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:
Before diving into the rules, let's briefly recall the number system hierarchy:
Divisibility rules mainly apply to integers and natural numbers. Mastering these rules will sharpen your number sense and speed up calculations.
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.
| 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:
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.
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.
| 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:
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.
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.
One popular method:
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]Calculate the alternating sum of digits:
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.
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.
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:
Step 4: Check if 39 is divisible by 13.
Step 5: 39 / 13 = 3, which is an integer.
Answer: 2730 is divisible by 13.
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.
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.
When to use: When you want to avoid long division and quickly test numbers.
When to use: Useful for quick elimination or confirmation in multiple-choice questions.
When to use: When you know the number is even but unsure about 3, or vice versa.
When to use: When dealing with numbers with many digits to avoid lengthy division.
When to use: For large numbers where direct division is time-consuming.
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