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Divisibility

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Introduction to Divisibility

Divisibility is a fundamental concept in arithmetic that helps us understand how numbers relate to each other through division. When we say a number a is divisible by another number b, it means that when a is divided by b, the result is a whole number without any remainder.

For example, 20 is divisible by 5 because 20 / 5 = 4, which is a whole number. But 20 is not divisible by 3 because 20 / 3 = 6 remainder 2.

Understanding divisibility is important because it helps us simplify calculations, factor numbers, and solve problems quickly - skills that are especially useful in competitive exams where time is limited.

Key Terms

  • Dividend: The number being divided (e.g., 20 in 20 / 5).
  • Divisor: The number by which the dividend is divided (e.g., 5 in 20 / 5).
  • Quotient: The result of division (e.g., 4 in 20 / 5 = 4).
  • Remainder: The leftover part after division if the dividend is not perfectly divisible (e.g., 2 in 20 / 3 = 6 remainder 2).

Divisibility Rules

Dividing large numbers directly can be time-consuming. To speed up the process, mathematicians have developed simple rules to quickly check if a number is divisible by another without performing full division. These are called divisibility rules.

Let's explore the divisibility rules for some common numbers:

Divisor Divisibility Rule Example
2 The number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8). 124 is divisible by 2 (last digit 4).
3 The 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 divisible by 3.
4 The number is divisible by 4 if the last two digits form a number divisible by 4. 316 -> last two digits 16, 16 / 4 = 4, so divisible by 4.
5 The number is divisible by 5 if its last digit is 0 or 5. 145 ends with 5, so divisible by 5.
6 The number is divisible by 6 if it is divisible by both 2 and 3. 114 is divisible by 2 (last digit 4) and 3 (1+1+4=6), so divisible by 6.
7 Double the last digit and subtract it from the remaining leading number. Repeat if needed. If the result is divisible by 7, so is the original number. 203: Double last digit 3x2=6; 20-6=14, 14 divisible by 7, so 203 divisible by 7.
8 The number is divisible by 8 if the last three digits form a number divisible by 8. 17,216 -> last three digits 216, 216 / 8 = 27, so divisible by 8.
9 The number is divisible by 9 if the sum of its digits is divisible by 9. 729 -> 7+2+9=18, 18 divisible by 9, so 729 divisible by 9.
11 Find the difference between the sum of digits in odd positions and the sum of digits in even positions. If the difference is 0 or divisible by 11, the number is divisible by 11. 2728: (2+2) - (7+8) = 4 - 15 = -11, divisible by 11, so 2728 divisible by 11.

Why Are These Rules Useful?

These rules help you quickly decide if a number can be divided evenly by another without performing the entire division. This is especially helpful in exams where time is limited and accuracy is important.

Divisibility

\[n \equiv 0 \pmod{d}\]

A number n is divisible by d if the remainder when n is divided by d is zero.

n = Number to test
d = Divisor

Worked Examples

Example 1: 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.

Since 15 is divisible by 3 (15 / 3 = 5), 12345 is divisible by 3.

Step 2: Check divisibility by 5 by looking at the last digit.

The last digit of 12345 is 5, so it is divisible by 5.

Answer: 12345 is divisible by both 3 and 5.

Example 2: Largest Number Less Than 200 Divisible by 6 Medium
Find the largest number less than 200 that is divisible by 6.

Step 1: Since 6 = 2 x 3, the number must be divisible by both 2 and 3.

Step 2: Start from 199 and check downwards.

199: Last digit 9 (odd), not divisible by 2.

198: Last digit 8 (even), so divisible by 2.

Sum of digits of 198 = 1 + 9 + 8 = 18, which is divisible by 3.

Therefore, 198 is divisible by both 2 and 3, hence divisible by 6.

Answer: 198 is the largest number less than 200 divisible by 6.

Example 3: Simplify Fraction 84/126 Medium
Simplify the fraction \(\frac{84}{126}\) using divisibility rules.

Step 1: Find the greatest common divisor (GCD) of 84 and 126.

Check divisibility by 2:

  • 84 ends with 4 (even), divisible by 2.
  • 126 ends with 6 (even), divisible by 2.

Divide both by 2:

\(\frac{84}{2} = 42\), \(\frac{126}{2} = 63\)

Step 2: Check divisibility by 3:

  • Sum of digits of 42 = 4 + 2 = 6, divisible by 3.
  • Sum of digits of 63 = 6 + 3 = 9, divisible by 3.

Divide both by 3:

\(\frac{42}{3} = 14\), \(\frac{63}{3} = 21\)

Step 3: Check divisibility by 7:

  • 14 / 7 = 2 (exact)
  • 21 / 7 = 3 (exact)

Divide both by 7:

\(\frac{14}{7} = 2\), \(\frac{21}{7} = 3\)

Answer: \(\frac{84}{126} = \frac{2}{3}\) after simplification.

Example 4: Divisibility by 11 Medium
Check if the number 2728 is divisible by 11 using the divisibility rule.

Step 1: Identify digits in odd and even positions from right to left.

