Tag Archives: Subtraction

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?

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.

[To return to text above, click here.]

Digit Add-On Problem

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

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

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

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

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

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

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

XY + increase = XYZ

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

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

XYZ – XY = increase

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

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

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

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

(10n + d) – n = i

Solving this equation gives us

n = (i – d)/9

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

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

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


Additional Comments

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

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

But notice this:

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

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

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

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

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

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

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

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

Kaprekar-6174

No, that is not someone’s telephone number up there in the title of this piece. It is the name of a numerical puzzle guaranteed to spark wonder and amazement in the minds of your students. It is called Kaprekar’s (pronounced kuh-PREE-kur) Constant*. This little, mysterious math activity is one of a family of math procedures called Recurrent Operations.

Recurrent operations are repetitive procedures wherein each intermediate result (i.e. sum, difference, etc.) is used in turn until a specific outcome is ultimately obtained. The procedure to work the Kaprekar sort of “magic” is as follows:

   1.	Select a four-digit number, for example, 5634.

   2.	Form a new number from this number by rearranging the
	digits in decreasing order, e.g. 6543.

   3.	Reverse the number obtained in Step 2, then subtract the
	smaller number from the larger one.

			  	  6543
				- 3456
			  	  3087 

   4.	Take the difference just obtained and repeat the procedure
	in Step 2 and 3; that is, order the digits, reverse, and
	subtract.  This should be continued until the surprise hits
	you.

		  8730		  8532		  7641
		- 0378		- 2358		- 1467
		  8352		  6174		  6174

Do you notice that with 6174 the process comes to a “stand-still”? That is, 6174 just yields itself!

But the real surprise is yet to come.


Try the procedure on any other four-digit number. It always ends with 6174. This is a famous number in the field of recreational mathematics; it’s called Kaprekar’s Constant.


Students find this discovery quite fascinating and unexpected, so much so that they don’t even realize that one of your instructional objectives is to provide some practice in subtraction skills. But more importantly you can now use this opportunity as a springboard for further exploration by posing the following questions to them:

   1.	Who can find a number that requires the greatest number
	of subtractions to obtain 6174?

   2.	What happens if the Kaprekar process is applied to three-
	digit or five-digit numbers?  (Dare we ask about still
	larger numbers?)

   3.	What happens if numbers in bases other than ten are used?
	For example, 4-digit numbers in "base 5".

   4.	Why does the game not work on such numbers as 3,333 or 888?

Many things could be said in favor of such activities from a pedagogical viewpoint, but only one will be mentioned here. Due to the strangeness of the whole situation and the interest created thereby, it causes the student to generate his own “problem set” of several exercises, based from only one or two starting numbers. That is to say, by asking him to do “only one or two problems”, he eventually does quite a few individual exercises — and enjoys it all the while.

One final note: if experience in systematic discovery is of more value to your instructional goals than merely subtraction practice, Questions #1 and #2 above make excellent investigations for pocket calculators.


*Discovered by Shri Dattathreya Ramachandra Kaprekar (1905-86?), a mathematician from India. He devised this activity in 1946 or 1949. See his photo on the right.

Additional websites that contain information about this topic:

  1. The Kaprekar Number and more playings with numbers
  2. Mudd Math Fun Facts: Kaprekar’s Constant
  3. Mathematical Black Holes
  4. Scientists of India – D.R. Kaprekar
  5. Sort, Reverse, Subtract
  6. Kaprekar Numbers
  7. More on Kaprekar’s Contants
  8. Mathews: Kaprekar Process (Highly recommended!)

Taken from T. Trotter, “Kaprekar”. Math Lab Matrix, Fall 1976, p. 8.


Divisibility Tests

Knowing quickly whether one number is divisible by another is a basic math skill which is very useful and serves many purposes as one studies mathematics at all levels, high or low. For many readers, this page will contain much information that you already know, but I think if you will bear with me, you will learn something that is a bit unusual. At least, I’ve never seen it in my school books on math.

I divide my presentation into three parts: primary, secondary, and the “unusual”. The primary section gives the divisibility tests, or rules for the small primes (2, 3, 5, 7, and 11); in fact, that’s why it is called the “primary” group. The secondary group contains the rules for the other small numbers (4, 6, 8, 9, 10, and 12) or special ones (25, 50, and 100). As for the unusual, well, I’ll wait on that to be a little surprise.


So let’s begin.

Primary Group

A number is divisible by

2: if it is an even number; that is, it ends with 0, 2, 4, 6, or 8.
3: if the sum of its digits is a multiple of three.

