Rule: Sum of all digits is divisible by 3. Reasoning: 10 ≡ 1 (mod 3) (since 10 = 3×3+1), so 10^k ≡ 1 (mod 3) for every k. Hence a number d_n d_(n-1)… d_1 d_0 = d_i · 10^i ≡ d_i (mod 3) — the number and its digit sum leave the same remainder when divided by 3.

Example 1: 45,231 → digit sum = 4+5+2+3+1 = 15, divisible by 3 → number divisible by 3. Example 2: 8,72,410 → digit sum = 8+7+2+4+1+0 = 22, not divisible by 3 → number not divisible by 3. Example 3: Find the remainder when 6,71,345 is divided by 3. Digit sum = 6+7+1+3+4+5 = 26 → 26 ÷ 3 leaves remainder 2. So the number leaves remainder 2.

Example 4 (edge case — multiple valid digits): Find the smallest digit x that can replace the blank in 5x2364 to make the number divisible by 3. Digit sum without the blank = 5+2+3+6+4 = 20. We need 20+x divisible by 3. Checking x=0,1,2,…: x=1 gives 21 (divisible by 3) — the smallest such value. In fact x=1,4,7 all work (giving digit sums 21, 24, 27), since divisibility by 3 repeats every 3 units. This shows that unlike divisibility-by-9 missing-digit problems (which usually pin down one digit), a divisibility-by-3 blank very often has three valid digits in the 0–9 range — always check whether the question wants “the smallest,” “the largest,” or “how many” values before answering. Smallest valid x = 1 (number 512364; check: 512364 ÷ 3 = 170788 exact ✓).

Example 5 (large number, quick remainder): What is the remainder when 9,87,65,43,210 (i.e., 9876543210) is divided by 3? Digit sum =9+8+7+6+5+4+3+2+1+0=45, which is divisible by 3. Remainder = 0 (the number is exactly divisible by 3), confirming that the digit-sum shortcut works just as fast for a 10-digit number as for a 3-digit one — the whole point of the rule.