Cracking the Four-Digit Lock

Worked solution

Start with 7295, which has two digits in the right place and two digits that are not in the code at all. Suppose 2 were in the code (so, in place, in position 2). Then 6932 and 3621 each already hold their single allowed code digit, so 6, 9, 3, 1 are absent; but 8753 needs three code digits from 8, 7, 5, 3, so 8, 7 and 5 would all be present, and then 7295 would contain three code digits (2, 7, 5), not two. Suppose 9 were in the code: the same argument (6, 3, 2 absent, 8753 forcing 8, 7, 5) gives three code digits 9, 7, 5 in 7295, again a contradiction. So 2 and 9 are absent, and the two correct digits of 7295 are 7 in position 1 and 5 in position 4. Now consider 8354: the 5 is in the code but misplaced (it sits in position 3 there), so its one right-place digit must be 8 in position 1, 3 in position 2 or 4 in position 4. Positions 1 and 4 of the code are already 7 and 5, so it is the 3, in position 2: the code is 7 3 _ 5. The guess 8753 then has code digits 7, 5, 3 (all misplaced), so 8 is absent; 8354 has code digits 3 and 5, so 4 is absent; 6932 has only the 3, so 6 is absent; 3621 has only the 3, so 1 is absent. The last digit has not been excluded only if it is 0. The code is 7305, and a computer search over all 5040 codes confirms it is the only one.