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 = $24 + 24 = 48$.