Tag Archives: addition

Guidelines for Writing Math Solutions

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


Readability

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

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

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

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

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

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

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

Getting Off To a Good Start

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

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

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

BAD ==> Let x = apples

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


Use of Guess-and-Check Procedure

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

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

One of our mentors advises students in the following way:

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


Writing a Complete Answer

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

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

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

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

Again, a thought from the ElemPoW service:

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

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


E-Mail Notation

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

Exponents

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

Square roots

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

     

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

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

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

Fractions

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

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

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

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

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

Subscripts

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

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

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

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

Quadratic Formula

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

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

Determinants

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

 

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

 


 

Alvin’s Theorem

While Ms. Powers was leading a class discussion about square numbers, Absent-minded Alvin was in another “world”, looking for interesting patterns in the topic. Shortly, he raised his hand and said, “Ms. Powers, I’ve found something rather nice. Look. If I take 2 consecutive squares and subtract them, the difference is always the sum of 2 consecutive integers.””Show the class what you mean by that, Alvin,” said the teacher.

Alvin wrote the following on the board:

49 – 36 = 13 and 13 = 6 + 764 – 49 = 15 and 15 = 7 + 8

Turning to the class, he shyly said, “I call this ‘Alvin’s Theorem‘.”

Ms. Powers smiled and said, “Very good, but if you want to call it a theorem, you must be able to prove it is always true for all numbers, using algebra.”

Alvin replied, “Oh yes, I can do that too. Here’s how.”

What did Alvin write on the board now?


Later, Alvin investigated the matter of the difference of consecutive “even” squares. What do you imagine he discovered this time?

Chocolate Chips

     Everyone knew that Charlie had a “sweet tooth”, ailment especially for anything with chocolate in it. So no one was very surprised to hear of his recent project. The only surprise was the manner he conducted it.

     He decided to eat one chocolate chip on the first day, search then 2 chips on the second day, 3 on the 3rd day. And so on. Each day he ate one more than the day before.

     By the last day of this current month, he will have consumed a total of 1035 chips.

     When did Charlie start this sweet journey?


Extra: On what date did Charlie reach the half-way point in his project? And which chip (1st, 2nd, 3rd, etc.) was it that day?


[Note: Since this is an “algebra” problem, it is expected that you will use normal algebra to solve it. This means that no brute force methods, such as spreadsheets or using a calculator to merely add up the numbers 1, 2, 3, 4, and so on, will be accepted. There is a famous formula that you should know about that automatically finds the sum of an expression like:

1 + 2 + 3 + … + n = ?

We hope you can find and use it.]

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.

Levy Expressions

In July of 2003, WTM had the good fortune to make acquaintance with a very creative mathematician by the name of Jerry Levy. Mr. Levy was attracted by a certain problem based on a popular theme in recreational math, namely pandigitality. Now, pandigitality is a new word not likely to be in the dictionaries yet, as WTM has just invented it here for use in this webpage. At its simplest, it merely means using all the digits 1 to 9 (or in its purest form, 0 to 9) once and only once in some unique manner.

[WTM already has presented some material in this regard. See Digital Diversions and Pandigital Diversions.]

Jerry’s original interest arose from the following situation:

Given the expression

A/BC + D/EF + G/HI

where adjoining letters, such as BC, indicate a 2-place number.

Objective #1: replace the letters with the digits 1 to 9, one digit per letter, in such a way that a integer (i.e. whole number) is the value of the expression.

Objective #2: find the replacements that result in the smallest and largest possible integer values.

[Note: To learn more about this problem and see its solution, click HERE.]

It is here where the creative juices in Jerry’s mind began to flow. You see, he then thought to himself, “What if I change the denominator to mean regular algebra notation?” This means that AB now indicates multiplication, as every school algebra student learns. To avoid confusion however, Jerry chose to write the expression this way:

A/(B*C) + D/(E*F) + G/(H*I)

Nonetheless, the two objectives remain the same: to obtain integer values and find the extreme solutions.

Not being content with that interesting situation, he pushed on where no man has gone before, namely these expressions:

(A/B)*C + (D/E)*F + (G/H)*I

(A/B)^C + (D/E)^F + (G/H)^I

Jerry proceeded to work on these problems by hand, or as he wrote in an email, “intuitive trial and error”. Eventually, he made contact with Patrick De Geest, whose website, World!OfNumbers, is a gold mine for number enthusiasts. Soon help was forthcoming from a French school math teacher, Jean-Claude Rosa. Patrick and JC turned their programming skills loose on the problem.

A virtual flood of information soon began to come to light and be offered to WTM. Very interesting it is, too. In fact, so much good data was obtained for the 3rd variant that WTM created a school-level math problem based on it. (See Power-full Fractions.)

