Hi everyone!
As my first contribution, I hope this one will be curious enough for you.
Do you remember about one of my inquiries in class regarding the easiest thing that anyone can do in Maths?
It
was, of course, TO COUNT. Therefore, if you could reduce in some way
another kind of problem to a counting-problem, it would be easier. The
purpose of this entry is to put an example about this way of reasoning:
"What is the sum of all the numbers from 1 to 500?"
Whenever
I mentioned this arithmetic problem in class, it seemed to take too
long due to the quantity of sums it required. So let's have a look at
another way to figure out the result of the sum:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + ... (here go all natural numbers between) ... + 497 + 498 + 499 + 500.
If
you were counting some things (red dots, for example) that have to do
with that situation, they could be displayed as groups by rows, with
each group standing for each one of these numbers, and all groups
aligned at their beginning, like this:
• (1)
• • (2)
• • • (3)
• • • • (4)
• • • • • (5)
(etc.)
It
is very easy to realize that all groups form a right triangle -with
same height as length in dots, so it is isosceles too-, no matter how
many numbers you want to consider, in our case 500. And that is very
near to be half square with 500 dots on each side.
Now,
imagine every dot inside a square box, and add more boxes on each row
to complete a big square. It is obvious that we have 500 · 500 boxes,
the only thing left to decide is how many of them have a red dot. It
will help if we colour diagonal boxes in blue, and other boxes with dots
in yellow, like in the picture below.
Boxes in both colours match to those with a dot. Notice that
blue boxes must be exactly
500.
And what about
yellow ones?. Notice that there are same number of
yellow boxes as no-color boxes, so they are
(500 · 500 - 500) : 2 = 124 750 .
Hence, the red dots are
125 250.
By the way... ...Could you apply this to other examples?
