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

Question

Check if 7,524 is divisible by 2, 3, 4, 6, 8, 9, and 11 without performing actual division. State the rule used each time.

(CBSE Class 6 — number system)

Solution — Step by Step

Divisor	Rule	Check for 7,524
2	Last digit is even	4 is even → Yes
3	Digit sum divisible by 3	$7+5+2+4 = 18$ , $18 \div 3 = 6$ → Yes
4	Last two digits divisible by 4	$24 \div 4 = 6$ → Yes
5	Last digit is 0 or 5	Last digit is 4 → No
6	Divisible by both 2 and 3	Yes to both → Yes
8	Last three digits divisible by 8	$524 \div 8 = 65.5$ → No
9	Digit sum divisible by 9	$18 \div 9 = 2$ → Yes
10	Last digit is 0	Last digit is 4 → No
11	Alternating sum divisible by 11	$7-5+2-4 = 0$ , and $0$ is divisible by 11 → Yes

For 7: double the last digit and subtract from the rest. Repeat until small enough.

$7524$ : double $4 = 8$ , subtract from $752$ : $752 - 8 = 744$

$744$ : double $4 = 8$ , subtract from $74$ : $74 - 8 = 66$

$66 \div 7 = 9.43\ldots$ → Not divisible by 7.

Why This Works

Each rule exploits number theory. For instance, divisibility by 3 works because $10 \equiv 1 \pmod{3}$ , so $10^n \equiv 1 \pmod{3}$ , meaning the digit sum has the same remainder as the number when divided by 3.

graph TD
    A["Is n divisible by d?"] --> B{"Which d?"}
    B -->|"2, 5, 10"| C["Check LAST DIGIT only"]
    B -->|"4"| D["Check last TWO digits"]
    B -->|"8"| E["Check last THREE digits"]
    B -->|"3, 9"| F["Check DIGIT SUM"]
    B -->|"6"| G["Check both 2 AND 3"]
    B -->|"11"| H["Check ALTERNATING SUM<br/>(odd position - even position)"]
    B -->|"7"| I["Double last digit,<br/>subtract from rest, repeat"]

Alternative Method — Factor Tree Approach

For composite divisors, check the prime factors: 6 = 2 times 3 (check both), 12 = 4 times 3 (check both), 15 = 3 times 5 (check both). This approach works for any composite number.

The alternating sum rule for 11 is easy to mess up with larger numbers. Always start from the rightmost digit: subtract the second-to-last, add the third-to-last, and so on. For $7524$ : starting from the right: $4 - 2 + 5 - 7 = 0$ . Since $0$ is divisible by $11$ , so is $7524$ .

Common Mistake

Students confuse the rules for 4 and 8. For 4, check the last two digits. For 8, check the last three digits. The pattern: for $2^n$ , check the last $n$ digits. So for 2 (which is $2^1$ ), check the last 1 digit; for 4 ( $2^2$ ), last 2 digits; for 8 ( $2^3$ ), last 3 digits. Mixing up 4 and 8 is the most common error in CBSE Class 6 tests.

Question

Solution — Step by Step

Why This Works

Alternative Method — Factor Tree Approach

Common Mistake

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