5: if the ending digit (in the units place) is 0 or 5.

7: if, after subtracting twice the units digit from the number formed by the remaining digits, you obtain a result that is easily seen to be a multiple of 7, then the original number was also divisible by 7.

[Since that rule sounds a bit confusing, we’ll take a moment to show an example. Test for 308:

The double of 8 is 16. 30 – 16 = 14. And since it’s easy to know that 14 is a multiple of 7, then we know 308 is likewise.

Proof: 308 ÷ 7 = 44

There’s more on this idea later in the unusual section.]

11: if a 0 or multiple of 11 is obtained by subtracting the sum of the digits in the “odd” places with the sum of the digits in the “even” places.

[Again, to help you, I’ll show a couple of examples.

Test for 4367 and 1848:

4367 –> (4 + 6) – (3 + 7) = 1010 = 0

1848 –> (8 + 8) – (1 + 4) = 165 = 11

Proof: 4367 ÷ 11 = 397

1848 ÷ 11 = 168]


Secondary Group

A number is divisible by

	4: if the number formed by the final pair of digits
	   is likewise divisible by 4.  Example:

		8672 --> 72 ÷ 4 = 18

		so 8672 ÷ 4 = 2198

	8: if the number formed by the final three digits
	   is likewise divisible by 8.  Example:

		73,104 --> 104 ÷ 8 = 13

		so 73,104 ÷ 8 = 9138

	9: if the sum of the digits is a multiple of 9.
	   Example: 4,617.

		4,617 --> 4 + 6 + 1 + 7 = 18

		since 18 = 9 × 2,

		then 4,617 ÷ 9 = 513

	6: if the number obeys the rules for 2 and 3; that is,
	   first it must be an even number, then has a digit
	   sum which is a multiple of 3.

       12: if the number obeys the rules for 3 and 4; that is,
	   it must have a digit sum that is a multiple of 3,
	   and its final digit pair is a multiple of 4.

[NOTE: The previous two rules combine earlier rules for smaller numbers because 6 = 2 × 3 and 2 and 3 are relatively prime; and 12 = 3 × 4 and 3 and 4 are likewise relatively prime numbers. And that merely means that their greatest common factor is 1. For more on the concept of relatively prime numbers, go here.]

We finish this section with some easy and obvious rules, so that you can get ready for the wild stuff to follow in the final section.

	10: if the digit in the units place is 0.

	25: if the number formed by the final pair of digits
	    is 25, 50, 75, or 00.

	50: if the number formed by the final pair of digits
	    is 50 or 00.

       100: if the number ends with a "00" pair.

The Unusual Stuff

Now we really begin the fun part, the part that usually is not included in most, if any, modern school math books. It actually may have begun for you above with the rule for 7; not too many books even

discuss that one either. They should, I feel, because it’s small and prime and at times useful. So, as the girl is pointing to it with her stick, let’s take another, closer look at the whole matter.

Some years ago I was reading an old and old-fashioned math book in Spanish. It had a chapter on divisibility tests that I found rather interesting because not only did it cover the easy basic numbers given above, but also such larger primes as 13, 17, and 19. Nowadays, if divisibility by 13, 17, or 19, or some larger prime were needed, we’d just probably get out our calculators and punch a few keys and be done with it. Afterall, memorizing a lot of rules for many numbers is not the best way to go about such things, knowing how easy it is to forget things or mix up the details of a given rule. But to learn about these other rules did prove interesting, nonetheless.

Upon looking at them carefully, I noted the procedure for each was very much like the rule for 7: multiply the units digit by something, then subtract that product from the number formed by the digits to the left. For example, and for fun, I’ll state the rule for 13 just as it was given in the book — in Spanish!

Un número es divisible por 13, cuando separando la primera
cifra de la derecha, multiplicándola por 9, restando este
producto de lo que queda a la izquierda y así sucesivamente,
da cero o múltiplo de 13.

Got that? You’ll understand when I show you the book’s example.

	See if 1456 is divisible by 13.

		145'6 × 9 = 54
	       - 54
	        09'1 × 9 = 9
	       - 9
		 0

	So, yes, 1456 is divisible by 13.

By now you could probably translate the Spanish as follows:

A number is divisible by 13, when separating the first digit
to the right, multiplying it by 9, subtracting that product
from what remains on the left and continuing in this manner,
[until] it gives zero or a multiple of 13.

