perhaps at first glance this appears to be sort of a hard problem, or at the very least a time consuming one. Especially when you realize that it is the 22nd item on a 25-item contest exam for which there is a 40-minute time limit.

But on closer examination it soon becomes apparent that there is a cyclic pattern involved in the terms that result from applying the three rules. The sequence begins this way:

98, 49, 44, 22, 11, 6, 54, 27, 22, …

Do you see that very quickly one of the numbers repeats itself? The number 22 arose in the 4th and 9th positions. This means there is a repeating cycle of five terms.

Our goal is to determine the 98th term. But if we remove the first three, the problem reverts to one which asks:

What is the 95th term of a string of terms that occur in groups of five?

Of course, it would be the 5th term of the cycle! Hence, the correct response to the question above is 27. What once seemed a lengthy sort of solution turned out to be incredibly short.


This looks a lot like something that was discussed in this website in an earlier article, the Ulam problem. In that article we had cycles of numbers occurring also. And this resembles a famous class of problems that make the rounds of many math contests, ones like

What is the 1000th digit in the decimal form of 1/7?

What is the unit’s digit of 31998?

and so on. [See The 100th Letter.]


But there’s more, much more to this problem than just getting the answer of 27. Recall that in the Ulam article we analyzed what happened when the rules given there were applied to many different numbers — specifically all those from 1 to 100. By way of refreshing your memory, we will state the general results here. For the first set of rules, all numbers sooner or later arrived to a simple three-term cycle of 4, 2, 1 (which returned to 4). For the second set of rules, all numbers entered one of three distinct cycles of varying numbers of terms (lengths of 2, 5, and 18).

Well, what about this new set of rules? Do all numbers flow into the cycle shown above? Or are there other cycles? One more? Two? How many? You see, after solving the original contest item, the real fun begins; now it’s time to do the real mathematics! I leave this matter in your hands now. I don’t want to spoil your pleasure at discovering the results for yourself.


Extensions

Surprise! Surprise! It’s not over yet. (You thought you were through with your homework, didn’t you?)

You see, problems like this one for me are like eating popcorn, once I start asking questions, I just can’t stop. The original contest problem from the AJHSME seems to me to be just one of many possible scenarios. The three rules as given are just begging to be adapted slightly to produce new problems. Here is an example of what we mean:

What cycle or cycles result from these rules?

  1. If the integer is less than 10, multiply it by 9.
  2. If the integer is even and greater than 9, divide it by 2.
  3. If the integer is odd and greater than 9, subtract 3 from it.

Did you notice the small change in the third rule? Now we subtract 3 instead of 5 from all odd numbers greater than 9. Small change, you say? True enough. But that little change is probably going to change our cycles quite a bit, don’t you think? Investigate this one to see what happens.

Then after that, try this:

What cycle or cycles result from these rules?

  1. If the integer is less than 10, multiply it by 7.
  2. If the integer is even and greater than 9, divide it by 2.
  3. If the integer is odd and greater than 9, subtract 5 from it.

This time we retained the subtraction of 5 in Rule #3, but changed Rule #1 to “multiply by 7″. Again, this subtle change makes a big difference in the outcomes, as I’m sure you’re beginning to suspect. But just what are the resulting cycles? That’s what I leave to you to find out. In general, we have before us a whole family of problems, similar in basic structure, but each one with its own distinct character. Very probably you can now begin to create your own variations of the original problem. You could change the subtraction value in Rule #3 to other odd numbers, like 1 or 7, or 9. [What would happen if an even number were used?] And you could change the multiplier in Rule #1 to some other value as well. But Rule #2 is probably better left as is, because not all even numbers are divisible by other numbers. To do so would bring up the nasty trouble of what to do with the “remainders”!


POSTSCRIPT

Send me the results of your research. I’d like to hear about what you have found.

For more activities of a simple cyclic nature, look at Happy & Dizzy Numbers, or Kaprekar.

For more information about the AJHSME [renamed as AMC 8], contact

Titu Andreescu, Director

American Mathematics Competitions

University of Nebraska-Lincoln

Lincoln, NE 68588-0658 U.S.A.

Tel: 402-472-6566, Fax: 402-472-6087

eMail: amcinfo@unl.edu