Key Concepts & Formulae at a Glance
- Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order of the factors.
- HCF by prime factorisation: Product of the smallest power of each common prime factor.
- LCM by prime factorisation: Product of the greatest power of each prime factor involved.
- Theorem 1.2: If p is a prime and p divides a², then p divides a.
- Proving irrationality: Standard method is proof by contradiction — assume the number is rational, then reach a contradiction using Theorem 1.2 or the known irrationality of a related number.
I. HCF(a, b) × LCM(a, b) = a × b (for any two positive integers a, b — does NOT extend to three or more numbers)
Exercise 1.1
7 Questions — Fundamental Theorem of Arithmetic, HCFLCM by prime factorisation.
1Express each number as a product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
(i) 140
140 = 2 × 70
140 = 2 × 2 × 35
140 = 2 × 2 × 5 × 7
140 = 2 × 2 × 35
140 = 2 × 2 × 5 × 7
140 = 2² × 5 × 7
(ii) 156
156 = 2 × 78
156 = 2 × 2 × 39
156 = 2 × 2 × 3 × 13
156 = 2 × 2 × 39
156 = 2 × 2 × 3 × 13
156 = 2² × 3 × 13
(iii) 3825
3825 = 3 × 1275
3825 = 3 × 3 × 425
3825 = 3 × 3 × 5 × 85
3825 = 3 × 3 × 5 × 5 × 17
3825 = 3 × 3 × 425
3825 = 3 × 3 × 5 × 85
3825 = 3 × 3 × 5 × 5 × 17
3825 = 3² × 5² × 17
(iv) 5005
5005 = 5 × 1001
5005 = 5 × 7 × 143
5005 = 5 × 7 × 11 × 13
5005 = 5 × 7 × 143
5005 = 5 × 7 × 11 × 13
5005 = 5 × 7 × 11 × 13
(v) 7429
7429 = 17 × 437
7429 = 17 × 19 × 23
7429 = 17 × 19 × 23
7429 = 17 × 19 × 23
2Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers:
(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
(i) 26 and 91
26 = 2 × 13
91 = 7 × 13
HCF(26, 91) = 13 (smallest power of the common prime factor)
LCM(26, 91) = 2 × 7 × 13 = 182 (greatest power of each prime factor involved)
Verification: HCF × LCM = 13 × 182 = 2366
26 × 91 = 2366 (matches)
91 = 7 × 13
HCF(26, 91) = 13 (smallest power of the common prime factor)
LCM(26, 91) = 2 × 7 × 13 = 182 (greatest power of each prime factor involved)
Verification: HCF × LCM = 13 × 182 = 2366
26 × 91 = 2366 (matches)
HCF = 13, LCM = 182 — verified: 13 × 182 = 26 × 91 = 2366
(ii) 510 and 92
510 = 2 × 3 × 5 × 17
92 = 2² × 23
HCF(510, 92) = 2 (smallest power of the only common prime, 2)
LCM(510, 92) = 2² × 3 × 5 × 17 × 23 = 23460
Verification: HCF × LCM = 2 × 23460 = 46920
510 × 92 = 46920 (matches)
92 = 2² × 23
HCF(510, 92) = 2 (smallest power of the only common prime, 2)
LCM(510, 92) = 2² × 3 × 5 × 17 × 23 = 23460
Verification: HCF × LCM = 2 × 23460 = 46920
510 × 92 = 46920 (matches)
HCF = 2, LCM = 23460 — verified: 2 × 23460 = 510 × 92 = 46920
(iii) 336 and 54
336 = 2⁴ × 3 × 7
54 = 2 × 3³
HCF(336, 54) = 2 × 3 = 6 (smallest powers of common primes 2, 3)
LCM(336, 54) = 2⁴ × 3³ × 7 = 3024
Verification: HCF × LCM = 6 × 3024 = 18144
336 × 54 = 18144 (matches)
54 = 2 × 3³
HCF(336, 54) = 2 × 3 = 6 (smallest powers of common primes 2, 3)
LCM(336, 54) = 2⁴ × 3³ × 7 = 3024
Verification: HCF × LCM = 6 × 3024 = 18144
336 × 54 = 18144 (matches)
HCF = 6, LCM = 3024 — verified: 6 × 3024 = 336 × 54 = 18144
3Find the LCM and HCF of the following integers by applying the prime factorisation method:
(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
(i) 12, 15 and 21
12 = 2² × 3
15 = 3 × 5
21 = 3 × 7
HCF(12, 15, 21) = 3 (only common prime, smallest power)
LCM(12, 15, 21) = 2² × 3 × 5 × 7 = 420 (greatest power of each prime involved)
15 = 3 × 5
21 = 3 × 7
HCF(12, 15, 21) = 3 (only common prime, smallest power)
LCM(12, 15, 21) = 2² × 3 × 5 × 7 = 420 (greatest power of each prime involved)
HCF = 3, LCM = 420
(ii) 17, 23 and 29
17, 23, 29 are all prime numbers, with no common factor other than 1.
