Long ago, the king of Madadia lost the combination of the safe where the secret Madadian Kebab recipe was held. The most prolific safe cracker in Madadia's prison system was sent for, with a royal pardon if he succeeded in opening the safe.
His name was Al Krakit, and after several attempts at bypassing the combination, he realised that the only way to open the safe safely was by trying every possible combination by hand.
The lock was a 4-character code, with 2 of them being letters of either case and the other 2 being numbers.
Not knowing what case the letters are, what would be the maximum number of attempts Al Krakit would need to make to find the correct code?
Note that the code is case sensitive.
AnswerThe maximum number of attempts required are 1,622,400.
There are 52 possible letters - a to z and A to Z, and 10 possible numbers - 0 to 9. Now, 4 characters - 2 letters and 2 numbers, can be selected in 52*52*10*10 ways. These 4 characters can be arranged in 6 different ways - the number of unique patterns that can be formed by lining up 4 objects of which 2 are distinguished one way (i.e. they must be letters) and the other 2 are distinguished another way (i.e. they must be numbers).
Consider an example : Let's assume that @ represents letter and # represents number. the 6 possible ways of arranging them are : @@##, @#@#, @##@, #@@#, #@#@, ##@@
Hence, the required answer is
= 1,622,400 attempts
= 1.6 million approx.
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