Challenge

Write down the numbers 1,2,...,2n for some positive integer n. Now out of these choose n+1 numbers. Prove that out of these n+1 numbers there exists numbers a and b such that a divides b.

demonstration/example.
1,2,3,4  n=2 so n+1=3.
The only set that doesn't contain 1 is {2,3,4} and 2|4.

1, 2.

|{1 / 3 = 4.}|

|Q.E.D.}

>>2

lol wut

Consider 1, 2, ..., 2n for some n in the natural numbers.
These may be labeled 1, a(2), a(3), ..., a(n)
1|a(i) where 1 < i < n+1

Consider 2, 3, 4, ..., 2n for some n > 1 in the natural numbers.
The number of odds between 2 and 2n is (2n - 2)/2 = n - 1
Selecting n+1 numbers ensures that at least 2 of the selected numbers are even.
We'll call these evens e(1) and e(2), e(1) being the smaller.
There are two cases
1: e(2) = e(1)*2^k, thus e(1)|e(2)

2: e(2) != e(1)*2^k, so e(1) does not divide e(2)
at least one of these evens divides, or is divisible by an odd.

sorry, i'm tired, the last line shouldnt have the even dividing an odd part.

>>4

Could you explain case 2 in more detail.

by "in detail" I mean why you can find an odd in the n+1 numbers you chose that divides one of these evens.

well, even if you only picked two evens, and youve picked all of the odds, one of the odds has to be half of one of the even numbers if one of the evens doesnt divide the other.

if the evens were powers of two, they would divide eachother, so if they dont, you can assume they contain odd prime factors, and so one of those odd factors would have been selected.

I think I can finish >>4's argument.  Let the chosen numbers consist of p evens and n+1-p odds.
Case 1: One even number divides another, easy.
Case 2: No even number divides another.  Thus, there are p even numbers whose odd prime factorizations are different.  n+1-p odd numbers were chosen.  [1,2n] contains n odd numbers.  Since (n+1-p)+p = n+1 > n, the sets {odd factors of chosen even #s} and {chosen odd #s} must overlap, and one of the odds has to divide one of the evens.  Amirite?

If you want us to help you with your homework, you could've just said so.

>>9
yeaaaah thats it =D
you'd have enough different factorizations containing odds in the evens, that even if you didnt pick all the odds you still would have had to have picked one of the ones that divides the even.

I'll do this easily

If R=4(X2-10)+R1(X-2*12)*(R+4X)\(x2+R12*X12-x14)+(R>13-X\x) = R(X\x)+45x(R-4*R4x)\(X2-R421+x24-X4+x-r) = N1N2N3N4 where N =x-X(R33-x/X) Where X=R-x and x=R+X/4