No solutions to these problems will be given here in this page. This is a deliberate decision, based on an underlying philosophy of the WTM website, namely to present challenging math materials to the school-aged student, or to the merely curious fan of recreational math, and let them play around with it, making their own discoveries. So dive in and start discovering.

However, what will be given here are some more expression ideas that have been formulated by Jerry, JC, and WTM. (Beware! Some of these cases have not been analyzed by a computer program yet, so proceed at your own risk.)

Variant #4:

A/(B+C) + D/(E+F) + G/(H+I)

Variant #5:

(A+B)/C + (D+E)/F + (G+H)/I

Variant #6:

A/(B-C) + D/(E-F) + G/(H-I)

Variant #7:

(A-B)/C + (D-E)/F + (G-H)/I

Variant #8:

(A + B)*C + (D + E)*F + (G + H)*I

Variant #9:

(A + B)^C + (D + E)^F + (G + H)^I

Variant #10:

(A – B)*C + (D – E)*F + (G – H)*I

Variant #11:

(A – B)^C + (D – E)^F + (G – H)^I

Variant #12:

A^B/C + D^E/F + G^H/I

Variant #13:

A^(B/C) + D^(E/F) + G^(H/I)

Variant #14:

(A + B)*C + (D – E)/F + G^H*I

Variant #15:

(you try one of your own)


What do you do now, you ask? Well, if you like to tinker around with numbers, why not begin by substituting the numbers from 1 to 9 in these expressions, and see if the results are interesting in some way? Like palindromes, primes, etc. What are the largest values, or smallest values, that you can obtain for a given expressiion? How many different distinct values can be found? Let your creativity flow. Numbers are also meant to be enjoyed, in addition to be admired for their utilitarian mode.

Alpha-Math

     The concept of connecting the letters of the alphabet with mathematical activities has a long and interesting history. One of the most popular and easy to understand is that of finding the numerical value of a word by assigning a number to each letter in that word (according to its position in the alphabet), then finding the sum of those values. For an elementary treatment of this idea, read the page Wordsworth in this very website.

     The purpose of this page is to investigate what happens when the input words are the word names of numbers. For example, what is the numerical value for zero, when the letters are replaced with their alphabetical position value (A = 1, B = 2, …, Z = 26), and then added?

     Z = 26      E = 5      R = 18      O = 15

26 + 5 + 18 + 15 = 64

     Got it now? It’s easy. Okay. Before we start with the big stuff, here are a couple of tasks to serve as warm-up.


     Task #1: Find the numerical values for all the numbers from 1 to 10, or more specifically, from one to ten. Then list the numbers from small to large according to the numerical values of their word names.

     Task #2: Find the number or numbers n, such that 10 < n < 100, whose number value(s) is/are exactly 100.


     Now we will present a most interesting idea, the one that prompted us to prepare this page in the first place. It’s clear by now that every number is associated with another, newer number, the sum of the letter values of its word name. So what would happen if you took that sum’s word name and found its numerical value? And then continued with the result as often as necessary for something to happen?

     Such a procedure has a name: recurrent operations. There are several pages in WTM that have activities along this line, including KAPREKAR, ULAM and
Happy & Dizzy Numbers.

     “Where do I begin?” do I hear you ask? That’s the beauty of this activity – anywhere you wish! Maybe start with your age, your house’s address, the last four digits of your phone number, whatever. You can even start at the beginning: 1, one. And go from there.

     The next question you’re likely to ask is, “when do I stop?” Ah, that one is harder to answer. The best WTM answer we can offer is “until you see something strange“. We recommend that you briefly examine those three webpages just mentioned to get some clues.


In the math class…

     This next section is really aimed at school teachers of math who would like to involve the concept discussed above while at the same time still teach a few of their standard objectives. So if you don’t care about this, you can skip it and go do other things, ok?

     The major algebraic objectives that can be taught or reviewed with this alpha-math topic are (1) writing multi-term expressions, involving numerical coefficients, and (2) evaluating these expressions by substituting appropriate number values for the variables. Here’s how WTM recommends that this be done.

  1. Select a number, not too large but not too small. Say, 132.
  2. Write the word name for the number. (Watch for spelling!)

    one hundred thirty-two

  3. Write the expression in alphabetical order, taking into consideration the number of times certain letters appear.

    2d + 2e + 2h + i + 2n + 2o + 2r + 3t + u + w + y

  4. Substitute the letter values into the expression.

    2*4 + 2*5 + 2*8 + 9 + 2*14 + 2*15 + 2*18 + 3*20 + 21 + 23 + 25

  5. Find the products.

    8 + 10 + 16 + 9 + 28 + 30 + 36 + 60 + 21 + 23 + 25

  6. Find the sum. (Here a calculator is recommended.)

    The sum is 266.

