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Divisibility rules

Introduction to Divisibility Rules

In mathematics, divisibility is a fundamental concept that helps us understand when one number can be divided by another without leaving any remainder. For example, 20 is divisible by 5 because 20 / 5 = 4, which is a whole number. However, 20 is not divisible by 3 because 20 / 3 = 6.66..., which is not an integer.

Divisibility rules are simple tests or shortcuts that allow us to quickly determine whether a number is divisible by another number without performing the full division. These rules are especially useful in competitive exams where speed and accuracy are crucial. Instead of long division, you can use these rules to save time and avoid errors.

Understanding divisibility rules also lays the foundation for other important topics like finding factors, multiples, simplifying fractions, and solving problems involving the Least Common Multiple (LCM) and Highest Common Factor (HCF).

Basic Divisibility Rules

Let's start with the simplest and most commonly used divisibility rules. These rules rely mostly on the digits of the number and can be applied quickly.

Divisor Divisibility Rule Example
2 Number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8). 124 (last digit 4 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
4 Number is divisible by 4 if the last two digits form a number divisible by 4. 316 (last two digits 16 divisible by 4) -> divisible by 4
5 Number is divisible by 5 if its last digit is 0 or 5. 145 (last digit 5) -> divisible by 5
8 Number is divisible by 8 if the last three digits form a number divisible by 8. 1,024 (last three digits 024 or 24 divisible by 8) -> divisible by 8
9 Number is divisible by 9 if the sum of its digits is divisible by 9. 729 (7+2+9=18, divisible by 9) -> divisible by 9
10 Number is divisible by 10 if its last digit is 0. 1,230 (last digit 0) -> divisible by 10

Notice how these rules often depend on the last digit(s) or the sum of digits. This is because of the way our number system is structured in base 10.

Advanced Divisibility Rules

Some divisibility rules are a bit more complex but still manageable with practice. These include divisibility by 6, 7, and 11.

Divisibility by 6

A number is divisible by 6 if it is divisible by both 2 and 3. This means:

  • The last digit is even (divisible by 2), and
  • The sum of the digits is divisible by 3.

Example: Check if 234 is divisible by 6.

  • Last digit is 4 (even) -> divisible by 2
  • Sum of digits = 2 + 3 + 4 = 9, which is divisible by 3
  • Therefore, 234 is divisible by 6.

Divisibility by 7

Divisibility by 7 is less straightforward. One common method is:

  1. Double the last digit of the number.
  2. Subtract this doubled value from the rest of the number (the number formed by all digits except the last one).
  3. If the result is divisible by 7 (including 0), then the original number is divisible by 7.
  4. If not, repeat the process with the new number.

Divisibility by 11

To check divisibility by 11:

  1. Add the digits in the odd positions (from right to left).
  2. Add the digits in the even positions.
  3. Find the difference between these two sums.
  4. If the difference is 0 or divisible by 11, then the number is divisible by 11.
graph TD    A[Start] --> B{Divisibility by 7?}    B --> C[Double last digit]    C --> D[Subtract from rest of number]    D --> E{Result divisible by 7?}    E -->|Yes| F[Number divisible by 7]    E -->|No| G[Repeat with result]    G --> C    H[Start] --> I{Divisibility by 11?}    I --> J[Sum digits at odd positions]    I --> K[Sum digits at even positions]    J --> L[Find difference]    K --> L    L --> M{Difference divisible by 11?}    M -->|Yes| N[Number divisible by 11]    M -->|No| O[Not divisible by 11]

Worked Examples

Example 1: Checking Divisibility by 3 and 9 Easy
Check if the numbers 729 and 123456 are divisible by 3 and 9.

Step 1: Find the sum of digits for 729.

7 + 2 + 9 = 18

Step 2: Check if 18 is divisible by 3 and 9.

18 / 3 = 6 (no remainder), 18 / 9 = 2 (no remainder)

Therefore, 729 is divisible by both 3 and 9.

Step 3: Find the sum of digits for 123456.

1 + 2 + 3 + 4 + 5 + 6 = 21

Step 4: Check if 21 is divisible by 3 and 9.

21 / 3 = 7 (no remainder), 21 / 9 = 2.33 (remainder exists)

Therefore, 123456 is divisible by 3 but not by 9.

Answer: 729 is divisible by both 3 and 9; 123456 is divisible by 3 only.

Example 2: Divisibility by 11 Medium
Determine if 2728 is divisible by 11.

Step 1: Identify digits at odd and even positions (from right to left):

  • Odd positions: 8 (1st), 7 (3rd) -> sum = 8 + 7 = 15
  • Even positions: 2 (2nd), 2 (4th) -> sum = 2 + 2 = 4

Step 2: Find the difference:

|15 - 4| = 11

Step 3: Since 11 is divisible by 11, 2728 is divisible by 11.

Answer: 2728 is divisible by 11.

Example 3: Using Divisibility Rules to Simplify Fractions Medium
Simplify the fraction \(\frac{462}{693}\) using divisibility rules.

Step 1: Check divisibility of numerator and denominator by 3.

Sum of digits of 462 = 4 + 6 + 2 = 12 (divisible by 3)

Sum of digits of 693 = 6 + 9 + 3 = 18 (divisible by 3)

Both divisible by 3, so divide numerator and denominator by 3:

\(\frac{462 / 3}{693 / 3} = \frac{154}{231}\)

Step 2: Check divisibility by 7.

For 154: Double last digit (4 x 2 = 8), subtract from rest (15 - 8 = 7), 7 is divisible by 7 -> 154 divisible by 7.

For 231: Double last digit (1 x 2 = 2), subtract from rest (23 - 2 = 21), 21 divisible by 7 -> 231 divisible by 7.

