Question

- this is solved correct - right ?

Answer 1

ok . so given the line of naturale numbers from 1 to 2002

let being the set of even numbers with name A and the set of odd numbers with name B

1: assume every numbers inside set A and inside set B
2. the smaler result substract from greater result

so in this way how many will get ?

Answer 2

1. than we try this above wrote with just the 4 numbers

A contain (2,4)
B cont. (1,3)

assumed from A result 2+4=6 and from B result 1+3=4
and 6-4=2 so there are 4 numbers and 4/2 = 2 the result is the number of used numbers halfed

2. A cont. (2,4,6) assumed A=12
and B cont. (1,3,5) assumed B=9

A-B=12-9=3 what is the used 6 numbers halfed

3. A cont (2,4,6,8) assumed A=20
and B cont. (1,3,5,7) assumed B=16

A-B=20-16=4 what is the used 8 numbers halfed

using this resulted ussefully way than given the nnumber line from 1 to 2002 we get 2002/2 = 1001

@Vocaloid what is your opinion about this please ?

Answer 3

@Nnesha ???

Answer 4

I know I am not who you requested, but it seems like a logical solution to me.

Answer 5

@jhonyy9

Answer 6

if you subtract two consecutive numbers you get a 1

Answer 7

@TheSmartOne any idea here on this - there exist an other way to solve it ?

Answer 8

Agreed with pokerface, it seems like by doing (even numbers - odd numbers), you're getting consecutive pairs so like
(2-1) + (4-3) + (6-5) + (8-7) + (10-9) + ...
so whatever result you get above is n which is how many pairs of even/odd numbers you have, essentially (total # of numbers)/2

Answer 9

So you could go way past 2002 I'm sure

If you use an eneven number of pairs like
A: 2, 4, 6
B: 1, 3, 5, 7

Then this won't work

12 - 16 = -4
7/2 = 3.5

Or like
A: 2, 4, 6, 8
B: 1, 3, 5

20 - 9 = 11
7/2 = 3.5

Answer 10

ty. @TheSmartOne