How many 4-digit positive integers divisible by 3 can be formed using only the…

Key idea: use divisibility by 3 (the sum of digits must be a multiple of 3) and the residues of the available digits modulo 3.

• Find residues modulo 3 of the digits: 1 → 1, 3 → 0, 4 → 1, 6 → 0, 7 → 1. Thus there are three digits with remainder 1 (1, 4, 7) and two digits with remainder 0 (3, 6).

• Let k be the number of chosen digits with remainder 1. The sum of the four digits is congruent to k (mod 3). To make the sum divisible by 3 we need k ≡ 0 (mod 3). With four digits, the possible k are 0 or 3.

• k = 0 is impossible because there are only two digits with remainder 0. Therefore k = 3: the three remainder-1 digits (1, 4, 7) must all be used, together with exactly one remainder-0 digit (either 3 or 6).

• Count the numbers: choose the remainder-0 digit in 2 ways. For each choice we have 4 distinct digits, which can be arranged in 4! = 24 ways. So the total number of valid 4-digit numbers is 2 × 24 = 48.

Answer: 48