# Sigma of P(n)

Let’s define a function over the non-negative integers in the following manner:

1. P(n) = n when n is a one-digit integer.
2. P(n) = the product of all the digits of n when n > 9.

Example: P(1729) = 126, because 1 × 7 × 2 × 9 = 126.

Evaluate the following:

Note: this problem was shared to WTM by Reza Kassai, of Shiraz, Iran.

# Calculating Expressions

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

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

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

### 12 [×] 5 [-] 8 [÷] 4 [+] 7 [×] 2 [=]

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

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

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

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

Elisa showed her this:

Dina then said, “Here’s mine.”

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

Calculating Expressions

# Trotter’s Treats

Legend has it that long ago, one of my ancestors was a master candy maker. His specialty was a unique item made of chocolate, fancy nuts, cream filling, and other ingredients that were a highly guarded family secret. He sold his product under the simple name of Trotter’s Treats, or simply TT’s. Demand was always high for his confection, and rumor is that he made a lot of money – for those times – selling it.

This ancestor also had an eccentric quality about himself and how he promoted his candy to the consumers. You see, he sold it only in little boxes of 4 treats or 7 treats per box. That’s right, just 4 or 7. It didn’t matter to him. Take it or leave it, he always said.

The people didn’t mind either, as long as they were buying for themselves and their immediate families. The trouble arose when someone who was planning a party for example, and wanted to buy an exact number of the candies in order to give each guest exactly one of the TT’s. They had to calculate carefully about how many of each size to buy.

For example, for a party of 30 persons, the host or hostess could buy 4 boxes of 4 treats and 2 boxes of 7 treats. The math looks like this:

4 boxes x 4 treats = 16 treats

2 boxes x 7 treats = 14 treats

And 16 + 14 = 30. That’s how they did it. Simple, don’t you agree?

Over the years the people showed great interest in calculating just how many boxes of each size would be needed to make any given number of TT’s. Sometimes they even found more than one way the purchase could be made, but other times, to their great puzzlement and wonder, they found certain numbers could not be exactly produced.

For example, no one doubted that exactly 10 treats could not be purchased. Two boxes of 4, giving 8, were too few, while any other additional box of 4 or 7 would be too many. Likewise, one box of 7 was too few, and an additional box of 4 gave 11.

So the challenge then became: what was the largest number of TT’s that could NOT be purchased? Can you find that number?

Later on, this wily old chap decided to change things a little. As the economy of his time was expanding, he decided to increase the number of treats in the small and large boxes to 5 and 8. The townspeople saw this as merely a new challenge. The debate then became: what was the largest number of TT’s that could NOT be purchased now?

Eventually, the math teachers in the town began to realize that there was a great opportunity here to involve their students in some good problem solving. So they began asking their students to try different box sizes, like 3 & 5, 4 & 9, 6 & 7, and so on. The students then prepared reports on their investigations. Very interesting results were found. Some students even found a formula!

So how about you? What can you do?

Good luck, and let me know.

# 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]

#### Summary

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.

# Number Line Land

Once upon a time and far away there was a wonderful place called Number Line Land. It was part of a larger world known as Mathematica. The creatures who lived in this place were like nothing you’ve ever seen before. There were two main types: heptoads and tetrogs. They lived in a strange habitat, strange that is, for you and me. For you see, they lived on number lines. (That’s why it’s called Number Line Land.) Each creature had his/her own number line to call home. They lived in nests positioned at the origin, the “0” point. The terrain to the right of their nests was positive, sunny and safe, while to the left all was negative, dark and dangerous.

Now the heptoads and tetrogs had a very unusual way to get around on their lines. Heptoads always jumped in 7’s, that is, each jump was 7 nuluns in length. [The nulun was their basic unit of length, just like the meter and foot are units of length for us.] You may be wondering: why 7? That’s because some Greeks once taught math in their schools, and hepta– is the Greek prefix for 7. So I’ll just bet you know how far the tetrogs would jump. Right, 4 nuluns per jump, because tetra– means 4 in Greek numbers.

When a heptoad and a tetrog got married and had offspring, there were two more types of individuals that could form a family: hepta-tets and tetra-heps. (Be patient, you’ll get used to it soon.) These individuals would combine the jumping skills of their parents. When going to the right, a young hepta-tet would always jump 7 nuluns at a time; yet when going left, it would always jump 4 nuluns. It’s only natural to assume that tetra-heps did it in the reverse way: 4 to the right and 7 to the left. Don’t ask me why they did it this way, it was just instinct with them.

