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?

We need to go from 0 cards showing '2' to 6 cards showing '2'.\n\n**Start:** 0 cards show '2'. (All 6 are '1').\n\n**Try 1:** We must flip 4 cards.\n- Since there are no '2's, we flip four '1's.\n- **Result:** Now **4** cards show '2' (and 2 show '1').\n\n**Try 2:** We are at 4. We want to reach 6.\n- To go from 4 to 6 directly, we would need to flip more '1's into '2's. Specifically, we'd need to flip 3 '1's and 1 '2' (Net change: +2). \n- *Constraint Check:* We only have 2 cards showing '1'. So we cannot flip 3 '1's. We are stuck and can't go up.\n- We must go down to prepare for a bigger jump. Let's flip 1 '1' (to make it '2') and 3 '2's (to make them '1's).\n- Net change: +1 - 3 = -2.\n- **Result:** 4 - 2 = **2** cards show '2'. (This means 4 cards now show '1').\n\n**Try 3:** We are at 2. We have 4 cards showing '1'.\n- We flip all those 4 cards showing '1'.\n- They all turn into '2's.\n- Total '2's = The 2 we didn't touch + The 4 we just flipped = 6.\n- **Result:** **6** cards show '2'. Target achieved.\n\nTotal tries: 3.