Re: [DEHAI] ...founder of Algebra

Solomon Tesfalidet (Solomon.Tesfalidet@CHEM.UMU.SE)
Sat, 22 Mar 1997 14:20:31 +0100

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Hej! Negga and seb dehai,

I like such type of questions where a mathematical model is needed to=20
express a word problem. When solved by other ways, such as using trial and=
=20
error, it can be time consuming. I am not a mathematician, but I would like=
=20
to see how your teacher answered the question which was originally posed by=
=20
your mom. There might be easier and much more better ways to solve the=20
problem. Anyway I solved it this way.

To sum the quintals:

You can simply add this way, 1+2+3+4+5+6+7+8+9 =3D 45. So far so good,=20
because it is only nine numbers. What about if it was=20
1+2+3+4=85=85=85=85998+999+1000 =3D X ?=20

Here, if a formula or a model is used the addition can be performed in a=20
simpler and faster way.

The simplest way is to add 1000+999+998=85=85=85=85=85=853+2+1 as follows:

1 + 2 + 3-----------998 + 999 + 1000 +=20
1000 + 999 + 998-------------3 + 2 + 1
______________________________________________________________________
1001 + 1001 + 1001----------1001 + 1001 + 1001=20
=20
=3D 1001X1000/2 =3D 500500.

The second row is added only to make the addition simpler. The only thing we=
=20
need to do after addition is to divide the sum by two.=20

We can now derive the following formula for solving this type of problems:

(N(f) + N(i))X N/2 where N(f), N(i) and N denote the final number, the=20
initial number and the number of figures to be added respectively.

Back to our original problem:

(9+1)9/2 =3D 45=20

45 quintals are to be shared by three merchants, which results in 15=20
quintals each.

To solve the problem on how the three men should share the camels equally,=
=20
do as follows.

Draw a square or a rectangle and divide the square or the rectangle into=20
nine smaller boxes of the same size.
Fill the boxes with the numbers 1-9 such that when you add column wise, row=
=20
wise and diagonally the sum =3D 15.

This can be done by placing the even numbers 2,4,6, and 8 in the four corner=
=20
boxes and number 5 in the central box. Then put the remaining numbers in=20
each box as follows:=20

2 9 4
7 5 3
6 1 8

This gives two different solutions to the problem i.e., the three rows and=
=20
the three columns.

Halsningar=20
Solomon Tesfalidet

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