UPSC 2005: There are 10 identical coins and each one of them has ‘H’ engraved on one face and ‘T’ engraved on the other face. These 10 coins are lying on a table and each one of them has the ‘H’ face as the upper face. In one attempt, exactly four (neither more nor less) coins can be turned upside down. What is the minimum number of attempts in which the ‘T’ faces of all the 10 coins can be brought to be the upper faces?

A minimum of four attempts suffices: 1) flip coins 1-4 → 4 T; 2) flip coins 5-8 → 8 T; 3) flip coins 9-10 + any two T-coins (say 1,2) → coins 1,2 revert to H (6 T total); 4) flip the two H-coins (1,2) and any other two T-coins → all ten show T. Fewer than four attempts cannot work because attempts×4 must be ≥10 and parity considerations force at least one coin to be flipped more than once, which demands at least four distinct 4-coin moves. Hence the minimum is 4 attempts. ([licchavilyceum.com](https://licchavilyceum.com/upsc-civil-services-prelims-2005-solved-question-paper/?utm_source=openai))