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Divisibility is a fundamental concept in mathematics that helps us understand when one number can be divided by another without leaving a remainder. For example, 12 is divisible by 3 because 12 / 3 = 4, which is a whole number. Divisibility plays a crucial role in number theory, simplifying fractions, finding factors, and solving many problems in competitive exams.
Understanding divisibility rules allows you to quickly check whether a number is divisible by another without performing long division. This saves time and effort, especially in exams where speed and accuracy are important. In this chapter, we will explore various divisibility rules, starting from the simplest ones and moving towards more advanced techniques.
Let's begin with some of the most common divisibility rules. These rules help you determine if a number is divisible by 2, 3, 5, 9, and 11 quickly and easily.
| Divisor | Rule | Example |
|---|---|---|
| 2 | Number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8). | 124 is divisible by 2 because last digit 4 is even. |
| 3 | Number is divisible by 3 if the sum of its digits is divisible by 3. | 123 -> 1 + 2 + 3 = 6, divisible by 3, so 123 is divisible by 3. |
| 5 | Number is divisible by 5 if its last digit is 0 or 5. | 145 ends with 5, so divisible by 5. |
| 9 | Number is divisible by 9 if the sum of its digits is divisible by 9. | 729 -> 7 + 2 + 9 = 18, divisible by 9, so 729 is divisible by 9. |
| 11 | Number is divisible by 11 if the difference between the sum of digits in odd positions and the sum of digits in even positions is divisible by 11. | 2728 -> (2 + 2) - (7 + 8) = 4 - 15 = -11, divisible by 11, so 2728 is divisible by 11. |
Each divisibility rule is based on properties of numbers in the decimal system. For example, the rule for 3 and 9 works because 10 ≡ 1 (mod 3) and (mod 9), so each digit contributes its face value modulo 3 or 9. The rule for 11 uses alternating sums because 10 ≡ -1 (mod 11), causing digits in odd and even positions to subtract.
Some divisors like 7, 13, and 17 do not have as simple rules as 2 or 3, but there are clever methods to check divisibility without full division. One popular method for 7 is the doubling and subtraction technique.
graph TD A[Start with number n] --> B[Separate last digit d and remaining number r] B --> C[Calculate new number = r - 2 x d] C --> D{Is new number divisible by 7?} D -->|Yes| E[Original number divisible by 7] D -->|No| F[Repeat process with new number]How to use this method: Take the last digit, double it, subtract from the rest of the number. Repeat until you get a small number easily checked for divisibility by 7.
Similar methods exist for 13 and 17, but often the remainder theorem (explained below) is more efficient for algebraic expressions or larger numbers.
The remainder theorem states that the remainder when a polynomial \( P(x) \) is divided by \( x - a \) is equal to \( P(a) \). In number theory, this can be used to check divisibility by plugging in values and evaluating expressions.
For example, to check if a number \( n \) is divisible by 7, express \( n \) in terms of powers of 10 and evaluate modulo 7 using the remainder theorem.
Step 1: Check divisibility by 2 by looking at the last digit.
Last digit of 1240 is 0, which is even.
Therefore, 1240 is divisible by 2.
Step 2: Check divisibility by 5 by looking at the last digit.
Last digit is 0, which is either 0 or 5.
Therefore, 1240 is divisible by 5.
Answer: 1240 is divisible by both 2 and 5.
Step 1: Sum the digits: 7 + 2 + 9 = 18.
Step 2: Check if 18 is divisible by 3 and 9.
18 / 3 = 6 (no remainder), so divisible by 3.
18 / 9 = 2 (no remainder), so divisible by 9.
Answer: 729 is divisible by both 3 and 9.
Step 1: Identify digits in odd and even positions from right:
Step 2: Calculate 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: Separate last digit (3) and remaining number (20).
Step 2: Double the last digit: 3 x 2 = 6.
Step 3: Subtract from remaining number: 20 - 6 = 14.
Step 4: Check if 14 is divisible by 7.
14 / 7 = 2 (no remainder), so 14 is divisible by 7.
Answer: Since 14 is divisible by 7, 203 is also divisible by 7.
Step 1: According to the remainder theorem, evaluate \( P(1) \).
\( P(1) = 1^3 + 2 \times 1^2 - 1 + 5 = 1 + 2 - 1 + 5 = 7 \).
Step 2: Since \( P(1) eq 0 \), the polynomial is not divisible by \( x - 1 \).
Answer: \( P(x) \) is not divisible by \( x - 1 \) because the remainder is 7.
When to use: Quickly check divisibility without performing division.
When to use: Fast elimination of numbers not divisible by 2 or 5.
When to use: Efficiently test large numbers for divisibility by 11.
When to use: When divisibility by 7 is required and direct division is cumbersome.
When to use: When dealing with algebraic expressions or number system problems involving polynomials.
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