The pace of life was pretty relaxed in Number Line Land. The creatures spent most of their time just eating during the day and sleeping at night. Each morning the Great Spirit of Material Sustenance would place baskets of food on the number line numbers and the owner of that line just had to go to the basket and eat all he or she wanted. But with that, life was still not so easy as you might think for them. They could only eat food that was found at the end of a jump or a sequence of jumps. This means that heptoads could only eat on a day in which food baskets were placed on a multiple of 7, while tetrogs feasted on baskets placed on multiples of 4.

Let’s do a little math here. For lines no longer than 50 nuluns to the right, a heptoad has only 7 points at which he could eat: 7, 14, 21, 28, 35, 42, and 49. On the same sized line a tetrog would have 12 points on which she could eat. What are they? And at which number could either a heptoad or a tetrog obtain nourishment? (Hard though it may have been, things were a little easier for tetrogs, I’m sure you can see why.)

But things were less difficult for hepta-tets and tetra-heps. While the others could not reach food placed on the number 10, hepta-tets could reach it by moving two 7-nulun jumps to the right, followed by one 4-nulun jump to the left. Here’s why:

7 + 7 – 4 = 10

Of course, another way would be 7-right, 4-left, and 7-right.

7 – 4 + 7 = 10

What about the tetra-heps? Well, these creatures could also reach food on number 10, it just required more jumping around. Six 4-nulun jumps to the right had to be combined with two 7-nulun jumps to the left.

6 × 4 – 2 × 7 = 10

Of course, a tetra-hep could mix up the jumps in different ways so as not to stray too far away from the food. Can you find how many different jump sequences there are? And which one(s) of those sequences do you feel best help the tetra-hep stay close to the food to protect it from poachers?

It’s time for some more math now. Help Hestor, the hepta-tet reach his daily meals if Monday it was found on 15, Tuesday on 20, Wednesday on 25, Thursday on 30, and Friday on 35. Then help Tessie, the tetra-hep, to reach her food if it were placed on the same numbers on those days? Finally, decide which one, Hestor or Tessie, had a harder time on a given day to get their food?

One final note: a special law must always govern the jumping behavior of the young. Until an individual reaches his/her 18th year, it is forbidden to jump into that scary territory of the NEGATIVE numbers!

Number Line Land:

<The Legend Continues>

Recent history in Number Line Land has turned a bit tragic, I’m sorry to relate.

Just as things are going recently on Planet Earth regarding the ozone layer in the upper atmosphere, the same thing began to happen there as well. The disappearance of this protective shield allowed greater and greater amounts of solar radiation to attack the flora and fauna (i.e. the plants and animals) in N.L.L. This caused some species to die out altogether while others just mutated (i.e. changed form).

Fortunately for our story, the heptoads, the tetrogs, and their mixed offspring were among the lucky ones: they just changed. And how they changed! Heptoads turned into octoads and tetrogs became pentrogs. This is because after 7 comes 8, and oct- is the Greek prefix for 8. Likewise, as 5 comes after 4, pent– is the Greek prefix for 5.

I’m sure you can figure out now how their jumping abilities were altered. Sure, octoads made leaps 8 nuluns long and pentrogs could go 5 nuluns; both could go either way, of course.

As before, children came out mixed in their jumping patterns if they had mixed parentage. We now have octa-pents (8 nuluns to the right and 5 to the left) and penta-octs (who have a 5-right/8-left jumping style).

In spite of longer jumps, reaching the food baskets was the same kind of problem that is was in times past. For example, if food were on Point #10, an octa-pent required at least eleven leaps to land on the basket; yet a penta-oct would need at least fifteen. (Can you figure out how many of each kind of jump each creature performed in order to dine on the delicious food?)

Pity the poor octoad! While the pentrog can eat from a 10’s basket in 2 simple steps, the octoad can never reach the goodies.

Teacher’s Guide

If you have already read the story about Apple Bobby and the Teacher’s Guide that is given there, you will recognize that this story is very much the same as the other one, yet at the same time, it’s an extension of it. Now it is more appropriate for the middle school student. This one admits the use of negative numbers, whereas the other did not. Yet the basic mathematical foundation still rests on the same number theory concept that was used earlier: equations of the form ax – by = c.

I have not given answers here, and that was a deliberate decision on my part. This encourages you and your students to work together to find them, and discuss among yourselves as to their correctness. However, if you wish to consult with me, please send me an e-mail.

# 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)

and the

##### (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)

# Back-to-Front Multiplication

#### Introduction

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 5 102,040,816,326,530,612,244,897,959,183,673,469,387,755 142,857 6 1,016,949,152,542,372,881,355,932,203,389,830,508,474,- 576,271,186,440,677,966 7 1,014,492,753,623,188,405,797 1,159,420,289,855,072,463,768 1,304,347,826,086,956,521,739 8 0,253,164,556,962 0,379,746,835,443 0,886,075,949,367 1,012,658,227,848 1,139,240,506,329 9 03,370,786,516,853,932,584,269,662,921,348,314,606,741,573 10,112,359,550,561,797,752,808,988,764,044,943,820,224,719

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
34
19
(a)                   (b)    	         37
31
(7)

(c)
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.)

