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Divisibility rules

Introduction to Divisibility Rules

In arithmetic, understanding whether one number can be divided evenly by another without leaving a remainder is a fundamental concept called divisibility. Divisibility rules are simple tests or shortcuts that help us quickly determine if a number is divisible by another number, without performing long division.

Why is divisibility important? It helps us:

  • Find factors of numbers quickly
  • Simplify fractions by identifying common factors
  • Solve problems efficiently in competitive exams where time is limited

Before diving into the rules, let's briefly recall the number system hierarchy:

  • Natural numbers: Counting numbers starting from 1, 2, 3, ...
  • Whole numbers: Natural numbers including 0
  • Integers: Whole numbers and their negatives, e.g., -3, -2, -1, 0, 1, 2, 3
  • Rational numbers: Numbers expressed as fractions of integers
  • Real numbers: All rational and irrational numbers

Divisibility rules mainly apply to integers and natural numbers. Mastering these rules will sharpen your number sense and speed up calculations.

Basic Divisibility Rules

We start with the simplest and most commonly used divisibility rules for 2, 3, and 5. These rules rely on patterns in the digits of numbers.

Basic Divisibility Rules Summary
Divisor Rule Example
2 Number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8) 2468 (last digit 8 is even) -> divisible by 2
3 Number is divisible by 3 if the sum of its digits is divisible by 3 123 (1+2+3=6, divisible by 3) -> divisible by 3
5 Number is divisible by 5 if its last digit is 0 or 5 145 (last digit 5) -> divisible by 5

Let's understand why these rules work:

  • Divisible by 2: Even numbers end with digits 0, 2, 4, 6, or 8. Since 10 is divisible by 2, the divisibility depends only on the last digit.
  • Divisible by 3: Because 10 ≡ 1 (mod 3), each digit's place value contributes the same remainder as the digit itself modulo 3. So, summing digits and checking divisibility by 3 works.
  • Divisible by 5: Since 10 ≡ 0 (mod 5), only the last digit determines divisibility by 5.
Example 1: Checking Divisibility by 3 and 5 Easy
Check if the number 12345 is divisible by 3 and by 5.

Step 1: Check divisibility by 3 by adding the digits: 1 + 2 + 3 + 4 + 5 = 15.

Step 2: Since 15 is divisible by 3 (because 15 / 3 = 5), 12345 is divisible by 3.

Step 3: Check divisibility by 5 by looking at the last digit, which is 5.

Step 4: Since the last digit is 5, 12345 is divisible by 5.

Answer: 12345 is divisible by both 3 and 5.

Intermediate Divisibility Rules

Now, let's move to slightly more complex rules for divisibility by 4, 6, and 9. These rules build on the basic ones and sometimes combine them.

Intermediate Divisibility Rules Summary
Divisor Rule Example
4 Number is divisible by 4 if the last two digits form a number divisible by 4 312 (last two digits 12, 12 / 4 = 3) -> divisible by 4
6 Number is divisible by 6 if it is divisible by both 2 and 3 234 (even and sum of digits 2+3+4=9 divisible by 3) -> divisible by 6
9 Number is divisible by 9 if the sum of its digits is divisible by 9 729 (7+2+9=18, 18 / 9 = 2) -> divisible by 9

Why these rules work:

  • Divisible by 4: Since 100 is divisible by 4, only the last two digits affect divisibility by 4.
  • Divisible by 6: 6 = 2 x 3, so a number must be divisible by both 2 and 3.
  • Divisible by 9: Similar to 3, because 10 ≡ 1 (mod 9), sum of digits determines divisibility.
Example 2: Divisibility by 4 and 6 Medium
Check if the number 1236 is divisible by 4 and by 6.

Step 1: For divisibility by 4, look at the last two digits: 36.

Step 2: Since 36 / 4 = 9 (an integer), 1236 is divisible by 4.

Step 3: For divisibility by 6, check divisibility by 2 and 3.

Step 4: Last digit is 6 (even), so divisible by 2.

Step 5: Sum of digits: 1 + 2 + 3 + 6 = 12, and 12 / 3 = 4, so divisible by 3.

Step 6: Since divisible by both 2 and 3, 1236 is divisible by 6.

Answer: 1236 is divisible by both 4 and 6.

Advanced Divisibility Rules

Divisibility rules for 7, 11, and 13 are more involved but very useful in competitive exams. These rules often use subtraction, addition, or alternating sums.

Divisibility by 7

One popular method:

  • Double the last digit of the number.
  • Subtract this from the rest of the number.
  • Repeat the process if needed.
  • If the result is divisible by 7 (including 0), the original number is divisible by 7.
graph TD    A[Start with number n] --> B[Double the last digit]    B --> C[Subtract from remaining leading part]    C --> D{Is result small enough?}    D -- No --> B    D -- Yes --> E{Is result divisible by 7?}    E -- Yes --> F[Number divisible by 7]    E -- No --> G[Number not divisible by 7]

Divisibility by 11

Calculate the alternating sum of digits:

  • Add digits in odd positions and subtract digits in even positions (from right to left or left to right consistently).
  • If the result is divisible by 11 (including 0), the number is divisible by 11.

Divisibility by 13

Multiply the last digit by 4 and add it to the remaining leading part. Repeat if necessary. If the result is divisible by 13, so is the original number.

Example 3: Divisibility by 7 and 11 Hard
Check if 203 is divisible by 7 and if 121 is divisible by 11.

Checking 203 for divisibility by 7:

Step 1: Double the last digit: last digit is 3, doubled is 6.

Step 2: Subtract 6 from the remaining number: 20 - 6 = 14.

