SOLUTION: Samir has a 23 link chain. He wants to cut and disengage the minimum number of links so that any number of links from 1 to 23 can be obtained by taking one or more of the resulting

(A)  First cut must be at the 3-rd link.

     Then Samir has this 3-rd link cut as a single link;
             he has also the block of 2 links (the 1-st and the 2-nd links connected together);
     and then he has 3 links as the union ((1st + 2nd together) U 3rd).



(B)  Next cut must be at the 8-th link.

     Then, in addition to the set (A), Samir has 

         - this 8-th link cut;
         - the block of 4 links (4th+5th+6th+7th all connected together),

         and then it is clear that combining this block of 4 links with what he just has in the set (A),
           he is able to have 
               a) 4 links; b) 5=4+1 links; c) 6=4+2 links; d) 7=4+3 links; and e) 8=4+3+1(8th) links.



By continuing with this pattern, it is clear that 

(C)  the next, third cut must be at the  8 + 8 + 1 = 17-th link.

     Then, in addition to the sets (A) and (B), Samir has 

         - this 17-th link cut;
         - the block of 8 links (9th+10th+11th+12th+13th+14th+15th+16th all connected together),

         and then it is clear that combining this block of 8 links with what he just has in the sets (A) and (B),
           he is able to have 
              a) 8 links;          b) 9=8+1 links;    c) 10=8+2 links;   d) 11=8+3 links;     e) 12=8+4 links;
              f) 13 = 8+4+1 links; g) 14=8+4+2 links; h) 15=8+4+3 links; i) 16=8+4+3+1(8th) links; k) 17=8+4+3+1(8th)+1(17-th) links.


(D)  the next (and the last), fourth cut must be at the  17 + 3 = 20-th link.

     Then it is OBVIOUS that having the sets (A), (B), (C) and additional 2 connected segments (18,19) and (21,22,23) and 1 cut link (20th)

          Samir will be able to present

              a) 18 = 17 +1;  b) 19 = 17+2;  c) 20 = 17+2+1;  d) 21=17+3+1;  e) 22=17+2+3; and, finally, all 23 links as well.


So, the problem is just solved and the ANSWER is:  the cuts should be made at 3-rd;  8-th;  17-th;  and  20-th links.


It is clear (intuitively, without formal proof) that it is the way to satisfy condition with the minimum number of cuts.


It is also clear, that if the chain is VERY LONG, then the cuts should be done in such a way to create the connected segments of the length

    2, 4, 8, 16, 32 . . . and so on by the degrees of 2 at each step . . .