Divisibility Rules
Imagine you have 156 sweets and want to share them equally among 3 friends. Can you do it without any leftovers? You could divide 156 by 3 — but there is a quicker way. Just add the digits: 1 + 5 + 6 = 12, and 12 is divisible by 3. So 156 is also divisible by 3. No long division needed.
These shortcuts are called divisibility rules (or divisibility tests). They let you quickly check whether a number can be divided exactly by 2, 3, 4, 5, 6, 7, 8, 9, 10, or 11 — without actually doing the division.
Divisibility rules are part of the Playing with Numbers chapter in Class 6 NCERT Maths. They save time and are very useful when finding factors, HCF, LCM, and simplifying fractions.
What is Divisibility Rules for 2 to 11 - Grade 6 Maths (Playing with Numbers)?
Definition: A number is divisible by another number if dividing leaves no remainder (remainder = 0).
A divisibility rule is a quick test to check divisibility without performing full division.
Example: 24 is divisible by 6 because 24 ÷ 6 = 4 (no remainder). But 25 is NOT divisible by 6 because 25 ÷ 6 = 4 remainder 1.
Notation: When we say "a | b" (read as "a divides b"), it means b is divisible by a. So 3 | 12 means 12 is divisible by 3.
Divisibility Rules Formula
Divisibility Rules Summary:
Quick Reference Table
| Divisor | Rule | Example |
|---|---|---|
| 2 | Last digit is even (0, 2, 4, 6, 8) | 438 → last digit 8 (even) → divisible |
| 3 | Sum of all digits is divisible by 3 | 729 → 7+2+9=18 → 18÷3=6 → divisible |
| 4 | Last two digits form a number divisible by 4 | 2,316 → last two digits = 16 → 16÷4=4 → divisible |
| 5 | Last digit is 0 or 5 | 1,235 → last digit 5 → divisible |
| 6 | Divisible by BOTH 2 AND 3 | 312 → even AND 3+1+2=6 → divisible by both |
| 7 | Double the last digit, subtract from remaining. Repeat if needed. | 343 → 34−(2×3)=34−6=28 → 28÷7=4 → divisible |
| 8 | Last three digits form a number divisible by 8 | 5,320 → 320÷8=40 → divisible |
| 9 | Sum of all digits is divisible by 9 | 4,518 → 4+5+1+8=18 → 18÷9=2 → divisible |
| 10 | Last digit is 0 | 7,450 → last digit 0 → divisible |
| 11 | Difference of sums of alternate digits is 0 or divisible by 11 | 918,082 → (9+8+8)−(1+0+2)=25−3=22 → 22÷11=2 → divisible |
Derivation and Proof
Why do these rules work?
Rule for 2: Any number = (remaining part × 10) + last digit. Since 10 is divisible by 2, the remaining part × 10 is always divisible by 2. So the whole number is divisible by 2 only if the last digit is divisible by 2 (i.e., even).
Rule for 3: Consider 523: it equals 5×100 + 2×10 + 3. We can write 100 = 99+1, 10 = 9+1. So 523 = 5×(99+1) + 2×(9+1) + 3 = (5×99 + 2×9) + (5+2+3). The part (5×99 + 2×9) is always divisible by 3 (since 99 and 9 are). So 523 is divisible by 3 only if (5+2+3) is — that is, only if the digit sum is divisible by 3.
Rule for 4: Any number = (hundreds and above part × 100) + last two digits. Since 100 is divisible by 4, we only need to check the last two digits.
Rule for 9: Works exactly like the rule for 3, since 9, 99, 999, etc. are all divisible by 9.
Rule for 11: Uses the fact that 10 = 11−1, 100 = 99+1 = (11×9)+1, 1000 = (11×91)−1. The alternating +1 and −1 pattern leads to the alternating digit sum rule.
Types and Properties
Types of divisibility rule problems:
Type 1: Check if a number is divisible by a given divisor
- Is 4,572 divisible by 4? Check: last two digits = 72, and 72 ÷ 4 = 18. Yes.
Type 2: Find all divisors of a number from 2 to 11
- For 360: divisible by 2 (even), 3 (digit sum 9), 4 (last two digits 60÷4=15), 5 (ends in 0), 6 (by 2 and 3), 8 (360÷8=45), 9 (digit sum 9), 10 (ends in 0). Not by 7 or 11.
Type 3: Find the missing digit
- 64_2 is divisible by 3. Find the missing digit. Digit sum = 6+4+_+2 = 12+_. For divisibility by 3, 12+_ must be a multiple of 3. So _ can be 0, 3, 6, or 9.
