a set which consists of 30 distinct positive numbers

orl

Let $M = \{ x_1..., x_{30}\}$ a set which consists of 30 distinct positive numbers, let $A_n,$ $1 \leq n \leq 30,$ the sum of all possible products with $n$ elements each of the set $M.$ Prove if $A_{15} > A_{10},$ then $A_1 > 1.$

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K09 · Poincare Conjecture Offline Joined: 08 Jan 2005 Posts: 118

I have solution .It is very simply.
We have S(5).S(10)>S(15)>S(10) then S(5)>1.
Otherwise S(1)^5 >S(5) >1 ,we have done

**Posted:** Thu Feb 03, 2005 7:23 pm

cosinerburc · Hodge Conjecture Offline Joined: 23 Jan 2005 Posts: 87

Here’s another solution
Suppose that A(1)<=1 then A(n)>=A(n)A(1)
it is clear that A(n)A(1)>A(n+1) so A(n)>A(n+1)
which implies A(10)>A(15).Contradiction

**Posted:** Sat Feb 05, 2005 12:27 pm