PRODUCTSwithNUMBER SUMS

```	Here is an interesting problem that only involves basic arith-
metic, but with an unusual twist.

Find 5 numbers such that when each number is multiplied by
the sum of the remaining 4 numbers, one of the following values
will result:

44    63    95     108    128

WOW!  Did you get that?  I hope you didn't get too confused.
All it really says is: let's take five numbers -- say 1, 3, 4, 7, and
8 -- add any four of them, then multiply the sum by the remaining
number.  In this sample, try adding 1, 3, 7, & 8, getting 19; now
multiply 19 by 4, getting 76.

Now it seems easy enough, right?  But that's because we knew
the 5 numbers in advance.  This is another of those problems I call
"JEOPARDY" problems.  When you are given the "answer", you must find
the "question".

You know?  Perhaps in this case I'll give you some pretty heavy
hints, as it does seem to be a "tough nut to crack" this time.  We begin
by forming factor T-charts for each of the 5 products.

44            63            95           108          128
1   44        1   63        1   95        1  108       1  128
2   22        3   21        5   19        2   54       2   64
4   11        7    9                      3   36       4   32
4   27       8   16
6   18
9   12

Next locate those factor pairs that have the same sum; in this
case it should be 24.  The smaller members of the required pairs are
indicated in red.  They are the 5 numbers which solve the original
problem.

Perhaps you're wondering how I knew the common sum should be 24.
It's really quite simple if you look at the chart that has the fewest
entries: 95.  Since it only has two factor pairs, whose sums are 96 and
24, this shortens our search quite a bit.  Then since the first two
charts (for 44 and 63) only have factor sums less than 96, this leaves
only 24 to be considered.  The rest, as they say, is "easy as pie".

PROOF

The reason the procedure works out so well can be proved by
simple algebra.  Assume the five numbers are a, b, c, d, and e.  There-
fore, we have

a(b + c + d + e)  =  44

b(a + c + d + e)  =  63

etc.

Notice that the sum of the two factors in each equation is always the
same, namely a + b + c + d + e, the sum of the desired numbers.  Hence,
the procedure of finding the factor pairs possessing equal sums becomes
rather obvious.

Teacher Notes:

Here are some problems that can be presented to your math class
after the procedure has been explained.  Many more can be prepared as
needed.

1.	245   297   152   320   360
2.	220   328   468   490   360
3.	111   319   375   204   175
4.	522   742   816   570   882

Variations

1.  Of course, more (or fewer) numbers than five can
serve as the basis.  It only depends on student
interest, level, time, etc.

2.  A problem can be given with one of the products
missing.  The problem will be considered as solved
when that product has been found.  (This is easier
than it sounds at first.)

3.  A competition between 2 students can be set up.
Each player chooses, say 6 numbers, and prepares
the set of products.  A limit on the size of the
numbers should be agreed upon in advance.  Then
they trade number sets and the first to declare
the other's 6 numbers is the winner.

Final Notes

Since the method of solution is being directly presented to
the student from the start, the pedagogical objectives of this activity
are these:

a) Important practice in finding factor pairs of whole
numbers---a skill much needed in algebra courses;

b) Experience in following directions in a non-tradi-
tional problem setting; and

c) Drill in systematic-search techniques for problem
solving.

d) NOTE: Calculators should be considered as an option,
especially if large numbers are employed.

REFERENCE: Maurice Kraitchik, MATHEMATICAL RECREATIONS, Dover Pub., 1953.
p. 46.

Answers: (1) 4, 7, 9, 10, 12
(2) 5, 8, 9, 13, 14
(3) 3, 5, 6, 11, 15
(4) 9, 10, 14, 16, 18

tt(1/5/86)

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