Divisibility Rules (Grade 5)
Divisibility rules are shortcut tests that help you check whether a number can be divided evenly by another number — without actually performing the division. In Class 5, students learn divisibility rules for 2, 3, 4, 5, 6, 8, 9, 10, and 11.
These rules are essential for finding factors, simplifying fractions, and solving problems involving HCF and LCM quickly.
Imagine you need to check if 7,236 is divisible by 4. Instead of dividing 7,236 by 4 (which takes time), you can simply look at the last two digits: 36. Since 36 ÷ 4 = 9, the number is divisible by 4. This is much faster!
In Class 4, students learned the basic divisibility rules for 2, 3, 5, 9, and 10. In Class 5, the set expands to include rules for 4, 6, 8, and 11. These advanced rules involve looking at the last two or three digits, or using alternating sums of digits.
Why learn divisibility rules?
- Find factors of large numbers quickly
- Check if a fraction can be simplified
- Speed up prime factorisation
- Solve HCF and LCM problems faster
- Verify division results mentally
What is Divisibility Rules - Class 5 Maths (Factors and Multiples)?
A number is divisible by another number if the division leaves no remainder (remainder = 0).
A divisibility rule is a quick test applied to the digits of a number to check divisibility without doing long division.
Divisibility Rules (Grade 5) Formula
| Divisor | Rule | Example |
|---|---|---|
| 2 | Last digit is 0, 2, 4, 6, or 8 | 346 → last digit 6 → divisible |
| 3 | Sum of digits is divisible by 3 | 471 → 4+7+1 = 12 → divisible |
| 4 | Last two digits form a number divisible by 4 | 5,316 → 16 ÷ 4 = 4 → divisible |
| 5 | Last digit is 0 or 5 | 2,735 → last digit 5 → divisible |
| 6 | Divisible by both 2 and 3 | 834 → even, 8+3+4 = 15 (÷3) → divisible |
| 8 | Last three digits form a number divisible by 8 | 7,120 → 120 ÷ 8 = 15 → divisible |
| 9 | Sum of digits is divisible by 9 | 2,349 → 2+3+4+9 = 18 → divisible |
| 10 | Last digit is 0 | 5,830 → last digit 0 → divisible |
| 11 | Difference of sums of alternate digits is 0 or divisible by 11 | 2,728 → (2+2)−(7+8) = 4−15... see examples |
Types and Properties
Rules based on last digit(s):
- Divisibility by 2: Check only the last digit (must be even: 0, 2, 4, 6, 8).
- Divisibility by 4: Check the last two digits.
- Divisibility by 5: Check if the last digit is 0 or 5.
- Divisibility by 8: Check the last three digits.
- Divisibility by 10: Check if the last digit is 0.
Rules based on sum of digits:
- Divisibility by 3: Sum of all digits must be divisible by 3.
- Divisibility by 9: Sum of all digits must be divisible by 9.
Combined rules:
- Divisibility by 6: The number must pass both the tests for 2 and 3.
Alternating sum rule:
- Divisibility by 11: Find the difference between the sum of digits at odd positions and the sum of digits at even positions. If the difference is 0 or divisible by 11, the number is divisible by 11.
Solved Examples
Example 1: Example 1: Divisibility by 4
Problem: Is 3,748 divisible by 4?
Solution:
Step 1: Look at the last two digits: 48.
Step 2: Check: 48 ÷ 4 = 12 (exact division, no remainder).
Answer: Yes, 3,748 is divisible by 4.
Example 2: Example 2: Divisibility by 6
Problem: Is 5,274 divisible by 6?
Solution:
Step 1: Test for 2: Last digit = 4 (even). ✓ Divisible by 2.
Step 2: Test for 3: Sum of digits = 5 + 2 + 7 + 4 = 18. 18 ÷ 3 = 6. ✓ Divisible by 3.
Step 3: Since the number passes both tests, it is divisible by 6.
Answer: Yes, 5,274 is divisible by 6.
Example 3: Example 3: Divisibility by 8
Problem: Is 45,312 divisible by 8?
Solution:
Step 1: Look at the last three digits: 312.
Step 2: Check: 312 ÷ 8 = 39 (exact division).
Answer: Yes, 45,312 is divisible by 8.
Example 4: Example 4: Divisibility by 11
Problem: Is 85,184 divisible by 11?
Solution:
Step 1: Write the digits with positions: 8 (odd), 5 (even), 1 (odd), 8 (even), 4 (odd).
Step 2: Sum of digits at odd positions = 8 + 1 + 4 = 13.
Step 3: Sum of digits at even positions = 5 + 8 = 13.
Step 4: Difference = 13 − 13 = 0.
Step 5: Since the difference is 0, the number is divisible by 11.
Answer: Yes, 85,184 is divisible by 11.
Example 5: Example 5: Testing Multiple Rules
Problem: Check whether 7,230 is divisible by 2, 3, 5, 6, 9, and 10.
Solution:
By 2: Last digit = 0 (even) → ✓ Yes
By 3: Sum = 7 + 2 + 3 + 0 = 12; 12 ÷ 3 = 4 → ✓ Yes
By 5: Last digit = 0 → ✓ Yes
By 6: Divisible by both 2 and 3 → ✓ Yes
By 9: Sum = 12; 12 ÷ 9 = 1 remainder 3 → ✗ No
By 10: Last digit = 0 → ✓ Yes
Answer: 7,230 is divisible by 2, 3, 5, 6, and 10, but not by 9.
Example 6: Example 6: Finding the Missing Digit
Problem: The number 54_2 is divisible by 3. What digit can replace the blank?
