ZE
ZESTEXAM

IBPS Clerk Divisibility Rules

Study Material — 4 PYQs (2019–2019) · Concept Notes · Shortcuts

IBPS Clerk Divisibility Rules is a frequently tested subtopic — 4 previous year questions from 2019–2019 papers are included below with concept notes, key rules and shortcut tricks.

4 PYQs
2019–2019
11 Practice
MCQs
8 Key Points
to remember
Free
no login needed
Take Free Mock →Full Practice Set
Also for:IBPS POSBI POSBI ClerkRBI Gr B
PYQs
4
Practice
11
Key Points
8
Access
Free
Previous Year Questions

IBPS Clerk Divisibility Rules — Past Exam Questions

4 questions from actual IBPS Clerk papers · all shown free · click option to reveal solution

Exam Q 12019Previous Year Pattern

A number when divided by 8 leaves remainder 0. Which statement is definitely true about this number?

Exam Q 22019Previous Year Pattern

A number is divisible by both 4 and 6. Which of the following must also divide this number?

Exam Q 32019Previous Year Pattern

Which of the following numbers is divisible by 11?

Exam Q 42019Previous Year Pattern

How many numbers between 100 and 200 are divisible by 7?

Concept Notes

Divisibility Rules— Rules & Concept

1.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.

2.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.

3.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.

4.Exam Traps

Common mistakes students make — avoid these

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 to Remember

  • 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-Specific Tips

  • 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)
Practice MCQs

Divisibility Rules — Practice Questions

11graded MCQs · easy to hard · full solution & trap analysis

All MCQs →
Practice 1easy

A number is divisible by 9. Which of the following is definitely true?

Practice 2easy

Which of the following numbers is divisible by 3 but NOT divisible by 9?

Practice 3medium

How many numbers between 100 and 500 are divisible by 7 but not by 3?

Practice 4medium

A number is divisible by both 8 and 9. Which of the following must also divide this number?

Practice 5medium

A number when divided by 11 leaves remainder 7. What will be the remainder when the square of this number is divided by 11?

Practice 6hard

A number N is divisible by 7. When N is divided by 14, the remainder is 0. When N is divided by 21, the remainder is 0. What is the smallest positive value of N that satisfies all three conditions?

Practice 7hard

A six-digit number 5a4b8c is divisible by 11. The digits a, b, c are such that a + b + c = 15. If the number is also divisible by 3, what is the value of a + b + c?

Practice 8hard

A number is divisible by 8 and also divisible by 15. Which of the following must be true about this number?

Practice 9hard

A number when divided by 11 leaves remainder 7. Another number when divided by 11 leaves remainder 4. If the sum of these two numbers is divided by 11, what will be the remainder?

Practice 10hard

A five-digit number 3a7b2 is divisible by both 4 and 9. If a and b are single digits, what is the maximum value of a + b?

Practice 11hard

How many numbers between 100 and 500 are divisible by 6 but NOT divisible by 9?

60-Second Revision — Divisibility Rules

  • 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
←LCM and HCFNumber Series→
Studied the notes? Now test yourself
See how Divisibility Rules appears in the real IBPS Clerk paper
Full timed mock · Instant All-India percentile · Free
Start Free Mock →Practice Panel
✓ Free forever for basic prep✓ No app download✓ Real exam-pattern questions
Test Divisibility Rules under exam conditions
Free IBPS Clerk mock · instant rank · no login
Free Mock →