Six identical cards are placed on a table. Each card has number '1' marked on one side and number '2' marked on its other side. All the six cards are placed in such a manner that the number '1' is on the upper side. In one try, exactly four (neither more nor less) cards are turned upside down. In how many least number of tries can the cards be turned upside down such that all the six cards show number '2' on the upper side ?
- (a) 3 ✓ UPSC's answer
- (b) 5
- (c) 7
- (d) This cannot be achieved
Why the answer is (a)
• Let k be the number of cards showing 2; initially k = 0, and each try flips exactly 4 cards, so k changes by 4 - 2x, where x is the number of 2-cards flipped.
• In one try, at most 4 cards can be turned to show 2, so k = 6 cannot be reached in 1 try.
• In two tries, the first try must make k = 4; from k = 4, reaching k = 6 in the second try would require flipping 3 cards showing 1 and 1 card showing 2, but only 2 cards show 1, so 2 tries are impossible.
• In three tries: first flip 4 cards to get k = 4; second flip 3 of those 4 cards and 1 of the remaining 2 cards to get k = 2; third flip the 4 cards showing 1 to get k = 6.
• Therefore the least number of tries is 3, so option (a) follows.
Why the other options are wrong
- (b) 5
- Option (b) is wrong because 3 tries are sufficient, so 5 is not the least number.
- (c) 7
- Option (c) is wrong because 3 tries are sufficient, so 7 is not the least number.
- (d) This cannot be achieved
- Option (d) is wrong because the required state is achieved in 3 tries by flipping 4 cards, then 3 previously flipped cards plus 1 unflipped card, then the remaining 4 unflipped cards.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2014, held on 24 August 2014. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.