Chapter 1: Real Numbers

Key Formulas:
• Euclid's Division Lemma: a = bq + r, 0 ≤ r < b
• HCF(a,b) × LCM(a,b) = a × b
• Fundamental Theorem of Arithmetic: Every composite number is a unique product of primes

Exercise 1.1

Q1. Use Euclid's division algorithm to find the HCF of:
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

Solution

(i) HCF of 135 and 225

225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
∴ HCF(135, 225) = 45

(ii) HCF of 196 and 38220

38220 = 196 × 195 + 0
∴ HCF(196, 38220) = 196

(iii) HCF of 867 and 255

867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
∴ HCF(867, 255) = 51

Q2. Show that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5.

Solution

Let a be any positive integer and b = 6. By Euclid's algorithm, a = 6q + r where 0 ≤ r < 6.

So a = 6q, 6q+1, 6q+2, 6q+3, 6q+4, or 6q+5. Now 6q, 6q+2, 6q+4 are even (divisible by 2). So odd numbers must be of the form 6q+1, 6q+3, or 6q+5.

Q3. An army contingent of 1000 members has to march behind a band of 56 members. What is the maximum number of columns?

Solution

HCF(1000, 56):
1000 = 56 × 17 + 48
56 = 48 × 1 + 8
48 = 8 × 6 + 0
∴ Maximum columns = 8

Q4. Show that the square of any positive integer is either of the form 3m or 3m + 1.

Solution

Let x = 3q, 3q+1, or 3q+2.

If x = 3q: x² = 9q² = 3(3q²) = 3m
If x = 3q+1: x² = 9q²+6q+1 = 3(3q²+2q)+1 = 3m+1
If x = 3q+2: x² = 9q²+12q+4 = 3(3q²+4q+1)+1 = 3m+1

In all cases, x² is of the form 3m or 3m+1.

Q5. Show that the cube of any positive integer is of the form 9m, 9m+1, or 9m+8.

Solution

Let x = 9q + r where r ∈ {0,1,2,...,8}. Checking key cases:

(9q)³ = 729q³ = 9(×) = 9m
(9q+1)³ = 729q³+243q²+27q+1 = 9m+1
(9q+2)³ = 9(×)+8 = 9m+8
Similarly for r = 3..8, all results fall in {9m, 9m+1, 9m+8}

Exercise 1.2

Q1. Express as product of prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Solution

(i) 140 = 2² × 5 × 7
(ii) 156 = 2² × 3 × 13
(iii) 3825 = 3² × 5² × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23

Q2. Find LCM and HCF and verify LCM × HCF = product:
(i) 12, 15 (ii) 21, 28 (iii) 8, 12 (iv) 36, 48

Solution

(i) 12=2²×3, 15=3×5 → HCF=3, LCM=60. Check: 3×60=180=12×15 ✓
(ii) 21=3×7, 28=2²×7 → HCF=7, LCM=84. Check: 7×84=588 ✓
(iii) 8=2³, 12=2²×3 → HCF=4, LCM=24. Check: 4×24=96 ✓
(iv) 36=2²×3², 48=2⁴×3 → HCF=12, LCM=144. Check: 12×144=1728 ✓

Q3. Find LCM and HCF by prime factorisation:
(i) 12, 15, 21 (ii) 17, 23, 29 (iii) 8, 9, 25

Solution

(i) HCF=3, LCM=2²×3×5×7=420
(ii) All prime → HCF=1, LCM=17×23×29=11339
(iii) HCF=1, LCM=2³×3²×5²=1800

Q4. Given HCF(306, 657) = 9, find LCM(306, 657).

Solution

LCM = (306 × 657) / 9 = 306 × 73 = 22338

Q5. Check whether 6³ can end with digit 0.

Solution

6³ = 2³ × 3³. Since 5 is not a prime factor, 6³ cannot be divisible by 5 and thus cannot end with 0.

Q6. Explain why 7×11×13+13 and 7!+5 are composite numbers.

Solution

7×11×13+13 = 13(77+1) = 13×78 = composite
7!+5 = 5(7!/5+1) = 5×1009 = composite

Exercise 1.3

Q1. Prove that √5 is irrational.

Solution

Assume √5 = a/b (co-prime integers). Then a² = 5b², so 5|a. Let a=5c. Then 25c²=5b², so b²=5c² and 5|b. This contradicts that a and b are co-prime. ∴ √5 is irrational.

Q2. Prove that 3 + 2√5 is irrational.

Solution

Assume 3+2√5 = a/b. Then √5 = (a-3b)/2b = rational. But √5 is irrational. Contradiction. ∴ 3+2√5 is irrational.

Q3. Prove these are irrational: (i) 1/√2 (ii) 7√5 (iii) 6 + √2

Solution

(i) 1/√2 = a/b → √2 = b/a = rational. Contradiction. √2 is irrational.
(ii) 7√5 = a/b → √5 = a/7b = rational. Contradiction. √5 is irrational.
(iii) 6+√2 = a/b → √2 = (a-6b)/b = rational. Contradiction. √2 is irrational.

Exercise 1.4

Q1. Determine if 987/10500 is terminating or non-terminating repeating.

Solution

987/10500 = 333/3500
3500 = 2² × 5³ × 7
Since 7 is a prime factor other than 2 and 5, it is non-terminating repeating.

Q3. Decide rational or irrational: (i) 43.123456789 (ii) 0.120120012000120000...

Solution

(i) Terminating decimal → rational, denominator has factors of 2 and 5 only.
(ii) Neither terminating nor repeating → irrational.

Chapter 2: Polynomials →

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