A pass code is created by combining one-digit prime numbers. What is the probability that the resulting four-digit number is a multiple of 12 ?
1/16
1/16
There 4 one-digit prime numbers : 2, 3, 5, and 7
So, total number of four-digit codes = 4*4*4*4
Now, for the code to be a multiple of 12, the code must be divisible by both 3 and 4.
So, the sum of the prime numbers used to for the code must be divisible by 3 and number formed by the last 2 digits of the code must be divisible by 4.
So, the last two digits of the code is either 32 or 52 or 72.
XX32 : XX can be 22, 25, 52, 55, 37, 73 ---> 6 possibilities
XX52 : XX can be 23, 32, 35, 53, 77 ---> 5 possibilities
XX72 : XX can be 33, 27, 72, 57, 75 ---> 5 possibilities
A total (6 + 5 + 5) = 16 such codes possible.
So, required probability = 16/(4*4*4*4) = 1/16