Step 3: 14 is a small number; check if divisible by 7.

Step 4: 14 / 7 = 2, so 14 is divisible by 7.

Conclusion: 203 is divisible by 7.


Checking 121 for divisibility by 11:

Step 1: Identify digits: 1 (hundreds), 2 (tens), 1 (ones).

Step 2: Sum of digits at odd positions: 1 (hundreds) + 1 (ones) = 2.

Step 3: Sum of digits at even positions: 2 (tens) = 2.

Step 4: Difference: 2 - 2 = 0.

Step 5: Since 0 is divisible by 11, 121 is divisible by 11.

Answer: 203 is divisible by 7 and 121 is divisible by 11.

Example 4: Divisibility by 13 Hard
Check if 2730 is divisible by 13.

Step 1: Multiply the last digit by 4: last digit is 0, so 0 x 4 = 0.

Step 2: Add this to the remaining number: 273 + 0 = 273.

Step 3: Repeat the process for 273:

  • Last digit: 3, 3 x 4 = 12.
  • Remaining number: 27.
  • Add: 27 + 12 = 39.

Step 4: Check if 39 is divisible by 13.

Step 5: 39 / 13 = 3, which is an integer.

Answer: 2730 is divisible by 13.

Quick Reference: Divisibility Rules

  • Divisible by 2: Last digit even (0,2,4,6,8)
  • Divisible by 3: Sum of digits divisible by 3
  • Divisible by 4: Last two digits divisible by 4
  • Divisible by 5: Last digit 0 or 5
  • Divisible by 6: Divisible by 2 and 3
  • Divisible by 7: Double last digit, subtract from rest, repeat
  • Divisible by 9: Sum of digits divisible by 9
  • Divisible by 11: Difference of sums of digits at odd and even positions divisible by 11
  • Divisible by 13: Multiply last digit by 4, add to rest, repeat
Key Takeaway:

Use these rules to quickly test divisibility without long division.

Formula Bank

Divisibility by 2
\[ n \equiv 0 \pmod{2} \iff \text{last digit of } n \text{ is even} \]
where: \( n \) = any integer
Divisibility by 3
\[ \sum \text{digits of } n \equiv 0 \pmod{3} \]
where: \( n \) = any integer
Divisibility by 5
\[ n \equiv 0 \pmod{5} \iff \text{last digit of } n \in \{0,5\} \]
where: \( n \) = any integer
Divisibility by 4
\[ n \equiv 0 \pmod{4} \iff \text{last two digits of } n \equiv 0 \pmod{4} \]
where: \( n \) = any integer
Divisibility by 6
\[ n \equiv 0 \pmod{6} \iff n \equiv 0 \pmod{2} \text{ and } n \equiv 0 \pmod{3} \]
where: \( n \) = any integer
Divisibility by 9
\[ \sum \text{digits of } n \equiv 0 \pmod{9} \]
where: \( n \) = any integer
Divisibility by 11
\[ \sum \text{digits at odd positions} - \sum \text{digits at even positions} \equiv 0 \pmod{11} \]
where: \( n \) = any integer
Example 5: Divisibility by 2 and 5 Easy
Check if 24680 is divisible by 2 and 5.

Step 1: Last digit is 0.

Step 2: Since 0 is even, 24680 is divisible by 2.

Step 3: Since last digit is 0, 24680 is divisible by 5.

Answer: 24680 is divisible by both 2 and 5.

Example 6: Divisibility by 3 and 9 Easy
Determine if 123456 is divisible by 3 and 9.

Step 1: Sum of digits: 1 + 2 + 3 + 4 + 5 + 6 = 21.

Step 2: Check divisibility by 3: 21 / 3 = 7, so divisible by 3.

Step 3: Check divisibility by 9: 21 / 9 = 2 remainder 3, so not divisible by 9.

Answer: 123456 is divisible by 3 but not by 9.

Tips & Tricks

Tip: Use the sum of digits shortcut to quickly check divisibility by 3 or 9.

When to use: When you want to avoid long division and quickly test numbers.

Tip: Check the last digit instantly for divisibility by 2 or 5.

When to use: Useful for quick elimination or confirmation in multiple-choice questions.

Tip: For divisibility by 6, confirm divisibility by both 2 and 3.

When to use: When you know the number is even but unsure about 3, or vice versa.

Tip: Use the alternating sum method for divisibility by 11.

When to use: When dealing with numbers with many digits to avoid lengthy division.

Tip: Apply the doubling and subtracting method repeatedly for divisibility by 7.

When to use: For large numbers where direct division is time-consuming.

Common Mistakes to Avoid

❌ Checking divisibility by 4 using only the last digit
✓ Check the last two digits of the number for divisibility by 4
Why: Divisibility by 4 depends on the last two digits because 100 is divisible by 4, not just the last digit.
❌ Adding digits incorrectly when checking divisibility by 3 or 9
✓ Carefully sum all digits accurately before checking divisibility
Why: An incorrect sum leads to wrong conclusions about divisibility.
❌ Assuming divisibility by 6 if divisible by 2 or 3 alone
✓ Number must be divisible by both 2 and 3 to be divisible by 6
Why: Divisibility by 6 requires both conditions simultaneously, not just one.
❌ Misapplying the alternating sum rule for 11 by mixing odd and even positions
✓ Ensure correct identification of odd and even digit positions consistently
Why: Wrong positioning leads to incorrect alternating sums and wrong results.
❌ Forgetting to reduce numbers repeatedly in the divisibility by 7 method
✓ Repeat doubling and subtracting steps until a small number is obtained
Why: Large numbers need reduction to apply the rule effectively and avoid errors.
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