HCF(17, 23, 29) = 1
LCM(17, 23, 29) = 17 × 23 × 29 = 11339
HCF(17, 23, 29) = 1
LCM(17, 23, 29) = 17 × 23 × 29 = 11339
HCF = 1, LCM = 11339
(iii) 8, 9 and 25
8 = 2³
9 = 3²
25 = 5²
No prime is common to all three, so HCF(8, 9, 25) = 1
LCM(8, 9, 25) = 2³ × 3² × 5² = 1800
9 = 3²
25 = 5²
No prime is common to all three, so HCF(8, 9, 25) = 1
LCM(8, 9, 25) = 2³ × 3² × 5² = 1800
HCF = 1, LCM = 1800
4Given that HCF (306, 657) = 9, find LCM (306, 657).
For any two positive integers a and b: HCF(a, b) × LCM(a, b) = a × b (HCF×LCM identity, valid for two numbers)
So, LCM(306, 657) = (306 × 657) ÷ HCF(306, 657)
LCM(306, 657) = (306 × 657) ÷ 9
LCM(306, 657) = 201042 ÷ 9
LCM(306, 657) = 22338
So, LCM(306, 657) = (306 × 657) ÷ HCF(306, 657)
LCM(306, 657) = (306 × 657) ÷ 9
LCM(306, 657) = 201042 ÷ 9
LCM(306, 657) = 22338
LCM(306, 657) = 22338
5Check whether 6ⁿ can end with the digit 0 for any natural number n.
If 6ⁿ were to end with the digit 0, it would be divisible by 10 = 2 × 5, so its prime factorisation would have to include the prime 5.
6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ (prime factorisation of 6ⁿ)
The only primes in the factorisation of 6ⁿ are 2 and 3 — there is no 5.
By the uniqueness part of the Fundamental Theorem of Arithmetic, no other prime factors can appear (Fundamental Theorem of Arithmetic).
6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ (prime factorisation of 6ⁿ)
The only primes in the factorisation of 6ⁿ are 2 and 3 — there is no 5.
By the uniqueness part of the Fundamental Theorem of Arithmetic, no other prime factors can appear (Fundamental Theorem of Arithmetic).
6ⁿ can never end with the digit 0, for any natural number n.
6Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
7 × 11 × 13 + 13
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) (taking 13 common)
= 13 × (77 + 1)
= 13 × 78
= 13 × (77 + 1)
= 13 × 78
= 13 × 78, a product of two factors both greater than 1, so it is composite.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040, which already has 5 as a factor
5040 + 5 = 5 × (1008 + 1) (taking 5 common)
= 5 × 1009
5040 + 5 = 5 × (1008 + 1) (taking 5 common)
= 5 × 1009
= 5 × 1009, a product of two factors both greater than 1, so it is composite.
7There 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?
They will next be together at the starting point after a time that is a common multiple of 18 and 12 — the earliest such time is their LCM.
18 = 2 × 3²
12 = 2² × 3
LCM(18, 12) = 2² × 3² = 36 (greatest power of each prime involved)
18 = 2 × 3²
12 = 2² × 3
LCM(18, 12) = 2² × 3² = 36 (greatest power of each prime involved)
They will meet again at the starting point after 36 minutes.
Exercise 1.2
3 Questions — Proving irrationality by contradiction.
1Prove 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 = ab (assumption for contradiction)
Rearranging: b√5 = a
Squaring both sides: 5b² = a² ... (1)
So, a² is divisible by 5, and by Theorem 1.2, a is also divisible by 5 (Theorem 1.2, p = 5)
So we can write a = 5c for some integer c
Substituting in (1): 5b² = (5c)² = 25c²
So, b² = 5c²
This means b² is divisible by 5, and so b is also divisible by 5 (Theorem 1.2, p = 5)
So a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime.
