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Divisibility Rules

Introduction

Divisibility rules help us check whether a number is divisible by another number without performing long division. They are powerful shortcuts for solving aptitude problems quickly, especially when dealing with large numbers.

Pattern: Divisibility Rules

Pattern

Divisibility rules are simple checks (based on digits or digit sums) to decide whether a number is divisible by another number.

  • By 2: Last digit is 0, 2, 4, 6, 8 (even).
  • By 3: Sum of digits divisible by 3.
  • By 4: Last two digits divisible by 4.
  • By 5: Last digit is 0 or 5.
  • By 6: Number divisible by both 2 and 3.
  • By 8: Last three digits divisible by 8.
  • By 9: Sum of digits divisible by 9.
  • By 10: Last digit is 0.
  • By 11: (Sum of digits in odd places - Sum of digits in even places) divisible by 11.
  • By 12: Number divisible by both 3 and 4.

Step-by-Step Example

Question

Check whether 4,356 is divisible by 3, 4, and 11.

Solution

  1. Step 1: Divisibility by 3:

    Rule: A number is divisible by 3 if the sum of its digits is divisible by 3. Sum = 4 + 3 + 5 + 6 = 18 → 18 is divisible by 3. ✅ So, 4356 is divisible by 3.

  2. Step 2: Divisibility by 4:

    Rule: A number is divisible by 4 if its last two digits are divisible by 4. Last two digits = 56 → 56 ÷ 4 = 14 (exact). ✅ So, 4356 is divisible by 4.

  3. Step 3: Divisibility by 11:

    Rule: A number is divisible by 11 if the difference between the sum of digits in odd and even places is divisible by 11. Odd place digits (from left): 4 + 5 = 9 Even place digits: 3 + 6 = 9 Difference = 9 - 9 = 0 → divisible by 11. ✅ So, 4356 is divisible by 11.

  4. Final Answer:

    4356 is divisible by 3, 4, and 11.

  5. Quick Check:

    4356 ÷ 3 = 1452, 4356 ÷ 4 = 1089, 4356 ÷ 11 = 396 → All exact. ✅

Quick Variations

1. Divisibility by 7 → Double the last digit and subtract from the rest. Result divisible by 7.

2. Divisibility by 13 → Repeated subtraction/addition rules based on patterns.

3. Combined divisibility tests (like 15 → check both 3 and 5).

Trick to Always Use

  • Step 1: Use digit sum rules for 3 and 9.
  • Step 2: Use last two/three digits for 4 and 8.
  • Step 3: Use alternating digit sum for 11.
  • Step 4: For combined divisibility (e.g., 6, 12, 15), check multiple rules together.

Summary

Summary

In the Divisibility Rules pattern:

  • Digit sum rules work for 3, 9, and 11.
  • Last digit(s) rules work for 2, 4, 5, 8, 10.
  • Combined divisibility applies for 6, 12, 15, etc.
  • Special rules exist for 7, 13, 17, etc., but rarely used in exams.

Practice

(1/5)
1. Check if 248 is divisible by 2, 4, and 8.
easy
A. Yes, by all three
B. Only by 2
C. By 2 and 4 only
D. Not divisible by any

Solution

  1. Step 1: Divisibility by 2:

    Rule: Last digit even → 8 is even → divisible by 2.

  2. Step 2: Divisibility by 4:

    Rule: Last two digits divisible by 4 → 48 ÷ 4 = 12 → divisible by 4.

  3. Step 3: Divisibility by 8:

    Rule: Last three digits divisible by 8 → 248 ÷ 8 = 31 → divisible by 8.

  4. Final Answer:

    248 is divisible by 2, 4, and 8 → Option A.

  5. Quick Check:

    248 ÷ 2 = 124, 248 ÷ 4 = 62, 248 ÷ 8 = 31 ✅
Hint: Check last digit for 2, last two digits for 4, last three digits for 8.
Common Mistakes: Stopping after checking divisibility by 2 and not testing 4 or 8.
2. Which of the following is divisible by 9?
easy
A. 135
B. 224
C. 472
D. 811

Solution

  1. Step 1: Rule for 9:

    If sum of digits is divisible by 9, the number is divisible by 9.

  2. Step 2: Compute sums:

    135 → 1+3+5 = 9 (divisible by 9). 224 → 2+2+4 = 8 (not). 472 → 4+7+2 = 13 (not). 811 → 8+1+1 = 10 (not).

  3. Final Answer:

    Only 135 is divisible by 9 → Option A.

  4. Quick Check:

    135 ÷ 9 = 15 ✅
Hint: Use digit-sum test for 9 instead of long division.
Common Mistakes: Trying to divide each number instead of applying the digit-sum rule.
3. Which number is divisible by both 2 and 5?
easy
A. 125
B. 260
C. 473
D. 589

Solution

  1. Step 1: Rule for 10:

    Number divisible by both 2 and 5 must end with 0.

  2. Step 2: Check options:

    125 ends with 5 (no). 260 ends with 0 (yes). 473 ends with 3 (no). 589 ends with 9 (no).

  3. Final Answer:

    260 is divisible by 10 → Option B.

  4. Quick Check:

    260 ÷ 10 = 26 ✅
Hint: For divisibility by 10, check if the last digit is 0.
Common Mistakes: Confusing divisibility by 5 (ends with 0 or 5) with divisibility by 10 (must end with 0).
4. What is the remainder when 3,652 is divided by 11?
medium
A. 1
B. 2
C. 0
D. 3

Solution

  1. Step 1: Rule for 11:

    Compute (sum of digits in odd positions) - (sum of digits in even positions). If result is divisible by 11 (including 0), remainder is 0.

  2. Step 2: Apply to 3652:

    Digits (from left): 3, 6, 5, 2. Sum odd positions = 3 + 5 = 8. Sum even positions = 6 + 2 = 8. Difference = 8 - 8 = 0.

  3. Final Answer:

    Difference 0 → 3652 is divisible by 11 → remainder = 0 → Option C.

  4. Quick Check:

    3652 ÷ 11 = 332 with remainder 0 ✅
Hint: Use alternating digit-sum difference to test divisibility by 11 quickly.
Common Mistakes: Applying the rule incorrectly by using wrong digit positions or order.
5. Which of the following numbers is divisible by both 3 and 4 (i.e., by 12)?
medium
A. 125
B. 140
C. 148
D. 144

Solution

  1. Step 1: Rule for 12:

    Number must be divisible by both 3 and 4.

  2. Step 2: Check options:

    125 → digit sum 1+2+5 = 8 (not divisible by 3) → not divisible by 12. 140 → digit sum 1+4+0 = 5 (not divisible by 3) → not divisible. 148 → digit sum 1+4+8 = 13 (not divisible by 3) → not divisible. 144 → digit sum 1+4+4 = 9 (divisible by 3) and last two digits 44 ÷ 4 = 11 (divisible by 4) → divisible by 12.

  3. Final Answer:

    144 is divisible by 12 → Option D.

  4. Quick Check:

    144 ÷ 12 = 12 ✅
Hint: For 12, verify both 3 (digit sum) and 4 (last two digits) rules.
Common Mistakes: Checking only 3 or only 4 instead of both conditions for 12.

Mock Test

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