Number: 2 7 2 8

  • Odd positions (from right): 8 (1st), 2 (3rd) -> sum = 8 + 2 = 10
  • Even positions: 2 (2nd), 7 (4th) -> sum = 2 + 7 = 9

Step 2: Find the difference: 10 - 9 = 1

Since 1 is not 0 or divisible by 11, 2728 is not divisible by 11.

Note: The example in the table used a different approach; here, we used the standard method from right to left.

Answer: 2728 is not divisible by 11.

Example 5: Word Problem Using Divisibility Hard
A total amount of Rs.7290 is to be paid in equal installments, each divisible by 9. Find the number of installments if each installment is less than Rs.1000.

Step 1: Since each installment is divisible by 9, let each installment be 9k, where k is an integer.

Step 2: Total installments = \(\frac{7290}{9k} = \frac{810}{k}\).

Step 3: Each installment must be less than Rs.1000, so:

9k < 1000 -> k < \(\frac{1000}{9} \approx 111.11\)

So, k can be any integer less than or equal to 111.

Step 4: Number of installments = \(\frac{810}{k}\) must be an integer.

Find divisors of 810 less than or equal to 111:

  • 810 / 1 = 810 (too many installments)
  • 810 / 9 = 90 (valid)
  • 810 / 10 = 81 (valid)
  • 810 / 15 = 54 (valid)
  • 810 / 18 = 45 (valid)
  • 810 / 27 = 30 (valid)
  • 810 / 30 = 27 (valid)
  • 810 / 45 = 18 (valid)
  • 810 / 54 = 15 (valid)
  • 810 / 81 = 10 (valid)
  • 810 / 90 = 9 (valid)
  • 810 / 90 = 9 (valid)

Choose the number of installments so that each installment is less than Rs.1000 and the number of installments is reasonable.

Answer: Possible number of installments include 9, 10, 15, 18, 27, 30, 45, 54, 81, 90, etc., with each installment divisible by 9 and less than Rs.1000.

Formula Bank

Divisibility by 2
\[ n \equiv 0 \pmod{2} \iff \text{last digit of } n \text{ is even} \]
where: \(n\) = number
Divisibility by 3
\[ \sum \text{digits of } n \equiv 0 \pmod{3} \]
where: \(n\) = number
Divisibility by 4
\[ \text{Last two digits of } n \equiv 0 \pmod{4} \]
where: \(n\) = number
Divisibility by 5
\[ n \equiv 0 \pmod{5} \iff \text{last digit is } 0 \text{ or } 5 \]
where: \(n\) = number
Divisibility by 6
\[ n \text{ divisible by 2 and 3 simultaneously} \]
where: \(n\) = number
Divisibility by 7
\[ \text{Double last digit and subtract from remaining number; repeat if needed} \]
where: \(n\) = number
Divisibility by 8
\[ \text{Last three digits of } n \equiv 0 \pmod{8} \]
where: \(n\) = number
Divisibility by 9
\[ \sum \text{digits of } n \equiv 0 \pmod{9} \]
where: \(n\) = number
Divisibility by 11
\[ \sum \text{digits in odd positions} - \sum \text{digits in even positions} \equiv 0 \pmod{11} \]
where: \(n\) = number

Tips & Tricks

Tip: Quickly sum digits to check divisibility by 3 or 9.

When to use: When testing large numbers for divisibility by 3 or 9.

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

When to use: For quick checks without calculation.

Tip: Use the alternating sum method for divisibility by 11 to avoid long division.

When to use: When checking divisibility by 11 for multi-digit numbers.

Tip: Combine divisibility rules for composite numbers like 6 (which requires divisibility by both 2 and 3).

When to use: When testing divisibility by numbers that are products of primes.

Tip: Memorize the divisibility rules table for quick recall during exams.

When to use: Before and during competitive exams for speed and accuracy.

Common Mistakes to Avoid

❌ Adding digits incorrectly when checking divisibility by 3 or 9.
✓ Carefully sum all digits and double-check the total before concluding.
Why: Students often rush and miss digits or add incorrectly, leading to wrong answers.
❌ Checking only the last digit for divisibility by 4 instead of the last two digits.
✓ Always consider the last two digits to determine divisibility by 4.
Why: Misunderstanding the rule causes incorrect conclusions.
❌ Confusing divisibility by 6 with divisibility by 3 or 2 alone.
✓ Remember that divisibility by 6 requires the number to be divisible by both 2 and 3 simultaneously.
Why: Students forget the composite nature of 6 and apply only one rule.
❌ Incorrectly applying the divisibility by 11 alternating sum rule by mixing digit positions.
✓ Ensure correct positioning of digits (odd and even places) counting from right to left.
Why: Confusing digit positions leads to wrong sums and wrong conclusions.
❌ Using divisibility rules for numbers not covered by standard rules (like 13 or 17) without proper methods.
✓ Use long division or specific tests designed for such numbers instead of standard rules.
Why: Standard divisibility rules do not apply to all numbers, leading to errors.
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