Divisibility rules for 3 4 6 8 9 and 11 — examples

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

Example: 567 → $5 + 6 + 7 = 18$. Since 18 ÷ 3 = 6 (exact), 567 is divisible by 3.

Example (fail): 572 → $5 + 7 + 2 = 14$. 14 is not divisible by 3, so 572 is not.

Why: Every power of 10 leaves remainder 1 when divided by 3 ($10 \equiv 1 \pmod{3}$). So a number $= a_n \times 10^n + \cdots + a_0 \equiv a_n + \cdots + a_0 \pmod{3}$.

Rule: A number is divisible by 4 if its last two digits form a number divisible by 4.

Example: 1732 → last two digits are 32. Since 32 ÷ 4 = 8, yes, 1732 is divisible by 4.

Example (fail): 1734 → last two digits 34. 34 ÷ 4 = 8.5 — not exact. 1734 is not divisible by 4.

Why: 100 is divisible by 4 ($100 = 4 \times 25$). So for any number, only the last two digits determine divisibility by 4.

Rule: A number is divisible by 6 if it is divisible by both 2 and 3 simultaneously.

• Must be even (divisible by 2): last digit is 0, 2, 4, 6, or 8
• Sum of digits must be divisible by 3

Example: 732 → last digit 2 (even ✓); $7+3+2=12$ (divisible by 3 ✓). So 732 is divisible by 6.

Why: $6 = 2 \times 3$ and gcd(2,3) = 1. When two coprime numbers both divide $n$, so does their product.

Rule: A number is divisible by 8 if its last three digits form a number divisible by 8.

Example: 5,824 → last three digits 824. $824 \div 8 = 103$. Yes, 5824 is divisible by 8.

Example (fail): 5,826 → last three digits 826. $826 \div 8 = 103.25$. Not divisible.

Why: $1000 = 8 \times 125$, so any multiple of 1000 is divisible by 8. Only the last three digits determine divisibility.

Rule: A number is divisible by 9 if the sum of its digits is divisible by 9.

Example: 738 → $7+3+8=18$. $18 \div 9 = 2$ ✓. So 738 is divisible by 9.

Example (fail): 735 → $7+3+5=15$. $15 \div 9 = 1.67$. Not divisible.

Note: Every number divisible by 9 is also divisible by 3 (but not vice versa).

Rule: A number is divisible by 11 if the difference between the sum of digits at odd positions and the sum of digits at even positions (counting from the right) is 0 or divisible by 11.

Example: 1,43,264 Digits from right: 4(pos1), 6(pos2), 2(pos3), 3(pos4), 4(pos5), 1(pos6)

Sum of odd-position digits: $4 + 2 + 4 = 10$ Sum of even-position digits: $6 + 3 + 1 = 10$

Difference: $|10 - 10| = 0$. Divisible by 11. ✓

Simpler example: 121 → $1 - 2 + 1 = 0$. Divisible by 11 ✓ (since $121 = 11^2$).

Why: $10 \equiv -1 \pmod{11}$. So $10^k \equiv (-1)^k \pmod{11}$, which alternates +1 and -1 — giving the alternating sum rule.