Divisibility rules for 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 — quick checks

Question

Check whether 7,524 is divisible by 2, 3, 4, 6, 8, 9, and 11. Show the divisibility test for each.

Solution — Step by Step

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

Divisible by 5? Last digit is 4 (not 0 or 5). No.

Divisible by 10? Last digit is 4 (not 0). No.

Digit sum: $7 + 5 + 2 + 4 = 18$

Divisible by 3? Sum = 18, which is divisible by 3. Yes.

Divisible by 9? Sum = 18, which is divisible by 9. Yes.

Divisible by 6? Must be divisible by both 2 and 3. It is. Yes.

Divisible by 4? Check last two digits: 24. Since $24 \div 4 = 6$ , Yes.

Divisible by 8? Check last three digits: 524. Since $524 \div 8 = 65.5$ , No.

Divisible by 11? Alternating sum: $7 - 5 + 2 - 4 = 0$ . Since 0 is divisible by 11, Yes.

Why This Works

graph TD
    A["Divisibility Check"] --> B["By 2: last digit even?"]
    A --> C["By 3: digit sum ÷ 3?"]
    A --> D["By 4: last 2 digits ÷ 4?"]
    A --> E["By 5: last digit 0 or 5?"]
    A --> F["By 6: divisible by both 2 AND 3?"]
    A --> G["By 8: last 3 digits ÷ 8?"]
    A --> H["By 9: digit sum ÷ 9?"]
    A --> I["By 10: last digit 0?"]
    A --> J["By 11: alternating sum ÷ 11?"]
    A --> K["By 7: double last digit, subtract from rest"]

The digit sum test for 3 and 9 works because $10 \equiv 1 \pmod{3}$ , so $10^n \equiv 1 \pmod{3}$ for all $n$ . This means any number equals its digit sum modulo 3. Similarly, $10 \equiv -1 \pmod{11}$ , so $10^n$ alternates between $+1$ and $-1$ modulo 11 — giving us the alternating sum test.

The last-two-digits test for 4 works because $100$ is divisible by 4, so only the tens and units digits matter. Similarly, $1000$ is divisible by 8, so only the last three digits matter.

Alternative Method

Divisibility by 7 (the trickiest one): Take the last digit, double it, and subtract from the remaining number. Repeat until you get a small number. If that small number is divisible by 7, the original is too.

Example: Is 371 divisible by 7? Last digit = 1, double it = 2. Remaining = 37. $37 - 2 = 35$ . Since $35 = 7 \times 5$ , yes, 371 is divisible by 7.

This rule is rarely tested in CBSE exams, but knowing it makes you faster in competitive scenarios.

Common Mistake

Confusing the divisibility rule for 4 with the rule for 8. For 4, check the last TWO digits. For 8, check the last THREE digits. Students sometimes check only the last two digits for 8, or the last three for 4. A reliable way to remember: $10^2 = 100$ is the first power of 10 divisible by 4 (so check 2 digits), and $10^3 = 1000$ is the first power divisible by 8 (so check 3 digits).

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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