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?

To solve this problem, we track the number of cards showing '1' on the upper side. The goal is to have 0 cards showing '1' (meaning all 6 cards show '2').

Initial state: All 6 cards show '1'. (6 '1's, 0 '2's)

Try 1: We must flip exactly four cards. Since all cards are '1', we flip 4 cards that show '1'. These 4 cards change from '1' to '2'. Number of '1's remaining = 6 - 4 = 2. Number of '2's = 4. Current state: (2 '1's, 4 '2's)

Try 2: We have 2 cards showing '1' and 4 cards showing '2'. We must flip exactly four cards. To move towards the goal of 0 '1's, we want to flip as many '1's as possible and as few '2's as possible, or at least not increase the '1's significantly. Let x be the number of '1's we flip, and y be the number of '2's we flip. We know x + y = 4. Possible scenarios:

1. Flip 2 '1's and 2 '2's (x=2, y=2): The 2 '1's become '2's. The 2 '2's become '1's. New number of '1's = (2 - 2) + 2 = 2. (This brings us back to the same state as after Try 1, no progress).
2. Flip 1 '1' and 3 '2's (x=1, y=3): The 1 '1' becomes '2'. The 3 '2's become '1's. New number of '1's = (2 - 1) + 3 = 4.
3. Flip 0 '1's and 4 '2's (x=0, y=4): The 4 '2's become '1's. New number of '1's = (2 - 0) + 4 = 6. (This takes us back to the initial state).

The most effective choice to make progress is to aim for a state that allows us to reach the target in the next step. Choosing to have 4 '1's and 2 '2's (scenario 2) is the best option here. Current state: (4 '1's, 2 '2's)

Try 3: We have 4 cards showing '1' and 2 cards showing '2'. We must flip exactly four cards. To reach the target of 0 '1's, we need to flip all 4 cards that show '1'. We can choose to flip 4 '1's and 0 '2's (x=4, y=0). This is possible as we have 4 '1's available. The 4 '1's become '2's. New number of '1's = (4 - 4) + 0 = 0. Number of '2's = 6. Current state: (0 '1's, 6 '2's). All cards show '2'.

Thus, it takes a minimum of 3 tries to achieve the desired state.

The final answer is A) 3.