Celebrating
INDEPENDENCE DAY
Salvadorean Style

     September 15th is Independence Day in El Salvador, which just happens to be where the World of Trotter Math has its home base, as it were. So for those of us who live in this beautiful little Central American country, September 15th is our "4th of July"!

     Recently in my school where I work --- La Escuela Americana --- my principal sent out a memo asking the teaching staff to think up unique ways to celebrate this event during the month. Being the mathematician that I hope I am, I came up with a mathematical method for doing this. I utilized the popular activity of forming numerical expressions with certain sets of digits, like making expressions for all the integers from 1 to 100 using the digits of the current year. But that's already been done; consult the fine website of Math Forum from Swarthmore College. Also since I felt that was a bit time consuming for my students, I looked for a way to give it a special twist of my own. Here is what I came up with; I hope you like it.

     First, we begin by noting that the numerical form for the date is

9/15

     Next, I wanted to use the concept of consecutive digits. Students need to deal with this idea more than they often do in their math lessons. So, I said: "Let's form expressions that have the values of 9 and 15. But with this condition --- the digits used must be consecutive!" But how many digits should be used, you ask. Well, that's the next element of my plan: two, three, four, ... , nine consecutive digits! Do you get it? This would produce eight different expressions for each number of 9 and 15. (I would say that would be worthy of a project grade for most middle school students, the level I currently teach, wouldn't you?)

     To get my numerical scholars off and running, I presented some examples.

     For the month number 9, I gave these items:

two: 4 + 5 or 32

three: 2 + 3 + 4 or 1 + 3 + 5 or 12 - 3

four: 1 × 2 + 3 + 4 or 14 - 2 - 3

     [Notice that there are some other conditions illustrated here.

  1. The set of consecutive digits need not begin with 1.

  2. The set may even be consecutive odd digits as well.

  3. The digits need not appear in their natural "counting" order. It is permitted that they may be "mixed up". However, if the natural order can be maintained, the expression has a bit more "charm".

  4. A digit may be used as an exponent, but then it can not be used elsewhere, of course.

  5. Digits may be combined to form 2-place numbers.]

     Turning to the day number of 15, these items were given:

two: 3 × 5 or 7 + 8

three: 1 × 3 × 5 or 4 + 5 + 6 or 3 + 5 + 7

four: (4 - 1) × (2 + 3)

     Some additional construction hints are shown here.

  1. Adapt an earlier expression to make a new one:

    3 × 5 helps to make 1 × 3 × 5.

  2. Parentheses may be used.

My Students' Work

     Well, I think that's enough from me. You get the general idea now, I'm sure. Let's hear from my students and see what they could do. I think you'll find some of their expressions rather nice.

     [Remember: this is their first time doing such a task. Some expressions could have been made more "charming" (i.e. arranging the digits in descending or ascending order), but I have decided to leave their work just as it was turned in to me.]

     Here are expressions for the month number 9:


       Three:   8 + 7 - 6                       Stephanie S.
                (1 + 2) × 3                     Rebeca M.

        Four:   23 - 14                         Stephanie S.

        Five:   14 ÷ 2 + 5 - 3                  Vanessa H.
                34 × 1 - 25                     José Roberto V.

         Six:   8 - 7 + 6 - 5 + 4 + 3           Alexya C.
                6 + 5 + 4 - 3 - 2 - 1           Ricardo S.

       Seven:   3 + 6 + 2 + 7 - 5 - 4 × 1       Eduardo A.
                56 - 47 - 2 + 3 - 1             Liliana L.
                (74 - 65) + 3 - 2 - 1           Stephanie S.
                23 + 45 - 67 + 8                Gladys R.
                7 × 2 - 4 + 6 - 5 - 3 + 1       Alexya C.
                ((6 × 4) + (52)) ÷ + 3 - 1      Anna M.

