A pass code is created by combining one-digit prime numbers.

[varun289]

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

[Atekihcan]

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

[Brent@GMATPrepNow]

Excellent solution, Atekihcan

I should point out that, given the time-consuming brute force required here, this question is not indicative of official GMAT questions.

One of the great things about almost all GMAT math questions is that they can be solved using at least 2 different approaches. Typically, one approach is much faster than the other(s). Unless I'm missing something, there's no faster solution here.

Now if this were a true GMAT question, the answer choices might look something like this
A) 1/27
B) 1/16
C) 1/12
D) 1/10
E) 1/9

Here, once we determine that the denominator is 4^4 (as Atekihcan dis), we can immediately eliminate A, C, D, and E, since it's impossible for X/(4^4) to simplify to be any of those four fractions.

This is why it's important to always include the answer choices when posting practice questions.

Also, when posting questions, please use the spoiler function to hide the correct answer. This will allow others to attempt the question without seeing the final answer.

Cheers,
Brent