So, a pattern was beginning to emerge, especially when I saw that the rules for 17 and 19 were pretty much the same; they just differed in the multiplier of the units digit. For 17, you need to multiply by 5, and for 19 you use 17. Otherwise there was no difference in the procedures.

I liked the rule for 17; you just multiply by five. That was not so hard. But I didn’t much care for the rule for 19; there you had to multiply by 17, a sorta larger number, at least not a single digit as the rules for 7, 13, and 17 were. However, it was the example for that rule that set me on the trail of why those numbers were selected in the first place. To explain the rule for 19, the book used 171. The form of the work was this:

	171 is divisible by 19 because

			17'1 × 17 = 17
		      - 17
			 0

Ah ha! Look at that 1 in the units place. Of course, I reasoned, 17 times 1, subtracted from 17, naturally would produce 0. Observe how numbers that might be called “1-enders” give us clues for the other rules. Try 21 (a multiple of 7), 91 (a multiple of 13) and 51 (a multiple of 17):

For 7: 21 –> 1 × 2 = 2 and 2 – 2 = 0.

For 13: 91 –> 1 × 9 = 9 and 9 – 9 = 0.

For 17: 51 –> 1 × 5 = 5 and 5 – 5 = 0.

Armed with that knowledge, I was ready to find divisibility rules for more numbers, mainly primes. Take 23 as our next case. Just look for a reasonably small multiple of it that is a “1-ender”. That would be 161 (23 × 7). So to see if a number was divisible by 23, just multiply the units digit by 16, etc. Take this number: 782.

782 –> 2 × 16 = 32      AND      78 – 32 = 46.

Do you see that 46 is 23 is a multiple of 23? Perhaps not, at least right away. You see, that’s the down side of the rules for larger and larger numbers; the final multiple check is sometimes not as obvious as it was for the smaller cases.

Then I began wondering if there wasn’t an easier way, at least in the multiplying step. Let’s return to the example above for 19. What if we ADDED 2 instead of subtracted 17. Wouldn’t that work? Here’s how it would look.

	171 is divisible by 19 because

			17'1 × 2 = 2
		      +  2
			19

Since multiplying by 2 is easier than by 17, this would make some problems easier, especially when the vertical format is being used. Note: 475.

Show that 475 is divisible by 19.

		Subtracting			Adding

		 47'5 x 17 = 85			 47'5 x 2 = 10
	       - 85			       + 10
		  ?				 57

The subtracting style causes a small problem, especially for elementary students: the second number is larger than the first. The advanced student would just say “it’s negative 38″ and go on with it. Afterall -38 is in fact a multiple of 19. However, this is not common knowledge for many students yet. The addition process is much simpler and doesn’t mess with the negative concept at all. In fact, if one doesn’t recognize that 57 is a multiple of 19 just yet, why not apply the rule one more time? Watch:

				5'7 × 2 = 14
			      + 14
			        19

And there’s your 19, all nicely in front of you!


You should be wondering now that if there is an addition way for 19 to go with the regular subtraction way. Or what about 7, 13, 17, 23, and any future number we might work with? Well, I’m happy to say that there are numbers for adding the products for each of our situations. And it’s very easy to determine what they are. And our clue comes from our work with 19. Recall that 17 was our subtracting-multiplier and 2 was our addition-multiplier. Notice: that once we find the multiple that is the “1-ender”, we obtain our subtraction-multiplier. Then the addition-multiplier comes naturally.

19 – 17 = 2

And that’s all there is to it!!! Deceptively simple.

To get the general idea, here is a chart for the numbers discussed so far.

Divisor Subt. Add’n
7 2 5
13 9 4
17 5 12
19 17 2
23 16 7

Perhaps you’d like to extend it a bit further.

One closing comment: as you are beginning to test a certain number to see if it is divisible by something and you use the addition-multiplier, but on the next step (if there is a need for one) you see that the subtraction-multiplier might be better, feel free to switch. Yes, you can change horses in the middle of the stream! Isn’t that nice?


Update April 2002

If you are interested in the divisibility test for the palindromic year 2002, try this: Here is a test for the question if a number n is divisible by 2002

The number must be even and, combine three digits of the number in order (from right to left) alternating them with + then -. If the result is 0 or a multiple of 1001, then the number 2002 divides it. E.g. is 1976756782 divisible by 2002 ? 782 – 756 + 976 – 001 = 1001. Yes, it is !

For this, and more interesting trivia about 2002, go to the World!Of Numbers. You will be well rewarded!


Update: January 24, 2003

For more information, consult Ask Dr. Math in the famous Math Forum site.