Solution:
Step 1: Sum of known digits = 5 + 4 + 2 = 11.
Step 2: Total sum = 11 + missing digit. This must be divisible by 3.
Step 3: Try values: 11 + 1 = 12 ✓, 11 + 4 = 15 ✓, 11 + 7 = 18 ✓
Answer: The missing digit can be 1, 4, or 7.
Example 7: Example 7: Divisibility by 11 (Not Divisible)
Problem: Is 4,523 divisible by 11?
Solution:
Step 1: Digits with positions: 4 (odd), 5 (even), 2 (odd), 3 (even).
Step 2: Sum at odd positions = 4 + 2 = 6.
Step 3: Sum at even positions = 5 + 3 = 8.
Step 4: Difference = |6 − 8| = 2.
Step 5: 2 is not 0 and not divisible by 11.
Answer: No, 4,523 is not divisible by 11.
Example 8: Example 8: Word Problem
Problem: Aditi has ₹8,496 and wants to distribute it equally among her 4 cousins. Can she do it without any money left over?
Solution:
Step 1: Check if 8,496 is divisible by 4.
Step 2: Last two digits = 96. Check: 96 ÷ 4 = 24 (exact).
Step 3: Yes, 8,496 is divisible by 4.
Step 4: Each cousin gets ₹8,496 ÷ 4 = ₹2,124.
Answer: Yes, each cousin gets ₹2,124 with nothing left over.
Example 9: Example 9: Divisibility by 8 (Not Divisible)
Problem: Is 23,500 divisible by 8?
Solution:
Step 1: Look at the last three digits: 500.
Step 2: Check: 500 ÷ 8 = 62 remainder 4.
Step 3: Since there is a remainder, 500 is not divisible by 8.
Answer: No, 23,500 is not divisible by 8.
Real-World Applications
Uses of divisibility rules:
- Finding factors quickly: Test which numbers divide a given number without doing long division.
- Simplifying fractions: Check if both numerator and denominator share a common factor.
- Prime factorisation: Use divisibility rules to identify prime factors step by step.
- Checking calculations: Verify whether a division will be exact before computing.
- Money problems: Determine if a sum of money can be split equally among a group.
Key Points to Remember
- Divisibility by 2: Last digit is even (0, 2, 4, 6, 8).
- Divisibility by 3: Sum of all digits is divisible by 3.
- Divisibility by 4: Last two digits form a number divisible by 4.
- Divisibility by 5: Last digit is 0 or 5.
- Divisibility by 6: Divisible by both 2 AND 3.
- Divisibility by 8: Last three digits form a number divisible by 8.
- Divisibility by 9: Sum of all digits is divisible by 9.
- Divisibility by 10: Last digit is 0.
- Divisibility by 11: Difference between sums of alternate digits is 0 or a multiple of 11.
- A number divisible by 9 is always divisible by 3, but not vice versa.
- A number divisible by 4 is always divisible by 2, but not vice versa.
Practice Problems
- Test whether 6,732 is divisible by 2, 3, 4, 6, and 9.
- Is 81,048 divisible by 8? Show your working.
- Check if 7,15,341 (7,15,341) is divisible by 11.
- The number 72_4 is divisible by 9. Find the missing digit.
- Rahul has 5,436 cricket cards to arrange equally into packets of 6. Can he do it with none left over?
- Which of these numbers are divisible by 4: 2,316; 5,530; 8,124; 9,301?
- Find all numbers between 100 and 120 that are divisible by both 3 and 4.
- Is 93,456 divisible by both 8 and 11?
Frequently Asked Questions
Q1. What is a divisibility rule?
A divisibility rule is a shortcut to check if one number divides another exactly (with no remainder), by examining the digits rather than performing long division.
Q2. Why is there no simple divisibility rule for 7?
The rule for 7 exists but is complex — you double the last digit, subtract it from the remaining number, and check if the result is divisible by 7. It is not commonly taught in Class 5 because it is harder to apply.
Q3. If a number is divisible by 6, is it always divisible by 3?
Yes. Since divisibility by 6 requires divisibility by both 2 and 3, any number divisible by 6 is automatically divisible by 3 (and by 2).
Q4. Can a number be divisible by 4 but not by 8?
Yes. For example, 12 is divisible by 4 (12 ÷ 4 = 3) but not by 8 (12 ÷ 8 = 1 remainder 4). Divisibility by 8 is a stricter test.
Q5. How does the divisibility rule for 11 work?
Add up the digits at odd positions (1st, 3rd, 5th...) and even positions (2nd, 4th, 6th...) separately. If the difference between these two sums is 0 or a multiple of 11, the number is divisible by 11.
Q6. Is every number ending in 0 divisible by 5 and 10?
Yes. If the last digit is 0, the number is divisible by both 5 and 10. However, numbers ending in 5 are divisible by 5 but not by 10.
Q7. How do divisibility rules help with fractions?
They help simplify fractions faster. If both the numerator and denominator are divisible by 2, 3, or another number, you can reduce the fraction quickly without trial-and-error.
Q8. What is the difference between the rules for 3 and 9?
Both use the sum of digits, but the divisibility by 3 requires the sum to be divisible by 3, while divisibility by 9 requires the sum to be divisible by 9. Every number divisible by 9 is also divisible by 3.
Q9. Is this topic covered in NCERT Class 5?
Yes. Divisibility rules for 2, 3, 4, 5, 6, 8, 9, 10, and 11 are part of the Factors and Multiples chapter in the NCERT/CBSE Class 5 Mathematics syllabus.
