Melody

there is nothing to solve.

This cannot be the whole question or else it simply is not a question.

You answer has no credibility whatsoever. So please justify it or go away.

Compute the sum 1 + 4 + 7 + ... + 2005 + 2008.

It is an arithmetic progression

a=1   d=3     L=2008

Tn=a+(n-1)d

2008=1+(n-1)*3

2007/3 = n-1

669=n-1

n=670

sn=n/2(a+L)

Sn=670/2    *(1+2008)

Sn=335  *2009

Sn=673015

My first tip is to draw it.

Have you done that?

Also, when someone talks about the distance of a point to a line they are referring to the perpendicular distance.

There are 2 possible points Cs.

Graph it to see what I mean

Hint.  One will be between the numbers, and the other will be outside.

Draw the number line and figure it out.

I have not worked it out properly.  I just drew the pic.

Looks like about 7

Nice logic Textot!   

That makes your way nice and simple.    

Why don't you tell us what shoelace theorum is?

I do not know how to do it either but I would be interested in seeing the answer.

Hi Textot,

Thanks for your contribution.

I really liked the way you started your answer.

You are of course correct,   

a+b+c = 22

where a,b, and c are ordered integers all greater or equal to 0

will have exactly the same number of solutions.

Your use of stars and bars was also interesting but unfortunately, it was not appropriate for this problem as it does not take into account that the triplet must be ordered.

This means that your answer is somewhere in the order of 3! or 6 times too big.

I used your simplification of  a+b+c=22

and made a probability contour map of it.

Instead of a,b, and c

I used

x,    y-x    and    22-y      (if you add these up you get 22)

Now these restraints come into play

\(x\ge0\qquad (1)\\~\\ y-x>x\\ y\ge2x\qquad(2)\\~\\ 22-y\ge y-x\\ -2y\ge -x-22\\ 2y\le x+22\\ y\le \frac{x}{2}+11\qquad (3)\\~\\ 22-y<22\\ y\ge0\qquad (4) \)