Type 4: Find the smallest/largest number divisible by a given number
- Find the largest 4-digit number divisible by 9. Largest 4-digit = 9999. Digit sum = 36, which is divisible by 9. So 9,999 is the answer.
Type 5: True or False statements
- "Every number divisible by 6 is also divisible by 3." True (because 6 = 2×3).
Solved Examples
Example 1: Example 1: Divisibility by 2 and 5
Problem: Check if 3,470 is divisible by 2 and by 5.
Solution:
- Last digit of 3,470 is 0.
- 0 is even → divisible by 2. Yes.
- 0 is either 0 or 5 → divisible by 5. Yes.
Answer: 3,470 is divisible by both 2 and 5.
Example 2: Example 2: Divisibility by 3
Problem: Is 8,247 divisible by 3?
Solution:
- Digit sum = 8 + 2 + 4 + 7 = 21
- Is 21 divisible by 3? 21 ÷ 3 = 7. Yes.
Answer: Yes, 8,247 is divisible by 3.
Example 3: Example 3: Divisibility by 4
Problem: Is 73,516 divisible by 4?
Solution:
- Last two digits of 73,516 are 16.
- 16 ÷ 4 = 4 (no remainder).
Answer: Yes, 73,516 is divisible by 4.
Example 4: Example 4: Divisibility by 6
Problem: Is 5,634 divisible by 6?
Solution:
A number is divisible by 6 if it is divisible by both 2 and 3.
- Divisibility by 2: Last digit = 4 (even). Yes.
- Divisibility by 3: Digit sum = 5 + 6 + 3 + 4 = 18. 18 ÷ 3 = 6. Yes.
Answer: Yes, 5,634 is divisible by 6.
Example 5: Example 5: Divisibility by 8
Problem: Is 41,720 divisible by 8?
Solution:
- Last three digits of 41,720 are 720.
- 720 ÷ 8 = 90 (no remainder).
Answer: Yes, 41,720 is divisible by 8.
Example 6: Example 6: Divisibility by 9
Problem: Is 32,454 divisible by 9?
Solution:
- Digit sum = 3 + 2 + 4 + 5 + 4 = 18
- 18 ÷ 9 = 2. Yes.
Answer: Yes, 32,454 is divisible by 9.
Example 7: Example 7: Divisibility by 11
Problem: Is 85,327 divisible by 11?
Solution:
- Number the digits from right: 7(odd), 2(even), 3(odd), 5(even), 8(odd)
- Sum of digits in odd positions (from right): 7 + 3 + 8 = 18
- Sum of digits in even positions (from right): 2 + 5 = 7
- Difference = 18 − 7 = 11
- 11 is divisible by 11. Yes.
Answer: Yes, 85,327 is divisible by 11.
Example 8: Example 8: Finding the Missing Digit
Problem: The number 7_4 is divisible by 3. Find all possible values of the missing digit.
Solution:
- Digit sum = 7 + _ + 4 = 11 + _
- For divisibility by 3, (11 + _) must be a multiple of 3.
- Multiples of 3 near 11: 12, 15, 18
- 11 + _ = 12 → _ = 1
- 11 + _ = 15 → _ = 4
- 11 + _ = 18 → _ = 7
Answer: The missing digit can be 1, 4, or 7.
Example 9: Example 9: Testing All Rules on One Number
Problem: Which numbers from 2 to 10 divide 2,520 exactly?
Solution:
- 2: Last digit 0 (even). Yes.
- 3: Digit sum = 2+5+2+0 = 9. 9÷3=3. Yes.
- 4: Last two digits = 20. 20÷4=5. Yes.
- 5: Last digit 0. Yes.
- 6: Divisible by 2 and 3. Yes.
- 7: 2520÷7 = 360. Yes.
- 8: Last three digits = 520. 520÷8 = 65. Yes.
- 9: Digit sum = 9. 9÷9=1. Yes.
- 10: Last digit 0. Yes.
Answer: 2,520 is divisible by all numbers from 2 to 10.
Example 10: Example 10: Word Problem — Equal Sharing
Problem: A school has 1,296 students. Can they be divided equally into groups of 4? Groups of 5? Groups of 9?
Solution:
- Groups of 4: Last two digits = 96. 96÷4=24. Yes, divisible by 4.
- Groups of 5: Last digit = 6 (not 0 or 5). No, not divisible by 5.
- Groups of 9: Digit sum = 1+2+9+6 = 18. 18÷9=2. Yes, divisible by 9.
Answer: Students can be divided into groups of 4 and 9, but not 5.
Real-World Applications
Real-life uses of divisibility rules:
- Sharing equally: Before dividing 252 sweets among 6 children, quickly check: 252 is even (÷2) and digit sum 9 (÷3), so it is divisible by 6. Each child gets 42 sweets.
