The remainder when (672 + 541 + 383 + 295 + 101 + 86) is divided by 3…

The correct answer is Option (3) → 2

The remainder of a sum is the sum of the remainders, reduced again modulo 3. By the divisibility test for 3, a number leaves the same remainder on division by 3 as the sum of its digits:

    \(672:\ 6 + 7 + 2 = 15,\quad 672 \equiv 0 \pmod{3}\)
    \(541:\ 5 + 4 + 1 = 10,\quad 541 \equiv 1 \pmod{3}\)
    \(383:\ 3 + 8 + 3 = 14,\quad 383 \equiv 2 \pmod{3}\)
    \(295:\ 2 + 9 + 5 = 16,\quad 295 \equiv 1 \pmod{3}\)
    \(101:\ 1 + 0 + 1 = 2,\quad 101 \equiv 2 \pmod{3}\)
    \(86:\ 8 + 6 = 14,\quad 86 \equiv 2 \pmod{3}\)

Adding the remainders:

    \(0 + 1 + 2 + 1 + 2 + 2 = 8\)
    \(8 \equiv 2 \pmod{3}\)

Check: the sum is 2078, and \(2078 = 3 \times 692 + 2\). So the remainder is 2.