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I am noobie in number theory so if something exists better than this equation I did please don't bash me this is my first equation.

Today I was thinking of a way to count all even from 1 to n,so this way I thought about is like this.

Let me first write the equation,then I will explain the logic.

iNumber = ((iNumber + 1) * (iNumber / 2)) - (((iNumber + 1) * (iNumber / 2) - (iNumber / 2)) / 2)

Okay let me explain how I got this logic,so lets pick odd numbers from 1 to 10 and see how they form.

1 = 0 + 1;

3 = 2 + 1;

5 = 4 + 1;

7 = 6 + 1;

9 = 8 + 1;

So here we have 5 ones,so for example if we removed those ones from the a equation that you add 1 to 10 from it will be like this;

Normal numbers which you added 1 to 10 ->

1 + 2 + 2 + 1 + 4 + 4 + 1 + 6 + 6 + 1 + 8 + 8 + 1 + 10 -

1 + 1 + 1 + 1 + 1,so the numbers now become

0 + 2 + 2 + 4 + 4 + 6 + 6 + 8 + 8 + 10 = 50,but thats 2x,so if we got that number and divide by 2 it will be 25,thats how much odds add up,so if we got the total numbers from 1 to 10 and minus 25 we should get the even number we want;

So lets put that in the equation and see if its right;

((11) * (5)) - (((11) * (5) - (5)) / 2) = 55 - 25 = 30;

2 + 4 + 6 + 8 + 10 = 30;

So the equation is right I am sure this has been done by another mathmetician,but its good to think about it the logic of it is great.

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# Count all even from 1 to n

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