How many 4-digit positive integers divisible by 3 can be formed using only the digits {1, 3, 4, 6, 7}, such that no digit appears more than once in a number?

Divisibility Rule Check

A number is divisible by 3 if the sum of its digits is divisible by 3.

The available digits are {1, 3, 4, 6, 7}. The sum of these digits is $1+3+4+6+7=21$.

We need to form 4-digit integers using these digits without repetition. This requires selecting 4 digits from the given 5.

Let the sum of the 4 chosen digits be $S_4$. The sum of all 5 digits is $S_5=21$. If we exclude one digit $d$, the sum of the remaining 4 digits is $S_4=S_5-d=21-d$.

For $S_4$ to be divisible by 3, $21-d$ must be divisible by 3. Since 21 is divisible by 3, the excluded digit $d$ must also be divisible by 3.

Identifying Usable Digit Sets

From the set {1, 3, 4, 6, 7}, the digits divisible by 3 are 3 and 6.

- Case 1: Exclude digit 3. The remaining digits are {1, 4, 6, 7}. Their sum is $1+4+6+7=18$, which is divisible by 3.
- Case 2: Exclude digit 6. The remaining digits are {1, 3, 4, 7}. Their sum is $1+3+4+7=15$, which is divisible by 3.

Counting Permutations

For each set of 4 digits, we find the number of distinct 4-digit integers that can be formed. Since the order of digits matters and repetition is not allowed, this is a permutation calculation.

- Case 1 (Digits {1, 4, 6, 7}): The number of possible 4-digit integers is $4!$. $4!=4\times 3\times 2\times 1=24$.
- Case 2 (Digits {1, 3, 4, 7}): The number of possible 4-digit integers is $4!$. $4!=4\times 3\times 2\times 1=24$.

Total Count Calculation

The total number of 4-digit positive integers divisible by 3 is the sum of the counts from the valid cases.

Total Integers = (Count from Case 1) + (Count from Case 2)

Total Integers = $24+24=48$.