#### Conclusion

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!

References

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.]

# 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.]

#### NUMBER SEQUENCES

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

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

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

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

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, ???, ???, …,

TABLE OF ATTRIBUTE LISTS

 NUMBER OPERATORS NUMBER PROPERTIES 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.

TROTTER’S RULE

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.)

# The Math Price is Right

Perhaps some of the readers of this page will not appreciate the unique reference being made in the title of this activity to a famous TV game show, called “The Price Is Right“. If you are one of those, here is a brief description of that program, so that the math activity presented below will make sense.

On the program the contestants won nice prizes if they could guess the monetary value of the object in question: TV sets, stereos, or other valuable items. There were often three persons competing for the same prize simultaneously. Each would state his or her best estimate
of the price. The winner was the person whose estimate came the closest without going over! Simple idea, but effective. It depended highly on an individual’s number sense (a hot topic these days in the mathematics literature) and general good sense about the value of material objects
in today’s economy.

#### Now for the “Math” Price…

We can turn this basic idea of closest without going over into a math class activity that uses higher level thinking, calculators, and the concept of squaring a number (something very necessary when one enters algebra and advanced math). It goes like this:

The class is told that they’re going to play a game much like the TV show. They will do two things:

1. Choose a number.

2. Multiply it by itself. (This is the squaring idea.
And where the calculator comes in.)

If one’s result is the closest to some pre-set TARGET number, announced before the selection process of Step #1, then the goal has been accomplished.

Initially, only whole numbers would be used, as I am assuming that we are playing this game with say, 4th grade students. So a game may have gone something like this:

TARGET NUMBER: 500

1. Bob chooses 21 and Ralph chooses 22.

2. Bob’s square number is 441 whereas Ralph’s is 484.

Hence Ralph is the winner.

NOTE: if Ralph had chosen 23, his square of 529 would have been closer than Bob’s value, but it was over 500, hence could not win.

After play has gone on for some time, and the students are becoming more adept at playing it, it is recommended to start extending the game into other dimensions. One thing that can be done while still working with whole numbers is to use the concept of the “cube” of a number. This merely means that one uses the selected number three times as a factor in the multiplication step.

For example: 1728 is the cube of 12 because

### 12 × 12 × 12 = 1728

Obviously, larger target numbers need to be selected now. But that’s okay; the computation is not hard due to the use of the calculator. The hard part is the thinking! (Hmmm… but that’s good, too.)

A second thing that can be tried is the use of decimals. Even at the 4th grade level this should cause no great difficulty. We are, after all, talking about money here. And most primary school students are familiar with prices such as \$12.95 and the sort. Returning to the squaring version of the game, we can proceed in this way:

Let’s use Bob and Ralph again. In trying to come close to 500 again, Bob might try 22.3, whereas Ralph chooses 22.4. Now when Bob squares his number he gets “497.29“. (Very close.) But poor Ralph! His square of “501.76” went over the target this time. So, he loses. What is nice about this feature of the game is that the squares of numbers in the “tenths” are numbers in the “hundredths“, which merely resemble money amounts. It is also important for students to see a fundamental pattern here, namely,

ab.x2 = cde.yy

[The reader is to understand that my focus is on the “x” and “y” parts;

a number with one decimal place has a square with two decimal places. It’s shocking how many students don’t observe this.]

Finally, the game can be turned into a single-person activity in this way:

How close can you come to a given target number, using the squaring procedure, if you are allowed as many guesses as you wish?

This takes the idea away from its competitive setting and puts

it in a problem solving one. This brings us back to recording our investigations in our old friend, the “T-chart“. Let’s see how it might look for a target of 200.

```                         n   |    n2
----------|------------
14    |  196  too low
15    |  225  too high
--------|---------
14.2  | 201.64  too high
14.1  | 198.81  too low```

It is clear that 14.1 produces the winning value this time. If students can handle it, one could proceed to values of n that have 2 decimals places. The principal change here will be that the squares will have 4 decimal places, that’s all.

ONE FINAL COMMENT

Not to be overlooked in this work is that we are actually preparing the student for the concept of “square root” (and “cube root”) which will be confronted in the future, concepts that need careful development prior to their formal use in higher mathematics. If some groundwork is laid in the early years, then things will go more smoothly later on.

# Distinct Digit Squares

INTRODUCTION

A. When a number is multiplied by itself, the resulting product is called a SQUARE NUMBER, or simply a SQUARE.

 12 × 12 = 144 so 144 is a square number. 35 × 35 = 1225 so 1225 is a square number. 133 × 133 = 17,689 so 17689 is a square number.

B. Sometimes a square is made up of digits that are all different, that is, it has “no repeats”. Such a square is called a distinct-digit square (DDS).

Example: 13 × 13 = 169; there are no repeated digits in 169,
so it is a distinct-digit square.

But 21 × 21 = 441; since the 4 is repeated in 441, this is not
a distinct-digit square.

PROBLEM

You are to use your calculator to help you make a list of ten (10) distinct-digit squares. But–one more thing–they must all contain either 5 or 6 digits. That is, they should be “5-place” or “6-place” numbers.

Largest Number Squared

INTRODUCTION

