There are three keys to the three doors with different locks on them.
How many attempts would be needed to determine the exact keys for all the doors under all circumstances?
3 attempts let us say A B and C locks with a b and c keys Try two keys on A if the keys are b and c it will not open obviously the key not tried is its key a.Leave it aside. Again try one key of the left keys b and c (not known) if it opens lock B it is b if does not it is C. so total three attempts. It can also happen that the first key opens lock A. The second attempt, if it opens lock B it is b if it does not it is c.In this case only two attempt will determine the keys. It can also happen that the first key does not open lock A but the second does. Keep it aside and try the remainder of b and c on B. If it opens Lock B it is b and if does not it is c. In this case also total attempts are two. Hence 3 is the answer.
If you are only allowed to press the 1 and 0 keys on a calculator with any of the add, take, times, divide and equals keys, what is the fewest number of presses needed to make the number 9 on the calculator?
In a kitchen there are 11 cupboards some with one door and some with two. There are 15 doors altogether so how many of each cupboard are there?