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#79  10 Nov 92 15:07:52  [2] (PRIVATE) (RECV'D) $0.22
From: Clovis Lacerda, 4:808/1, American School of Reci, Boa Viagem PE
To:   Carol Hooper, 1:381/64
Subj: cycle 2 - grades 9-10
          Group;Roberto Astuto, Thomas Messier, Tamara Kano, Karen 
Araujo
PROBLEM GRADE LEVEL 9-10
1. In such problem instead of looking for a solution for each
   number, we found two formulas that work one with odds and the
   other with evens, except 2 and 4.
2. Trying to find solutions for each odd number, we noticed that
   there was an easy way to get the solution for any odd number.
   You simply have to add and subtract the numbers, so that
   groups of two numbers cancel each other, leaving the last
   number, the answer.
Ex;  3=3-3+3
     5=5-5+5-5+5
     x=x-x+x-x+x-x........... +x
       __________________
       even number of x's
3. For evens, we represented each number writing it as twice
   the number minus itself.
Ex;  8=16-8
   Now, we had to simplify 16 with 8's. Since 16 is 2 times 8, we
   then have;
     8=2x8-8
   The problem now is to write 2 in terms of 8's. Since 2 is 1+1,
   both 1's can be written as 8 divided by 8. Therefore:
     8=(8/8  +  8/8)8 - 8
   We see that we now have six 8's, leaving us two more 8's to
   include in the equation. Since our equation is already right,
   we just add and subtract the remaining 8's. Finally:
     8=(8/8  +  8/8)8 - 8  +8-8
   This method only works for even numbers greater or equal to 6.
   For an even number Y, the equation is:
     Y=(Y/Y  +  Y/Y)Y - Y    + Y - Y...........-Y
                            _____________________________
                            even number of remaining  Y's.
Obs: The only way we could get a solution for 2 and 4 was using a
     square root. Is there any problem in using square root
     operations considering the inplicit 2 as an index of the
root?
Ex: Square root of 2 times square root of 2 = 2.
    Square root of 4 + 4 + 4 + 4 = 4


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