A machine takes any number, doubles it and then takes 3 away. The number 5 is put in; the answer is put straight back into the same machine; and the answer to that is put back in once more. So the machine is used three times in all. What number comes out at the end?

Run the machine one pass at a time. First pass: $5\times 2=10$, then 10 − 3 = 7. Second pass: $7\times 2=14$, then 14 − 3 = 11. Third pass: $11\times 2=22$, then 22 − 3 = 19. The traps: 11 stops after two passes; 37 doubles three times first (5 → 10 → 20 → 40) and then subtracts 3 only once at the very end; 31 does the same three doublings first but subtracts 3 three times (40 − 9). Both of those last two forget that each answer has to go back in before the next doubling.