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?

On re-checking, my earlier conclusion was inconsistent. The backward method gives:

• D has 10 (smallest two-digit number). Verdict: given.
• C to D: C gave half of his coins and 2 more to D, so amount given = 10. Hence C/2 + 2 = 10 => C = 16. Verdict: correct.
• B to C: After receiving from A, B then gave half of his coins and 2 more to C, and C finally had 16. So B/2 + 2 = 16 => B = 28. Verdict: correct.
• A to B: A gave half of his coins and 2 more to B, and B had 28 before giving to C. So A/2 + 2 = 28 => A = 52. Verdict: correct.

Thus A initially had 52 coins, so the correct option is (d). My earlier response had a calculation-answer mismatch; the reasoning actually supports (d), not (a).