Divisibility Rules and Tests

The divisibility rules

These shortcuts let you check divisibility without actually dividing:

• By 2: Last digit is even (0, 2, 4, 6, 8).
• By 3: Sum of digits is divisible by 3.
• By 4: Last two digits form a number divisible by 4.
• By 5: Last digit is 0 or 5.
• By 6: Divisible by both 2 AND 3.
• By 8: Last three digits form a number divisible by 8.
• By 9: Sum of digits is divisible by 9.
• By 10: Last digit is 0.
• By 11: Alternating sum of digits is divisible by 11.

Step-by-step: Test 7,284

Divisible by 2? Last digit is 4 (even). Yes ✓

Divisible by 3? Sum of digits: $7 + 2 + 8 + 4 = 21$. Is 21 divisible by 3? $21/3 = 7$. Yes ✓

Divisible by 4? Last two digits: 84. $84/4 = 21$. Yes ✓

Divisible by 6? Divisible by 2 ✓ AND by 3 ✓. Yes ✓

Divisible by 8? Last three digits: 284. $284/8 = 35.5$. No ✗

Divisible by 9? Sum of digits: $21$. $21/9 = 2.33...$. No ✗

Summary for 7,284

| Divisor | Rule | Result | |---------|------|--------| | 2 | Last digit 4 (even) | ✓ | | 3 | Digit sum 21 ÷ 3 = 7 | ✓ | | 4 | Last two "84" ÷ 4 = 21 | ✓ | | 6 | By 2 ✓ and by 3 ✓ | ✓ | | 8 | Last three "284" ÷ 8 ≠ integer | ✗ | | 9 | Digit sum 21 ÷ 9 ≠ integer | ✗ |

Verification: $7284 / 2 = 3642$ ✓, $7284 / 3 = 2428$ ✓, $7284 / 4 = 1821$ ✓, $7284 / 8 = 910.5$ ✗.

Why the digit-sum rules work

The rule for 3 works because $10 \equiv 1 \pmod{3}$, so $10^k \equiv 1 \pmod{3}$ for all $k$. This means a number like $7284 = 7 \times 10^3 + 2 \times 10^2 + 8 \times 10 + 4 \equiv 7 + 2 + 8 + 4 = 21 \equiv 0 \pmod{3}$. The same reasoning works for 9.

Try it in the visualization

Enter any number. Each divisibility rule is tested with a pass/fail indicator. The calculation behind each rule is shown step by step. Highlight which rules are actually testing the same underlying property (e.g., the rule for 6 combines the rules for 2 and 3).