Ever wondered why division always leaves a tiny “left‑over” called a remainder? That’s not a mistake – it’s a neat rule called Euclid's Division Lemma.
💡 In Simple Words: When you divide any whole number by a positive whole number, you can always write the result as a perfect multiple plus a tiny leftover that is smaller than the divisor.
What is Euclid's Division Lemma?
In plain language, the lemma says: for any two integers a (the dividend) and b (the divisor, with b > 0), there are unique integers q (the quotient) and r (the remainder) such that
a = b·q + r where 0 ≤ r < b.
Think of a as a pile of candies, b as the size of each bag you want to fill, q as the number of completely filled bags, and r as the candies left over that don’t fill another bag.
How to Apply Euclid's Division Lemma
The steps are simple:
- Step 1: Identify the dividend a and the divisor b (b must be positive).
- Step 2: Find the largest whole number q such that b·q ≤ a. In other words, q = ⌊a/b⌋ (the floor of the exact division).
- Step 3: Compute the remainder r = a – b·q.
- Step 4: Verify that 0 ≤ r < b. If it holds, you’ve found the unique pair (q, r).
Worked Example for CBSE Class 10
Find q and r when a = 87 and b = 13.
- Compute q = ⌊87/13⌋ = 6 because 13×6 = 78 ≤ 87 and 13×7 = 91 > 87.
- Find r = 87 – 13×6 = 87 – 78 = 9.
- Check the condition: 0 ≤ 9
So, 87 = 13·6 + 9. The quotient is 6 and the remainder is 9.
Another Example with a Smaller Dividend
Let a = 5 and b = 12. Here the dividend is smaller than the divisor.
- q = ⌊5/12⌋ = 0 (because 12×0 = 0 ≤ 5 and any larger q would exceed 5).
- r = 5 – 12·0 = 5.
- 0 ≤ 5
Result: 5 = 12·0 + 5. The remainder can be the whole dividend when the divisor is larger.
Key Points Summary
| Symbol | Meaning |
|---|---|
| a | Dividend – the number you start with |
| b | Divisor – the positive number you divide by |
| q | Quotient – how many whole times b fits into a |
| r | Remainder – the leftover part, always smaller than b |
| 0 ≤ r < b | Condition that guarantees uniqueness |
Common Mistakes to Avoid
- Using a negative divisor – the lemma requires b > 0.
- Forgetting the condition r . If r equals b, you can increase q by 1 and set r to 0.
- Assuming the remainder is always zero for any division. Only exact multiples give r = 0.
Why Euclid's Division Lemma Matters for CBSE Exams
The lemma is the foundation of the Division Algorithm and appears in many CBSE questions: finding GCD (greatest common divisor) using the Euclidean algorithm, proving properties of integers, and simplifying algebraic expressions. Knowing the lemma helps you write clear, step‑by‑step solutions that earn full marks.
📝 Likely Exam Questions
- State Euclid's Division Lemma and explain each term.
- Find the quotient and remainder when 154 is divided by 17.
- Using Euclid's Division Lemma, show that 23 = 5·4 + 3 and verify the conditions.
- Explain why the remainder is always less than the divisor.
- Apply the lemma to prove that any integer can be expressed as 7q + r where 0 ≤ r
Model Answers (brief):
- Lemma: For any integers a and b (b>0) there exist unique integers q and r such that a = bq + r and 0 ≤ r
- 154 ÷ 17 → q = 9, r = 1 because 17·9 = 153 and 154‑153 = 1.
- 23 = 5·4 + 3; here q=4, r=3 and 0 ≤ 3
- The remainder is the part left after taking out as many whole copies of b as possible; by definition it cannot reach b.
- Write 23 = 7·3 + 2, so q=3, r=2, and 0 ≤ 2