So we can find coprime integers a and b (b ≠ 0) such that √5 = ab (assumption for contradiction)
Rearranging: b√5 = a
Squaring both sides: 5b² = a² ... (1)
So, a² is divisible by 5, and by Theorem 1.2, a is also divisible by 5 (Theorem 1.2, p = 5)
So we can write a = 5c for some integer c
Substituting in (1): 5b² = (5c)² = 25c²
So, b² = 5c²
This means b² is divisible by 5, and so b is also divisible by 5 (Theorem 1.2, p = 5)
So a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction arose from assuming √5 is rational. So √5 is irrational.
2Prove that 3 + 2√5 is irrational.
Let us assume, to the contrary, that 3 + 2√5 is rational.
So we can find coprime integers a and b (b ≠ 0) such that 3 + 2√5 = ab
Rearranging: 2√5 = ab − 3 = a − 3bb
√5 = a − 3b2b
Since a and b are integers, a − 3b2b is rational, so √5 would be rational (quotient of integers)
But this contradicts the fact that √5 is irrational (proved in Q1).
So we can find coprime integers a and b (b ≠ 0) such that 3 + 2√5 = ab
Rearranging: 2√5 = ab − 3 = a − 3bb
√5 = a − 3b2b
Since a and b are integers, a − 3b2b is rational, so √5 would be rational (quotient of integers)
But this contradicts the fact that √5 is irrational (proved in Q1).
This contradiction arose from assuming 3 + 2√5 is rational. So 3 + 2√5 is irrational.
3Prove that the following are irrationals:
(i) 1√2 (ii) 7√5 (iii) 6 + √2
(i) 1√2
Let us assume, to the contrary, that 1√2 is rational.
So we can find coprime integers a and b (b ≠ 0) such that 1√2 = ab
Rearranging: √2 = ba
Since a and b are integers, ba is rational, so √2 would be rational.
But this contradicts the known fact that √2 is irrational (Theorem 1.3).
So we can find coprime integers a and b (b ≠ 0) such that 1√2 = ab
Rearranging: √2 = ba
Since a and b are integers, ba is rational, so √2 would be rational.
But this contradicts the known fact that √2 is irrational (Theorem 1.3).
So 1√2 is irrational.
(ii) 7√5
Let us assume, to the contrary, that 7√5 is rational.
So we can find coprime integers a and b (b ≠ 0) such that 7√5 = ab
Rearranging: √5 = a7b
Since 7, a and b are integers, a7b is rational, so √5 would be rational.
But this contradicts the fact that √5 is irrational (proved in Q1).
So we can find coprime integers a and b (b ≠ 0) such that 7√5 = ab
Rearranging: √5 = a7b
Since 7, a and b are integers, a7b is rational, so √5 would be rational.
But this contradicts the fact that √5 is irrational (proved in Q1).
So 7√5 is irrational.
(iii) 6 + √2
Let us assume, to the contrary, that 6 + √2 is rational.
So we can find coprime integers a and b (b ≠ 0) such that 6 + √2 = ab
Rearranging: √2 = ab − 6 = a − 6bb
Since a and b are integers, a − 6bb is rational, so √2 would be rational.
But this contradicts the known fact that √2 is irrational (Theorem 1.3).
So we can find coprime integers a and b (b ≠ 0) such that 6 + √2 = ab
Rearranging: √2 = ab − 6 = a − 6bb
Since a and b are integers, a − 6bb is rational, so √2 would be rational.
But this contradicts the known fact that √2 is irrational (Theorem 1.3).
So 6 + √2 is irrational.
Frequently Asked Questions
It states that 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.
That identity only holds for exactly two positive integers. For three numbers p, q, r, HCF(p,q,r) × LCM(p,q,r) is generally not equal to p × q × r — so always find HCF and LCM of three numbers directly from prime factorisation rather than via that shortcut.
Proof by contradiction: assume the number is rational and can be written as ab in lowest terms, then use Theorem 1.2 (if a prime p divides a², then p divides a) to show a and b share a common factor — contradicting the lowest-terms assumption.
Assume the expression is rational, then rearrange the equation to isolate the irrational part (e.g. √3) on one side. Since the rest of the equation is built only from rational numbers, this would force √3 to be rational too — contradicting its known irrationality.
Yes — a printable Class 10 Maths Formula Card covering the full syllabus, including Real Numbers, is available. See the Formula Card page for the current price.
Continue to the Next Chapter
Chapter 2 (Polynomials) builds directly on the number properties covered here.