Machine A takes a number, multiplies it by 5 and then subtracts 8. Machine B takes a number, adds 4 and then divides by 3. A number is fed into A, whatever comes out of A is fed straight into B, and B finally gives out 12. What number was fed into A at the start?

Walk backwards, undoing the last step first. B divided by 3 to reach 12, so before that it held $12\times 3=36$; B had added 4 to reach 36, so it received 36 − 4 = 32. That 32 is what A gave out. A had subtracted 8 to reach 32, so before that it held 32 + 8 = 40; A had multiplied by 5 to reach 40, so it received 40 ÷ 5 = 8. Checking forwards: $8\times 5=40$, 40 − 8 = 32, 32 + 4 = 36, 36 ÷ 3 = 12. The traps: 32 is the number that left A, one machine short of the start; 40 is the value inside A after only its multiplication step; 20 comes from undoing the 'subtract 8' first, before B has been undone at all, and so applies A's step to B's output.