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Divisibility Rules

SBI SO · Quantitative Aptitude

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Core Concept

Read this first — the foundation of the topic

💡Key Divisibility Rules

Rule for 2: A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8)

🔑Rule for 3

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

✏️Example

234 → 2+3+4 = 9, which is divisible by 3

🔑Rule for 4

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

✏️Example

1316 → 16 is divisible by 4

🔑Rule for 5

A number is divisible by 5 if it ends in 0 or 5

🔑Rule for 6

A number is divisible by 6 if it's divisible by both 2 and 3

🔑Rule for 8

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

🔑Rule for 9

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

🔑Rule for 11

A number is divisible by 11 if the alternating sum of its digits is divisible by 11.

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Exam Patterns

What examiners ask — read before attempting PYQs

SSC CGL frequently asks direct divisibility questions, number of factors problems, and divisibility in data sufficiency questions. Questions often combine multiple rules or ask for remainders. Shortcut for 11: Instead of alternating sum, use this trick - for any 4-digit number abcd, calculate (a+c)-(b+d). If result is 0 or divisible by 11, the number is divisible by 11.

✏️

Worked Example

Solve this step-by-step before moving on

1

Step 1

Apply alternating sum rule Start from right: 3 - 7 + 8 - 4 = 0

2

Step 2

Since the result is 0, which is divisible by 11, the number 4873 is divisible by 11. Verification: 4873 ÷ 11 = 443 (exact division) Alternate method using shortcut: (4+7) - (8+3) = 11 - 11 = 0 Since result is 0, number is divisible by 11. Advanced Trick: For checking divisibility by 7, 11, 13 simultaneously, divide the number into groups of 3 digits from right. Apply alternating sum. If divisible by 7, 11, or 13, the original number follows the same divisibility. Common Mistake: Students often confuse the divisibility rule for 4 with checking only the last digit. Remember, you need to check the last TWO digits, not just one. Also, for rule of 11, maintain the correct alternating pattern - start from the rightmost digit. Practice Tip: Create a mental checklist of these rules. In exams, quickly eliminate options using these rules rather than performing lengthy divisions.

🔑 Key Points

Divisibility by 2: last digit must be even (0,2,4,6,8)
Divisibility by 3: sum of all digits must be divisible by 3
Divisibility by 4: last two digits must form a number divisible by 4
Divisibility by 5: number must end in 0 or 5
Divisibility by 6: number must be divisible by both 2 and 3
Divisibility by 9: sum of all digits must be divisible by 9
Divisibility by 11: alternating sum of digits must be divisible by 11
Divisibility by 8: last three digits must form a number divisible by 8

📌 Exam Facts

A number divisible by both 4 and 9 is always divisible by 36
For divisibility by 12, the number must be divisible by both 3 and 4
The divisibility rule for 7 involves grouping digits in threes from right and applying alternating sum
Any number divisible by 8 is automatically divisible by 2 and 4
The sum of digits rule works for 3 and 9 because 10 ≡ 1 (mod 3) and 10 ≡ 1 (mod 9)
For divisibility by 25, the last two digits must be 00, 25, 50, or 75
A number is divisible by 15 if it's divisible by both 3 and 5
The alternating sum for 11 works because 10 ≡ -1 (mod 11)

🚀 60-Second Revision

Remember: Check last digit for 2 and 5, last two digits for 4, last three digits for 8
Formula: Sum of digits rule applies to 3 and 9 only
Trick: For 11, use alternating sum starting from rightmost digit
Pattern: Composite divisibility requires checking all prime factors
Trap: Don't confuse divisibility by 4 (last two digits) with divisibility by 2 (last digit)
Speed tip: Use elimination method in MCQs by applying easiest rules first
Quick check: For 6, 12, 15 - always verify both component divisibilities

🎯

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