       Eight:   3 × 5 + 7 - 8 - 6 + 4 - 2 - 1     Claudia T.
                9 × 8 - 7 - 65 + 4 + 3 + 2        Gladys R.
                8 + 7 - 6 - 5 - 4 + 3(2 + 1)      Alexandra C.
                8 - 7 + 6 + 5 + 4 - 3 × 2 - 1     Adriana J.

        Nine:   23 + 4 × 5 + 6 - 7 - 8 - 9 - 1     Daniela S.


     Now here are expressions for the day number 15:


       four:    1 × 3 + 5 + 7                   Alexya C.
                5 × 6 - 7 - 8                   Marinés P.

       five:    4 × 2 + 5 + 3 - 1               Eduardo A.

        six:    32 - 15 + 4 - 6                 Liliana L.
                45 - 6 - 7 - 8 - 9              Vanessa H.
                9 + 8 - 7 + 6 - 5 + 4           Gladys R.
                7 × 6 - 5 × 8 + 9 + 4           Camila B.

       seven:   7 × 6 - 42 + 5 × 3 × 1		Jessica A.
                56 - 47 + 1 + 2 + 3		Liliana L.
                2 + 3 - 4 + 5 - 6 + 7 + 8	Julio S.
                12 + 3 + 4 - 5 - 6 + 7		Adriana J.
                57 - 46 + 3 + 2 - 1		Alexya C.
                2 × 3 + 4 + 5 + 6 - 7 + 1	Carmen C.
               (6 × 5 ÷ 3) + (7 - 4) + (2 × 1)	Cori J.

        eight:  8 × 7 - 56 + 3 × 4 + 2 + 1              Jessica A.
                -2 - 3 + (4 × 5) + 6 - 7 - 8 + 9        Julio S.
                3 + 5 + 7 + 8 - 6 - 4 + 2 × 1           Claudia T.

        nine:   9 + 8 + 7 + 6 - 5 - 4 - 3 - 2 - 1       Vanessa H.
                1 + 23 + 4 + 5 + 6 - 7 - 8 - 9          Daniela S.
                98 - 76 - 5 - 4 + 3 - 2 + 1             Cristina F.


Closing Comments

     Sometimes when one is confused and does something contrary to the rules of the game, new and interesting discoveries can be made. Just such an event took place with this activity with one of my students. Marinés did not fully understand the terms "consecutive digits" and wound up making a different kind of expression problem. She understood it to mean "consecutive numbers", a natural mix-up by many students, as any experienced teacher knows.

     She created the following expressions. As you examine them, you will see where her confusion lies, yet, I believe, be impressed with her effort and creativíty.

15 = 9 + 10 + 11 + 12 - 13 - 14

15 = 7 + 9 + 11 - 13 + 5 - 3 - 1

     While of course the second one can be arranged in ascending or descending order, it was the first one that caught my eye. It started me to wondering if there was a pattern here for longer (or shorter) strings of consecutive numbers that could be made to equal 15. By applying a little elementary algebra, I quickly came up with these expressions:


	three:   16 + 17 - 18

	five:    19 + 20 + 21 - 22 - 23

	seven:   24 + 25 + 26 + 27 - 28 - 29 - 30

	nine:    31 + 32 + 33 + 34 + 35 - 36 - 37 - 38 - 39

	eleven:  40 + ... + 45 - 46 - ... - 50

     And I think you might begin to see a pattern here "in the patterns". Examine closely the first terms in each expression, and how they increase in a regular manner. [Nice, huh!]

     Now for some expressions with an even number of terms:

ten: 15 + 16 + ... + 20 - 21 - ... - 24

fourteen: 25 + ... + 31 - 32 - ... - 37

     Now, if my algebra didn't fail me, I claim that if the number of terms is a multiple of four (4, 8, 12, ...), you cannot form an expression of this type. However, for the other even numbers an expression is possible --- as Marinés and I have shown you.

     We leave it up to you to prove us wrong.

     We also leave it up to you to look for similar expressions that have a value of 9.

     In sum, WTM wishes to "take our hat off" to the fine work done by these young, budding mathematicians.

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