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November 23rd, 2016, 09:18 AM  #11  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
$1(0) + 2(0) + 3(1) = 3.$ First solution. $1(1) + 2(1) + 3(0) = 3.$ Second solution. $1(3) + 2(0) + 3(0) = 3.$ Third solution. The first solution satisfies the condition that each element of the triplet be either 0 or 1. As I pointed out before, FOR ANY n $x_k = 0\ if\ 1 \le k < n\ and\ x_n = 1 \implies$ $\displaystyle \sum_{i=1}^n(i *x_i) = \left (\sum_{i=1}^{n1}(i * 0) \right) + (n * 1) = 0 + n = n.$ The third solution satisfies the condition that each element of the triplet except the last either be positive or be followed by a zero element because the first element is both positive and followed by a zero element and the second element is followed by a zero element. As I pointed out before, FOR ANY n $x_k = 0\ if\ 1 < k \le n\ and\ x_1 = n \implies$ $\displaystyle \sum_{i=1}^n(i *x_i) = 1(n) + \sum_{i=2}^n(i * 0) + (n * 1) = n + 0 = n.$ The second solution satisfies both conditions. So two solutions satisfy the first condition, and two solutions satisfy the second condition. So the number of solutions that satisfy the first condition equals the number of solutions that satisfy the second condition. I admit that checking out n = 4 seems a bit daunting because there are 625 cases, but it is not that hard to do with a computer.  
November 23rd, 2016, 11:32 AM  #12 
Newbie Joined: Nov 2016 From: Slovenia Posts: 24 Thanks: 0 
yes, thanks, I am trying to prove it now, vainly as yet

November 24th, 2016, 05:13 AM  #13 
Newbie Joined: Nov 2016 From: Slovenia Posts: 24 Thanks: 0 
maybe have you found a proof ?

November 25th, 2016, 05:52 AM  #14 
Newbie Joined: Nov 2016 From: Slovenia Posts: 24 Thanks: 0 
or a hint?


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equation, fixed, integer, positive 
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