If the number is even, and its digits are all 3s and 4s, it must end in 4. If the number is divisible by 4, its last two digits must form a multiple of 4, so must be "44" (and not "34"). If the number is divisible by 3, the sum of its digits is a multiple of 3, and we can see that "444" satisfies all of the criteria in the question, so A is the answer.

Though I think a test taker could justifiably assume, from the question's wording, that the number should contain at least one digit that is a "3", and at least one digit that is a "4". Under that assumption, the answer would be "four", since 3444 satisfies all of the conditions.

Gmat_mission wrote:Suppose x is a positive even number, all of whose digits are either 3 or 4. If x is also divisible by both 3 and 4, how many digits are in the smallest possible value of x?

A. Three
B. Four
C. Five
D. Six
E. Seven

[spoiler]OA=A[/spoiler]

Source: Veritas Prep

To be divisible by 4, the final two digits of the number must be divisible by 4; to be divisible by 3, the sum of all the digits of the number must be divisible by 3. The smallest number that is divisible by 3 and 4 with digits of 3 and/or 4 is 444. So it has three digits.

Answer: A