AMC 10 Daily Practice Round 1

In the worst-case scenario, you would try the first key with 3 locks before finding the correct one, then set that key and lock aside. For the remaining 3 keys and locks, you would try the second key with 2 locks before finding the correct match, then set them aside as well. For the remaining 2 keys and locks, you would only need to try 1 lock before matching the key. The last key and lock do not require testing since they are the only pair left. Therefore, the maximum number of attempts needed is 3+2+1 = 6 attempts. Thus, the answer is .