A problem from the British Columbia Colleges Senior High School Mathematics Contest, 2000:
Not all of the nine members on the student council are on speaking terms. This table shows their relationships — 1 means two members are speaking to each other, and 0 means they’re not:
A B C D E F G H I
A  0 0 1 0 0 1 0 0
B 0  1 1 1 1 1 1 1
C 0 1  0 0 0 1 1 0
D 1 1 0  1 0 1 0 1
E 0 1 0 1  0 1 0 0
F 0 1 0 0 0  0 0 1
G 1 1 1 1 1 0  0 0
H 0 1 1 0 0 0 0  0
I 0 1 0 1 0 1 0 0 
Recently councilor A started a rumor, and it was heard by each councilor once and only once. Each councilor heard it from, and passed it to, another councilor with whom she was on speaking terms. If we count councilor A as zero, then councilor E was the eighth and last councilor to hear the rumor. Who was the fourth?

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We know that A is at the start of the sequence, and E is at the end. Of the others, F and H are each speaking to only two other councilors. So F must pass the rumor between B and I (in one direction or the other, BFI or IFB), and H must pass it between B and C (BHC or CHB). Putting these fragments together, we must have either CHBFI or IFBHC. Also, this larger fragment can’t connect directly to either of the endpoints A or E, because neither C nor I is speaking to either of them. That means that the remaining councilors, D and G, must go on either end of our fragment before we can attach the endpoints. So the extended fragment is either GCHBFID or DIFBHCG, and with the endpoints A and E added we get the full sequence AGCHBFIDE or ADIFBHCGE. In either case, if A is counted as zero then the fourth person is B.
(Crux Mathematicorum 27:3 [April 2001], 194.):
