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).
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.
Some divisibility rules are a bit more complex but still manageable with practice. These include divisibility by 6, 7, and 11.
A number is divisible by 6 if it is divisible by both 2 and 3. This means:
Example: Check if 234 is divisible by 6.
Divisibility by 7 is less straightforward. One common method is:
To check divisibility 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]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.
Step 1: Identify digits at odd and even positions (from right to left):
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.
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 \(\frac{2}{3}\).
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.
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:
Answer: The smallest number divisible by 2, 3, 5, and 7 is 210.
When to use: Quickly determine if a large number is divisible by 3 or 9 without division.
When to use: Instantly decide divisibility by these numbers by looking at the last digit.
When to use: Avoid confusion caused by negative differences.
When to use: Quickly check divisibility by 7 without long division, especially for large numbers.
When to use: Simplify checks for numbers that are products of primes.
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