     And there you have it. Algebra meets Alpha-Math !

     [By the way, in presenting the procedure given above, the reader should not come away with the idea that it was necessarily the best way to find the number values. Its purpose is to bring together standard algebra techniques and this concept of number words. A more efficient way to research this topic in depth would be to make a list of all the necessary words and find their values in advance. There are not so many words. Thirty-two, in fact, will serve you nicely up to numbers with 15 digits.]


Some interesting trivia

     Take the number 15551. The first thing obvious about it is that it’s a palindrome. What is not so obvious is that it is also a prime number. Hence, it is called a palindromic prime, or palprime, for short. Now find its number word value. It is likewise another palprime. Can you find it?

     Of a lesser nature, we can make the following observations:

     Palindromic sums occur for the following numbers: 13, 50, 21, 61, 91, and 57. In fact, one of those “alpha-sums” is another palprime. Which one is it?

     If we take groups of numbers and find the sums of their alpha-sums, we can see other interesting results. For example, the sum of all the alpha-sums for the nine numbers from 61 to 69 is likewise a palindrome, a rather special one as well. Nice, huh?

(more coming later.)

Square-Cube Ages

The article that you are about to read in this page was written by the staff of the MATHCOUNTS Foundation and published in their Winter 1995 edition of their newsletter MATHCOUNTS NEWS. It is is reprinted here by permission.


Mild Turns Wild

Every once in a while, tucked deep within a MATHCOUNTS School Handbook or MATHCOUNTS competition, is a problem that seems somewhat mild, but after all
is said and done, could make a person go wild!

Last year my age was a perfect square. Next year it will be a perfect cube. How old am I?

This particular MATHCOUNTS problem isn’t a difficult one, until another dimension is added, as Terrel Trotter, Jr. of Escuela Americana in San Salvador, El Salvador did. He asked his students, “if we drop the idea of age and use larger numbers, can we find other examples of squares and cubes that have a difference of two?” At first glance, this seems like an easy answer, but it’s anything but.

The problem appeared a few months ago in Trotter’s monthly classroom tabloid, Trotter Math News. Accompanying the problem was a reminder: “problem solving is not merely computing the sum of two fractions or the product of two decimals, but rather doing whatever is necessary to answer a big question, using whatever method that may help, guessing, looking for patterns, using calculators — the whole works.”

That’s great advice, but even with the use of calculators and spreadsheet software, the students came up empty-handed. When they reported their findings, Trotter confessed, “I don’t know if there is indeed an answer!”

Not ready to accept defeat, Trotter went to the MATHCOUNTS head office in search of some answers.

MATHCOUNTS Curriculum Coordinator Scott Stull recalled, “when the problem was first written, we didn’t have a strong argument that the answer was unique for all integers. However we did convince ourselves that the answer was unique for all reasonable integral ages for a human being.”

To answer Trotter’s question, Stull enlisted the help of Richard Case, P.E., director of strategic development for IBM Corporation and MATHCOUNTS national judge, and also Harold Reiter, Ph.D., professor of mathematics at The University of North Carolina at Charlotte and MATHCOUNTS question writer.


The two initially agreed that another solution seemed possible but how they went about finding the alternate solution was quite different.

Case developed an algorithm to test the first n integers. He reduced the problem to finding solutions of the form {x, y} for the equation y3 – 2 = x2. His algorithm involved taking an integer, cubing it, subtracting two, taking the square root, and evaluating whether or not the result was also an integer. He was able to confirm that the solution {5, 3} is unique for the first 10,201 positive integers (that’s as high as his computer could go).


Reiter’s approach was slightly different. He developed his argument through number theory. Referencing an article by John Stillwell titled, What Are Algebraic Integers and What Are They For?, Reiter found an argument by Euler verifying that the original solution is the only solution:

A rigorous proof of this argument relies on abstract algebra, and establishing the existence of prime factorization for the ring of algebraic integers Z[-2].


However, Stull warned, “this cannot be generalized to say that there is only one solution to any problem of the form y3j = x2.” Take this problem for instance:


Two years ago the age of a certain tortise and that tortise’s child were both perfect squares. In two years both of the ages will be perfect cubes. How old is each tortise?

Now that the first problem has been tamed, what about this one? Is there a solution? If there is, is it unique? A reward is offered to the first team (of four students) who sends in a solution and an argument verifying or disproving the uniqueness of the solution. To collect the bounty, send your solution to MATHCOUNTS, 1420 King St., Alexandria, VA 23314-2794.

Good luck!


P.S. (March 1999) Don’t rush to send in your solution; the problem has already been solved and the reward claimed. But it’s still a nice problem about the mommy and baby tortise, don’t you think?

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.

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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.


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?

Answers

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.