Divide numerator and denominator by 7:

\(\frac{154 / 7}{231 / 7} = \frac{22}{33}\)

Step 3: Check divisibility by 11.

For 22: (2 - 2) = 0 -> divisible by 11

For 33: (3 - 3) = 0 -> divisible by 11

Divide numerator and denominator by 11:

\(\frac{22 / 11}{33 / 11} = \frac{2}{3}\)

Answer: The simplified fraction is \(\frac{2}{3}\).

Example 4: Divisibility by 7 Using the Double and Subtract Method Hard
Check if 203 is divisible by 7 using the rule: double the last digit and subtract from the rest.

Step 1: Last digit is 3; double it: 3 x 2 = 6.

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

Step 3: Check if 14 is divisible by 7.

14 / 7 = 2 (no remainder), so 14 is divisible by 7.

Step 4: Since 14 is divisible by 7, 203 is also divisible by 7.

Answer: 203 is divisible by 7.

Example 5: Application in Entrance Exam Problem Hard
Find the smallest number divisible by 2, 3, 5, and 7.

Step 1: To find a number divisible by multiple numbers, find their Least Common Multiple (LCM).

Given numbers: 2, 3, 5, 7

Step 2: Since all are prime numbers, LCM is their product:

LCM = 2 x 3 x 5 x 7 = 210

Step 3: Verify divisibility:

  • 210 / 2 = 105 (no remainder)
  • 210 / 3 = 70 (no remainder)
  • 210 / 5 = 42 (no remainder)
  • 210 / 7 = 30 (no remainder)

Answer: The smallest number divisible by 2, 3, 5, and 7 is 210.

Formula Bank

Divisibility by 2
\[ n \equiv 0 \pmod{2} \iff \text{last digit of } n \text{ is even} \]
where: \( n \) = given number
Divisibility by 3
\[ n \equiv 0 \pmod{3} \iff \sum \text{digits of } n \equiv 0 \pmod{3} \]
where: \( n \) = given number
Divisibility by 4
\[ n \equiv 0 \pmod{4} \iff \text{last two digits of } n \equiv 0 \pmod{4} \]
where: \( n \) = given number
Divisibility by 5
\[ n \equiv 0 \pmod{5} \iff \text{last digit of } n = 0 \text{ or } 5 \]
where: \( n \) = given number
Divisibility by 6
\[ n \equiv 0 \pmod{6} \iff n \equiv 0 \pmod{2} \text{ and } n \equiv 0 \pmod{3} \]
where: \( n \) = given number
Divisibility by 7
\[ n \equiv 0 \pmod{7} \iff \text{double last digit subtracted from rest is divisible by 7} \]
where: \( n \) = given number
Divisibility by 8
\[ n \equiv 0 \pmod{8} \iff \text{last three digits of } n \equiv 0 \pmod{8} \]
where: \( n \) = given number
Divisibility by 9
\[ n \equiv 0 \pmod{9} \iff \sum \text{digits of } n \equiv 0 \pmod{9} \]
where: \( n \) = given number
Divisibility by 10
\[ n \equiv 0 \pmod{10} \iff \text{last digit of } n = 0 \]
where: \( n \) = given number
Divisibility by 11
\[ n \equiv 0 \pmod{11} \iff \left| \sum \text{digits at odd positions} - \sum \text{digits at even positions} \right| \equiv 0 \pmod{11} \]
where: \( n \) = given number

Tips & Tricks

Tip: Use the sum of digits shortcut for checking divisibility by 3 and 9.

When to use: Quickly determine if a large number is divisible by 3 or 9 without division.

Tip: Check only the last digit for divisibility by 2, 5, and 10.

When to use: Instantly decide divisibility by these numbers by looking at the last digit.

Tip: For divisibility by 11, always take the absolute difference between sums of alternate digits.

When to use: Avoid confusion caused by negative differences.

Tip: Use the double last digit and subtract method to test divisibility by 7 efficiently.

When to use: Quickly check divisibility by 7 without long division, especially for large numbers.

Tip: Combine rules for composite numbers like 6 by verifying divisibility by 2 and 3 separately.

When to use: Simplify checks for numbers that are products of primes.

Common Mistakes to Avoid

❌ Checking divisibility by 4 using only the last digit.
✓ Check the last two digits for divisibility by 4.
Why: Divisibility by 4 depends on the last two digits, not just one.
❌ Assuming the sum of digits rule applies to all divisors.
✓ Remember sum of digits rule applies only to 3 and 9, not to 7 or 11.
Why: Different divisors have different rules; applying sum of digits incorrectly leads to errors.
❌ Forgetting to take the absolute difference in divisibility by 11.
✓ Always take the absolute value of the difference between sums of alternate digits.
Why: Negative differences can mislead; absolute difference is required.
❌ Not applying divisibility rules iteratively for large numbers.
✓ Apply rules stepwise or reduce the number progressively for accurate checks.
Why: Large numbers may require multiple iterations for accurate divisibility checks.
❌ Mixing up divisibility rules for 6 and 9.
✓ Remember 6 requires divisibility by 2 and 3, while 9 requires sum of digits divisible by 9.
Why: Confusing these leads to wrong conclusions about divisibility.

General Divisibility Concept

\[n \equiv 0 \pmod{d} \iff \text{Divisibility rule for } d \text{ applies}\]

A number n is divisible by d if it satisfies the specific divisibility rule for d.

n = Given number
d = Divisor

Quick Divisibility Tips

  • Check last digit for 2, 5, 10
  • Sum digits for 3 and 9
  • Use difference of sums for 11
  • Double last digit and subtract for 7
  • Combine rules for 6 (2 and 3)
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