Solution
Ten Islanders and the Count of Knaves
Show the problem again
On an island every inhabitant is either a knight (always tells the truth) or a knave (always lies). Ten islanders stand in a line, numbered 1 to 10. Islander number $k$ says: "At least $k$ of the ten of us are knaves." How many of the ten are knights? Give a single integer.
Worked solution
Suppose exactly $x$ of the ten are knaves. Then the claim "at least $k$ are knaves" is true precisely for $k = 1, \dots, x$. Truthful speakers are knights, so the knights are exactly islanders $1, \dots, x$, which gives $x$ knights. The ten inhabitants split into $x$ knights and $x$ knaves, so $2x = 10$ and $x = 5$. Check: islanders 1 to 5 say "at least $k$ knaves" with $k \le 5$ and there are 5 knaves, so they speak truthfully; islanders 6 to 10 claim at least 6 knaves, which is false. The configuration is consistent, and a brute-force search over all $2^{10}$ assignments finds no other. Answer: 5 knights.
Source: Counting-statement puzzle in the knights-and-knaves tradition of Raymond Smullyan. Statement written for AxiomIQ.