A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more to D. The number of coins D has now, is the smallest two-digit number. How many coins does A have in the beginning?

To solve this elegantly, we use the **Backtracking Strategy** [1] [7], starting from the end and working backwards.\n\n1. **Identify the Final State**: The question states D has the 'smallest two-digit number'. From our knowledge of number properties, this is **10** [35] [40].\n\n2. **Reverse the Operation**: The forward operation is 'Half + 2'. \n - To go back, we must do the **Inverse Operation** [16]: Subtract 2 first, then Double (multiply by 2).\n\n3. **Step-by-Step Backtracking**:\n - **From D to C**: D has 10. \n - Reverse: $(10 - 2) \\times 2 = 8 \\times 2 = 16$. So, C had 16.\n - **From C to B**: C has 16.\n - Reverse: $(16 - 2) \\times 2 = 14 \\times 2 = 28$. So, B had 28.\n - **From B to A**: B has 28.\n - Reverse: $(28 - 2) \\times 2 = 26 \\times 2 = 52$. So, A had 52.\n\nThis approach avoids complex algebra and relies on intuitive arithmetic.