solve the puzzle how many liars? Each boy in a group of $20$ boys either always tells thuth or always tells a lie. These boys are sitting around a table. Each boy says that his neighbours are liars. Prove that at least $7$ out of $20$ must be truth tellers. 
 A: If there are $\le 6$ truth tellers, then there exist $3$ liars sitting together by pigeon-hole principle. 
Now the one in the middle of these three is telling the truth when he/she says "my neighbours are liars", a contradiction!
A: A liar tells that both his neighbours are liars.
Therefore at least 1 neighbour of each liar is truthful.
Therefore no three successive boys can be liars.
Each truth teller's both neighbour are liars, so no two successive boys are truthful.
Suppose there are $k$ truth tellers. Around the table, between pairs of successive truth tellers there are $k$ gaps. In each gap, there must be at least one liar (otherwise, the successive boys would be truthful) and at most two liars (otherwise, there would be $3$ successive liars).
Therefore the number of liars is at least $k$ and at most $2k$.
So total number of boys  is at least $k + k =2k$ and at most $k+2k=3k$.
If $k\leq 6$, then $3k \leq 18$ but we have $20$ boys.
Therefore $k$ is greater than or equal to $7$. 
Hence proved.
A: No three consecutive boys can be liars, or else the middle one would not say both his neighbours are liars. So every boy either tells the truth or has a neighbour that tells the truth (since the table is round, everybody has two neighbours). If there were at most $6$ truth tellers, that would account at most for those $6$, plus their $12$ neighbours; their might be overlap, but in any case this cannot cover all $20$ boys, contradiction