```If you multiply 142 by itself, what is the product?  _________
If you multiply 781 by itself, what is the product?  _________
The first one was a 5-place number, and the second one was a
6-place number, right?
(If not, you made a mistake somewhere.  Do the wrong one(s) again.)```

PROBLEM I

You now see than when you multiply a 3-place number by itself, you might get a 5-place or a 6-place product.

Your problem is to use your calculator to find the largest 3-place number that when multiplied by itself gives just a 5-place product.

(Hint: The number is greater than 142.)

PROBLEM II

Compute these two products:

 1022 × 1022 = ________ 7803 × 7803 = ________

Do you see that the first product is a 7-place number, and the second one is an 8-place number? (If not, check your work as before.)

This time you are to find the largest 4-place number which when multiplied by itself will still only make a 7-place product.

(HINT: It is greater than 1022.)

PROBLEM III

Compute these two products:

 17 × 17 = _______ 83 × 83 = _______

Do you see that the first product is a 3-place number, and the second one is a 4-place number?

This time you are to find the largest 2-place number which when multiplied by itself will still only make a 3-place product.

(HINT: It is greater than 17.)

PROBLEM IVThe Brainbuster

You have done three problems with your calculator that were almost the same. Each time you had to find the largest number which
when multiplied by itself gave a product with an odd number of places,
right?

Now you will be asked to do the whole thing one more time–this is the BRAINBUSTER!

Find the largest five-place number which when multiplied by itself gives only a nine-place product.

But unfortunately, this time your calculator will not be able to help you; a 9-place number is too big for the calculator’s display area.

However, things are not so bad if you will look at the answers you found for the first three problems. There is an important clue there that will tame this tough problem. Do you see it?
CLUE PATTERN:

 The largest 3-place product came from ______; The largest 5-place product came from ______; The largest 7-place product came from ______.

Same-Digit Pairs of DDSs

INTRODUCTION

In first section you found several squares that we called DDSs. (Remember: these are squares whose digits are “all different, no repeats”.)

In this section, we will explore something interesting about certain of those DDSs. Look at these squares:

37² = 1369 and 44² = 1936

Both 1369 and 1936 are DDSs, of course. BUT, there is one more thing that is strange: they both contain the same digits, just arranged in a different order.

There are many more cases like this. Before you start the exercise below, make sure you understand this idea by finding the squares for these two numbers: 32 and 49.

EXERCISES

In the groups of numbers below, two of them will give DDSs with the same digits, but arranged in a different order. The other numbers also produce DDSs, but do not have the same digits. Find the correct
pair in each group.

 144, 175, 174 305, 153, 198 136, 228, 267, 309 233, 193, 305, 172 152, 142, 118, 179, 147

Below is given a large group of numbers that will give “same-digit pairs”, like you found above; some will not. Find the numbers that make this type of pair and put them together.

 267 281 186 273 224 213 282 286 226 214

Once in a while we can find three or more DDSs that use the same digits. Look at this example:

36² = 1296     54² = 2916     96² = 9216

Do you see that all three squares contain the same digits, only in a different order. Now this is strange indeed! And it does not happen as oiften as was true for the same-digit pairs. But, as we will see, it can happen several times, if we are patient enough to look.

The following eleven numbers will produce DDSs that can be grouped into three same-digit families. Each family will have at least three members in it, maybe more. Can you separate all of them into their proper families?

 181 148 154 128 209 203 269 196 191 302 178 .

So far, all of our DDSs have been only 5-place numbers. But the same thing can happen with 6-place DDSs, too. And, would you believe it? There are even more pairs and family-sized groups than you saw before.

Here are several numbers that will produce DDSs pairs or families. Can you separate them as you did before?

 324 353 364 375 403 405 445 463 504 509 589 645 661 708 843 905

### NOTE:

This piece was written by me and published in The Oregon Mathematics Teacher, Sept. 1978. At that time calculators with a 10-digit display were not the common models available to students at the elementary or middle school levels. So the “Brainbuster” problem above needs to be adjusted to take that into account, or only permit the use of 8-digit models while
doing this activity.