Exercise 1.1 Solutions
1. Express each number as a product of its prime factors:
(i) 140
140 = 2 × 70
= 2 × 2 × 35
= 2 × 2 × 5 × 7
= 2² × 5 × 7
(ii) 156
156 = 2 × 78
= 2 × 2 × 39
= 2 × 2 × 3 × 13
= 2² × 3 × 13
(iii) 3825
3825 = 3 × 1275
= 3 × 3 × 425
= 3 × 3 × 5 × 85
= 3 × 3 × 5 × 5 × 17
= 3² × 5² × 17
(iv) 5005
5005 = 5 × 1001
= 5 × 7 × 143
= 5 × 7 × 11 × 13
(v) 7429
7429 = 17 × 437
= 17 × 19 × 23
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
26 = 2 × 13
91 = 7 × 13
HCF = 13
LCM = 2 × 7 × 13 = 182
Verification: LCM × HCF = 182 × 13 = 2366
Product of numbers = 26 × 91 = 2366
Hence, LCM × HCF = Product of the two numbers
(ii) 510 and 92
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23 = 2² × 23
HCF = 2
LCM = 2² × 3 × 5 × 17 × 23 = 23460
Verification: LCM × HCF = 23460 × 2 = 46920
Product of numbers = 510 × 92 = 46920
Hence, LCM × HCF = Product of the two numbers
(iii) 336 and 54
336 = 2 × 2 × 2 × 2 × 3 × 7 = 2⁴ × 3 × 7
54 = 2 × 3 × 3 × 3 = 2 × 3³
HCF = 2 × 3 = 6
LCM = 2⁴ × 3³ × 7 = 16 × 27 × 7 = 3024
Verification: LCM × HCF = 3024 × 6 = 18144
Product of numbers = 336 × 54 = 18144
Hence, LCM × HCF = Product of the two numbers
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
12 = 2 × 2 × 3 = 2² × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3
LCM = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420
(ii) 17, 23 and 29
17 = 17 (prime)
23 = 23 (prime)
29 = 29 (prime)
HCF = 1
LCM = 17 × 23 × 29 = 11339
(iii) 8, 9 and 25
8 = 2 × 2 × 2 = 2³
9 = 3 × 3 = 3²
25 = 5 × 5 = 5²
HCF = 1
LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
We know that: HCF × LCM = Product of the two numbers
So, 9 × LCM = 306 × 657
LCM = (306 × 657) / 9
LCM = 201042 / 9 = 22338
Therefore, LCM (306, 657) = 22338
5. Check whether 6ⁿ can end with the digit 0 for any natural number n.
If a number ends with digit 0, it must be divisible by 5 and 2.
6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ
The prime factorization of 6ⁿ contains only primes 2 and 3.
It does not contain the prime factor 5.
By the Fundamental Theorem of Arithmetic, the prime factorization is unique.
Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.
6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
First expression: 7 × 11 × 13 + 13
= 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78
= 13 × 2 × 3 × 13
= 2 × 3 × 13²
Since it has factors other than 1 and itself, it is a composite number.
Second expression: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
Since it has factors other than 1 and itself, it is a composite number.
7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
To find when they meet again at the starting point, we need to find the LCM of their times.
Sonia's time = 18 minutes
Ravi's time = 12 minutes
18 = 2 × 3²
12 = 2² × 3
LCM = 2² × 3² = 4 × 9 = 36
Therefore, they will meet again at the starting point after 36 minutes.
Exercise 1.2 Solutions
1. Prove that √5 is irrational.
Let us assume, to the contrary, that √5 is rational.
So, we can find coprime integers a and b (b ≠ 0) such that √5 = a/b.
Then, b√5 = a.
Squaring both sides: 5b² = a².
This means 5 divides a², and by Theorem 1.2, 5 divides a.
So, we can write a = 5c for some integer c.
Substituting: 5b² = (5c)² = 25c², so b² = 5c².
This means 5 divides b², and by Theorem 1.2, 5 divides b.
Therefore, a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that √5 is rational.
So, we conclude that √5 is irrational.
2. Prove that 3 + 2√5 is irrational.
Let us assume, to the contrary, that 3 + 2√5 is rational.
Then, we can write 3 + 2√5 = a/b, where a and b are coprime integers (b ≠ 0).
Rearranging: 2√5 = a/b - 3 = (a - 3b)/b
So, √5 = (a - 3b)/(2b)
Since a and b are integers, (a - 3b)/(2b) is rational.
Therefore, √5 is rational.
But this contradicts the fact that √5 is irrational (proved in Question 1).
This contradiction has arisen because of our incorrect assumption that 3 + 2√5 is rational.
So, we conclude that 3 + 2√5 is irrational.
3. Prove that the following are irrationals:
(i) 1/√2
Let us assume, to the contrary, that 1/√2 is rational.
Then, we can write 1/√2 = a/b, where a and b are coprime integers (b ≠ 0).
Rearranging: √2 = b/a
Since a and b are integers, b/a is rational.
Therefore, √2 is rational.
But this contradicts the fact that √2 is irrational (proved in Theorem 1.3).
This contradiction has arisen because of our incorrect assumption that 1/√2 is rational.
So, we conclude that 1/√2 is irrational.
(ii) 7√5
Let us assume, to the contrary, that 7√5 is rational.
Then, we can write 7√5 = a/b, where a and b are coprime integers (b ≠ 0).
Rearranging: √5 = a/(7b)
Since a and b are integers, a/(7b) is rational.
Therefore, √5 is rational.
But this contradicts the fact that √5 is irrational (proved in Question 1).
This contradiction has arisen because of our incorrect assumption that 7√5 is rational.
So, we conclude that 7√5 is irrational.
(iii) 6 + √2
Let us assume, to the contrary, that 6 + √2 is rational.
Then, we can write 6 + √2 = a/b, where a and b are coprime integers (b ≠ 0).
Rearranging: √2 = a/b - 6 = (a - 6b)/b
Since a and b are integers, (a - 6b)/b is rational.
Therefore, √2 is rational.
But this contradicts the fact that √2 is irrational (proved in Theorem 1.3).
This contradiction has arisen because of our incorrect assumption that 6 + √2 is rational.
So, we conclude that 6 + √2 is irrational.
Key Concepts
Fundamental Theorem of Arithmetic
Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
Theorem 1.2
Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.
Finding HCF and LCM using Prime Factorisation
- HCF = Product of the smallest power of each common prime factor in the numbers
- LCM = Product of the greatest power of each prime factor involved in the numbers
Important Properties
- For any two positive integers a and b: HCF(a, b) × LCM(a, b) = a × b
- This relationship does not hold for three or more numbers
- A number ends with digit 0 only if its prime factorisation contains both 2 and 5
Proof by Contradiction
This is a technique used to prove that certain numbers are irrational:
- Assume the number is rational
- Show that this leads to a contradiction
- Conclude that the original assumption must be false, so the number is irrational