10.24.2011

Geometric approach for arithmetic problems

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?

18 comments:

  1. This comment has been removed by the author.

    ReplyDelete
  2. I don´t understand this , Rafa.
    I think that we must see this in class for understand better.
    Bye :)

    ReplyDelete
  3. Hello!
    The sum of all the numbers (from 1 to 250)is:

    (250x250):2=31.250

    I think it's correct.

    Bye!

    ReplyDelete
  4. I was thinking and these isn't the solution.

    I have made an easy example:

    1+2+3+4+5=15 (5x5+5):2=15

    So:(250x250+250):2=31375

    ReplyDelete
  5. Hi!:)

    I don't understand the question...

    Bye! :)

    ReplyDelete
  6. Hello!

    The sum of all the numbers from (1 to 110)

    (110·110-110):2= (12100-110):2= 11990:2=5995

    Bye.

    ReplyDelete
  7. Hello class!

    The sum of all the numbers from 1 to 50.

    (50·50-50):2 = (2500-50):2 = 2450:2 = 1225

    The sum of all the numbers from 1 to 50 is 1225.

    Bye class!

    ReplyDelete
  8. This comment has been removed by the author.

    ReplyDelete
  9. This comment has been removed by the author.

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  10. You got it (or almost)!

    Rafa Martinez and I were discussing about it was a too complex question for 2º ESO students. But some of you are very near, and even correct, as I see in examples you mentioned above.

    My trick consists on subtracting from the squared highest number in the sequence, that proper number, then obtaining the half of that, and after this adding the number again...

    But you inferred a way to simplify that a bit more!

    Regarding your examples:

    Sum from 1 to 5 is [my way]
    (5·5 - 5): 2 + 5 =
    (25 - 5):2 + 5 =
    20 : 2 + 5 = 10 + 5 = 15

    [Your way] (5·5 + 5): 2 = 15 too!

    Can anyone explain why are both ways driving to same result, no matter the number considered?

    Sum from 1 to 250 is essentially correct. Congrats J.A.!

    Rafa: The question is to sum consecutive numbers from one (1) to any you want to, in the simplest way you can find!

    Ana Maria and Javi: Try again with cases you selected! You're very close to! There's only a detail missing!

    ReplyDelete
  11. hello!

    Sum of 1 to 20

    (20.20-20):2+20=
    (400-20):2+20=
    380:2+20=
    190+20=210


    and...


    1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20=210
    This is ok!


    Bye!

    ReplyDelete
  12. Hello!

    1 to 30=
    =(30·30-30):2+30=
    =(900-30):2+30=
    =870:2+30=
    =435+30=
    =465

    Goodbye:)

    ReplyDelete
  13. Hello!

    1 to 110=
    (110·110-110):2+110= (12100-110):2+110= 11990:2+110= 5995+110= 6105

    I think it is correct!

    ReplyDelete
  14. Hello!

    1 to 45=
    (45·45-45):2+45=
    =(2025-45):2+45=
    =1980:2+45=
    =990+45=
    =1035

    ReplyDelete
  15. Ok! So I'll try to do now. What are the sum for all the numbers 1 to 150?

    (150·150+150)/2 = (22500+150)/2 = 22650/2 = 11325

    ReplyDelete
  16. I don't understand the question, if I remember I'm going to tell you in the class okay¿?

    ReplyDelete
  17. I think it's corret

    1 to 110=
    (110·110-110):2+110= (12100-110):2+110= 11990:2+110= 5995+110= 6105

    bye class

    ReplyDelete

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