- Simplifying fractions: To simplify 72/96, check common divisors. Both are divisible by 2, 3, 4, 6, 8. Divide by 24 to get 3/4.
- Finding HCF and LCM: Divisibility rules speed up prime factorisation, which is used to find HCF and LCM.
- Checking ISBN numbers: Book ISBN numbers use divisibility by 10 or 11 as error-checking codes.
- Money calculations: If ₹4,500 needs to be split into ₹500 notes, check divisibility by 5 — last digit 0, so yes. Number of notes = 9.
- Time tables: Is 168 hours exactly 7 days? Check: 168 ÷ 7. Using the rule for 7: 16 − 2×8 = 16 − 16 = 0. Yes, divisible by 7. So 168 hours = 24 × 7 = 7 days.
Key Points to Remember
- A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8).
- A number is divisible by 3 if the sum of its digits is divisible by 3.
- A number is divisible by 4 if its last two digits form a number divisible by 4.
- A number is divisible by 5 if its last digit is 0 or 5.
- A number is divisible by 6 if it is divisible by both 2 AND 3.
- A number is divisible by 8 if its last three digits form a number divisible by 8.
- A number is divisible by 9 if the sum of its digits is divisible by 9.
- A number is divisible by 10 if its last digit is 0.
- A number is divisible by 11 if the difference between the sum of digits at odd positions and even positions (from right) is 0 or a multiple of 11.
- If a number is divisible by 9, it is automatically divisible by 3 (but not vice versa).
Practice Problems
- Check if 67,890 is divisible by 2, 3, 4, 5, 6, 8, 9, 10.
- Is 4,95,726 divisible by 11? Use the alternating digit sum test.
- The number 52_8 is divisible by 9. Find the missing digit.
- Find the smallest 5-digit number that is divisible by both 4 and 5.
- Is 7,77,777 divisible by 7? By 3? By 11?
- A farmer has 1,764 mangoes. Can he pack them equally into boxes of 6? Of 8? Of 9?
- Find all numbers from 2 to 10 that divide 3,600 exactly.
- The number 3_45 is divisible by both 3 and 5. Find the missing digit.
Frequently Asked Questions
Q1. What does 'divisible by' mean?
A number is divisible by another if dividing it gives a remainder of 0. For example, 15 is divisible by 3 because 15 ÷ 3 = 5 with no remainder. But 15 is not divisible by 4 because 15 ÷ 4 = 3 remainder 3.
Q2. How do I check divisibility by 6?
A number is divisible by 6 if it passes BOTH the test for 2 (last digit is even) AND the test for 3 (digit sum is a multiple of 3). Both conditions must be true.
Q3. Why is there no simple rule for 7?
The rule for 7 exists (double the last digit, subtract from the rest), but it is more complex because 7 does not divide evenly into 10, 100, or 1000. For 2, 4, 5, 8, and 10, the rules are simpler because these numbers are factors of powers of 10.
Q4. If a number is divisible by 9, is it divisible by 3?
Yes, always. Since 9 = 3 × 3, any multiple of 9 is automatically a multiple of 3. For example, 81 is divisible by 9 (81÷9=9) and also by 3 (81÷3=27).
Q5. If a number is divisible by 2 and 3, is it divisible by 6?
Yes, always. Since 6 = 2 × 3 and 2 and 3 share no common factor, any number divisible by both 2 and 3 must be divisible by 6.
Q6. If a number is divisible by 2 and 4, is it divisible by 8?
Not necessarily. For example, 12 is divisible by 2 and 4, but 12 ÷ 8 = 1 remainder 4. So 12 is NOT divisible by 8. Being divisible by 2 and 4 does not guarantee divisibility by 8.
Q7. Can a number be divisible by both 3 and 5 but not by 15?
No. If a number is divisible by both 3 and 5, it is always divisible by 15 (since 3 and 5 share no common factors). For example, 45 is divisible by 3 and 5, and also by 15 (45÷15=3).
Q8. How does the divisibility rule for 11 work?
Add the digits in odd positions (1st, 3rd, 5th from right) and the digits in even positions (2nd, 4th, 6th from right) separately. If the difference between these two sums is 0 or a multiple of 11, the number is divisible by 11.
Q9. Are divisibility rules useful for large numbers?
Yes, especially for large numbers where long division would take time. For example, to check if 4,56,789 is divisible by 3, just add: 4+5+6+7+8+9=39, and 39÷3=13. Done in seconds.
Q10. Is every even number divisible by 4?
No. Even numbers are divisible by 2, but not all are divisible by 4. For example, 10 is even but 10÷4 = 2 remainder 2. Only even numbers whose last two digits form a multiple of 4 are divisible by 4.
