Exercise 1.1 — Division Lemma Problems

What Does Exercise 1.1 Cover?

Exercise 1.1 of Class 10 Mathematics — Chapter 1: Real Numbers — is entirely built around Euclid's Division Algorithm. Once you understand the Division Lemma (a = bq + r, where 0 ≤ r < b), this exercise trains you to apply it in two ways: finding the HCF of large numbers, and constructing algebraic proofs about the form of integers. Both types appear regularly in CBSE, Telangana, and Andhra Pradesh board exams, making this exercise one of the most important in the chapter.

Question 1 — Finding HCF Using Euclid's Algorithm

The first question asks you to find the HCF of three pairs of numbers. The key idea is to always divide the larger number by the smaller, then carry the remainder forward as the new divisor, repeating until the remainder is zero. The last non-zero divisor is the HCF.

• HCF of 900 and 270: Dividing 900 by 270 gives remainder 90. Then 270 ÷ 90 leaves remainder 0. So HCF = 90.
• HCF of 38220 and 196: 38220 ÷ 196 gives remainder 0 directly (196 divides 38220 exactly 195 times). So HCF = 196.
• HCF of 2032 and 1651: 2032 = 1651 × 1 + 381; then 1651 = 381 × 4 + 127; then 381 = 127 × 3 + 0. So HCF = 127.

Notice that for the pair 38220 and 196, no repeated steps are needed — this happens when one number is a multiple of the other. Always check for this first to save time in exams.

Questions 2 & 3 — Proving the Form of Odd Integers and Squares

These questions use the Division Lemma not to find an HCF, but to prove a general statement about all positive integers. The strategy is always the same: assume any positive integer a, choose a suitable divisor b, list every possible remainder, and show which cases are even and which are odd.

Question 2 asks you to prove that every positive odd integer must be of the form 6q + 1, 6q + 3, or 6q + 5. Taking b = 6, the six possible remainders are 0 through 5. The cases r = 0, 2, 4 produce even integers (each is a multiple of 2), while r = 1, 3, 5 produce odd integers — confirming the result.

Question 3 asks you to prove that the square of any positive integer is of the form 3p or 3p + 1. Taking b = 3 gives three possible remainders:

• If r = 0: a = 3q, so a² = 9q² = 3(3q²) — this is of the form 3p.
• If r = 1: a = 3q + 1, so a² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 — this is of the form 3p + 1.
• If r = 2: a = 3q + 2, so a² = 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1 — again 3p + 1.

The form 3p + 2 never appears, which is exactly what the question wants you to show.

Question 4 — Cubes Are of the Form 9m, 9m+1, or 9m+8

This is the most calculation-heavy proof in the exercise. Again take b = 3, giving three cases. When you cube each form and expand using the binomial theorem:

• If a = 3q: a³ = 27q³ = 9(3q³) — the form 9m.
• If a = 3q + 1: a³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1 — the form 9m + 1.
• If a = 3q + 2: a³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8 — the form 9m + 8.

The key step students often miss is factoring out 9 cleanly from the first three terms after expanding. Write out the full expansion before grouping.

Question 5 — Only One of n, n+2, n+4 Is Divisible by 3

This elegant problem uses the same approach. If n is divided by 3, the remainder is 0, 1, or 2. In each case, check all three numbers n, n+2, and n+4: exactly one of them simplifies to a multiple of 3, while the other two leave remainders of 1 or 2. Since this holds for every possible remainder, it holds for every positive integer n — proving that among any three consecutive odd numbers (or equivalently, n, n+2, n+4), exactly one is always divisible by 3.

Common Mistakes in Exercise 1.1

• Starting division with the smaller number as dividend — always divide the larger by the smaller in the first step.
• Stopping one step too early — the HCF is the divisor when the remainder first becomes zero, not the remainder in the previous step.
• In proof questions, forgetting to check all possible remainders — missing even one case makes the proof incomplete and costs marks.
• Algebraic errors while expanding cubes in Question 4 — expand step by step and verify each coefficient before factoring out 9.

What to Study Next

After completing Exercise 1.1, you are ready to move deeper into the chapter. The next major topics build directly on this foundation: the Fundamental Theorem of Arithmetic uses prime factorisation to find HCF and LCM more efficiently, and the section on proving irrationality of numbers like √2 and √3 uses the Division Lemma-style proof logic you practised here. You can also revisit the concepts behind this exercise in the Real Numbers Introduction.