Suppose x is a positive even number, all of whose digits

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