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Class 10 Maths Chapter 1 · Real Numbers Exercise 1.1 Easy Malayalam Explained

Use Euclid's division algorithm to find the HCF of: (i) 135 and 225  (ii) 196 and 38220  (iii) 867 and 255

NCERT Class 10 Maths · Chapter 1 Real Numbers · Exercise 1.1 · Question 1

(i) HCF(135, 225) = 45  |  (ii) HCF(196, 38220) = 196  |  (iii) HCF(867, 255) = 51

Step-by-Step Solution

We use Euclid's Division Lemma: For any two positive integers a and b, there exist unique integers q and r such that a = bq + r where 0 ≤ r < b. We keep dividing until the remainder = 0. The last non-zero remainder is the HCF.

Part (i): HCF of 135 and 225

1

Since 225 > 135, divide 225 by 135

225 = 1 × 135 + 90

Quotient = 1, Remainder = 90 (≠ 0, so continue)

2

Now divide 135 by 90

135 = 1 × 90 + 45

Quotient = 1, Remainder = 45 (≠ 0, so continue)

3

Now divide 90 by 45

90 = 2 × 45 + 0

Remainder = 0 ✓ — Stop here!

Final Answer — Part (i)
HCF (135, 225) = 45
The last non-zero remainder is 45

Part (ii): HCF of 196 and 38220

1

Since 38220 > 196, divide 38220 by 196

38220 = 195 × 196 + 0

Remainder = 0 immediately! ✓ Stop here.

Final Answer — Part (ii)
HCF (196, 38220) = 196
196 divides 38220 exactly — so 196 itself is the HCF

Part (iii): HCF of 867 and 255

1

867 > 255, so divide 867 by 255

867 = 3 × 255 + 102

Remainder = 102 (≠ 0, continue)

2

Divide 255 by 102

255 = 2 × 102 + 51

Remainder = 51 (≠ 0, continue)

3

Divide 102 by 51

102 = 2 × 51 + 0

Remainder = 0 ✓ Stop here!

Final Answer — Part (iii)
HCF (867, 255) = 51
The last non-zero remainder is 51

Theory: Euclid's Division Lemma

Statement: For any two positive integers a and b, there exist unique non-negative integers q (quotient) and r (remainder) such that:

a = bq + r, where 0 ≤ r < b

Algorithm for HCF: (1) Write a = bq + r. (2) If r = 0, then b is the HCF. (3) If r ≠ 0, apply the lemma again with b and r. (4) Continue until remainder = 0. The divisor at that step is the HCF.

Part (i): Find HCF(135, 225)

1

Apply Euclid's Lemma: a = 225, b = 135

225 = 1 × 135 + 90

Here a = 225, b = 135, q = 1, r = 90. Since r = 90 ≠ 0, we apply the lemma again with a = 135, b = 90.

2

Apply Euclid's Lemma: a = 135, b = 90

135 = 1 × 90 + 45

q = 1, r = 45 ≠ 0. Continue with a = 90, b = 45.

3

Apply Euclid's Lemma: a = 90, b = 45

90 = 2 × 45 + 0

r = 0. The process terminates. The HCF is the last non-zero remainder = 45.

∴ HCF (135, 225)
= 45

Part (ii): Find HCF(196, 38220)

1

Apply Euclid's Lemma: a = 38220, b = 196

38220 = 195 × 196 + 0

Since 196 × 195 = 38220 exactly, the remainder r = 0 immediately. Therefore HCF = divisor = 196.

This means 196 is a factor of 38220. The HCF equals the smaller number 196.

∴ HCF (196, 38220)
= 196

Part (iii): Find HCF(867, 255)

1

a = 867, b = 255

867 = 3 × 255 + 102

r = 102 ≠ 0. Continue with a = 255, b = 102.

2

a = 255, b = 102

255 = 2 × 102 + 51

r = 51 ≠ 0. Continue with a = 102, b = 51.

3

a = 102, b = 51

102 = 2 × 51 + 0

r = 0. Process terminates. HCF = last non-zero remainder = 51.

∴ HCF (867, 255)
= 51
Malayalam Explanation — മലയാളം വിശദീകരണം
Euclid's Division Algorithm (ഗണിത വിഭജന അൽഗോരിതം) ഉപയോഗിച്ച് HCF കണ്ടെത്തൽ:

HCF (ഹൈയർ കോമൺ ഫാക്ടർ) = രണ്ടു സംഖ്യകളുടെ ഏറ്റവും വലിയ പൊതു ഘടകം.

Formula: a = bq + r (0 ≤ r < b)

ചെയ്യേണ്ടത്:
1️⃣ വലിയ സംഖ്യ (a) ÷ ചെറിയ സംഖ്യ (b) → ഫോർമുല എഴുതുക
2️⃣ Remainder (r) = 0 ആകുമ്പോൾ നിർത്തുക
3️⃣ അവസാനത്തെ Non-zero remainder = HCF

Part (i) ഉദാഹരണം:
225 ÷ 135 → 225 = 1×135 + 90 (r=90)
135 ÷ 90 → 135 = 1×90 + 45 (r=45)
90 ÷ 45 → 90 = 2×45 + 0 (r=0 ✓)
∴ HCF = 45

📌 ഓർക്കുക: Remainder 0 ആകുമ്പോൾ, ആ ഘട്ടത്തിലെ divisor ആണ് HCF!

Frequently Asked Questions

What is Euclid's Division Algorithm for finding HCF?
Euclid's Division Algorithm states that for any two positive integers a and b, we can write: a = bq + r where 0 ≤ r < b. To find HCF, we repeatedly apply this: start with the larger number as a, apply the formula, then use the previous divisor as a and the remainder as b. Continue until the remainder becomes 0. The last non-zero remainder is the HCF.
What is the HCF of 135 and 225? Explain with steps.
HCF(135, 225) = 45. Steps: 225 = 1×135 + 90 → 135 = 1×90 + 45 → 90 = 2×45 + 0. Since remainder = 0, HCF = last divisor = 45.
Why is HCF(196, 38220) = 196?
Because 38220 ÷ 196 = 195 exactly (no remainder). This means 196 divides 38220 completely, so 196 is a factor of 38220. The HCF of a number and its multiple is the smaller number itself.
How do I remember Euclid's Division Lemma for CBSE board exams?
Think of simple division: Dividend = Divisor × Quotient + Remainder. That's exactly Euclid's formula (a = bq + r). The only extra condition is r must be less than b (the divisor). Our Xello tutors teach this with visual tricks and memory techniques in live 1-to-1 classes — so students never forget it in exams.

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