Consider the numbers \(4^{n}\), where n is a natural number. Check whether there is any value of n for which \(4^{n}\) ends with the digit zero.
If \(4^{n}\), for any n, were to end with the digit 0, it would have to be divisible by 5 — i.e. its prime factorisation would have to contain the prime 5.
The only prime in the factorisation of \(4^{n}\) is 2. By the uniqueness part of the Fundamental Theorem of Arithmetic, no other prime (in particular 5) can appear in this factorisation.
So there is no natural number n for which \(4^{n}\) ends with the digit zero.
Find the LCM and HCF of 6 and 20 by the prime factorisation method.
We have:
HCF(6, 20) = product of the smallest power of each common prime factor:
LCM(6, 20) = product of the greatest power of each prime factor involved:
Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.
The prime factorisation of 96 and 404 gives:
Therefore, HCF(96, 404) = \(2^{2}\) = 4.
Using HCF × LCM = product of the two numbers:
Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
We have:
Here \(2^{1}\) and \(3^{1}\) are the smallest powers of the common factors 2 and 3, respectively.
And \(2^{3}\), \(3^{2}\) and \(5^{1}\) are the greatest powers of the primes 2, 3 and 5 involved in the three numbers.
Remark: Notice that \(6 \times 72 \times 120 \ne \text{HCF}(6,72,120) \times \text{LCM}(6,72,120)\) — the product-of-two-numbers rule does not extend to three (or more) numbers.
Express each number as a product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
We divide each number successively by prime numbers, starting from the smallest:
(i) 140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7
(ii) 156 = 2 × 78 = 2 × 2 × 39 = 2 × 2 × 3 × 13
(iii) 3825 = 3 × 1275 = 3 × 3 × 425 = 3 × 3 × 5 × 85 = 3 × 3 × 5 × 5 × 17
(iv) 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 437 = 17 × 19 × 23
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 (ii) 510 and 92 (iii) 336 and 54
(i) 26 and 91: 26 = 2 × 13 and 91 = 7 × 13
HCF = product of the smallest power of each common prime factor = 13
LCM = product of the greatest power of each prime factor = 2 × 7 × 13 = 182
(ii) 510 and 92: 510 = 2 × 3 × 5 × 17 and 92 = 2² × 23
HCF = 2
LCM = 2² × 3 × 5 × 17 × 23 = 23460
(iii) 336 and 54: 336 = 2⁴ × 3 × 7 and 54 = 2 × 3³
HCF = 2 × 3 = 6
LCM = 2⁴ × 3³ × 7 = 3024
Find 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
The only prime common to all three is 3, so HCF = 3.
(ii) 17, 23 and 29: each number is prime, so they share no common factor other than 1.
HCF = 1
(iii) 8, 9 and 25: 8 = 2³, 9 = 3², 25 = 5²
The three numbers are pairwise co-prime, so HCF = 1.
Given that HCF (306, 657) = 9, find LCM (306, 657).
For any two positive integers a and b:
Therefore:
LCM (306, 657) = 22338
Check whether 6n can end with the digit 0 for any natural number n.
A number ends with the digit 0 only if it is divisible by 10, i.e. its prime factorisation must contain both 2 and 5.
The prime factorisation of 6n contains only the primes 2 and 3 — the prime 5 does not occur anywhere.
By the Fundamental Theorem of Arithmetic, the prime factorisation of every composite number is unique. So no other prime (in particular, 5) can ever appear in the factorisation of 6n.
Hence 6n can never end with the digit 0 for any natural number n.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
A composite number has at least one factor other than 1 and itself. We take out the common factor in each expression:
First number:
Since 13 and 78 are factors of 1014 (besides 1 and 1014), it is composite.
Second number:
Since 5 and 1009 are factors of 5045 (besides 1 and 5045), it is also composite.
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?
Both will be at the starting point together after a time that is a common multiple of their round times. The earliest such time is the LCM of 18 and 12.
Sonia and Ravi will meet again at the starting point after 36 minutes (by then Sonia completes 2 rounds and Ravi completes 3 rounds).
Prove that \(\sqrt{3}\) is irrational.
Proof by contradiction: Let us assume, to the contrary, that \(\sqrt{3}\) is rational. That is, we can find integers a and b (\(b \ne 0\)) such that:
Suppose a and b have a common factor other than 1. Then we divide by the common factor and assume that a and b are co-prime.
Squaring both sides and rearranging, we get \(3b^{2} = a^{2}\). Therefore \(a^{2}\) is divisible by 3, and by Theorem 1.2 (if a prime p divides \(a^2\), then p divides a), it follows that a is also divisible by 3.
So we can write \(a = 3c\) for some integer c. Substituting for a:
This means \(b^{2}\) is divisible by 3, and so b is also divisible by 3 (using Theorem 1.2 with p = 3).
Therefore a and b have at least 3 as a common factor — this contradicts the fact that a and b are co-prime.
This contradiction has arisen because of our incorrect assumption that \(\sqrt{3}\) is rational. So we conclude that \(\sqrt{3}\) is irrational. ∎
Show that \(5 - \sqrt{3}\) is irrational.
Proof by contradiction: Let us assume, to the contrary, that \(5 - \sqrt{3}\) is rational. That is, we can find co-prime a and b (\(b \ne 0\)) such that:
Rearranging this equation:
Since a and b are integers, \(\dfrac{5b-a}{b}\) is rational, and so \(\sqrt{3}\) would be rational. But this contradicts the fact that \(\sqrt{3}\) is irrational (Example 5).
This contradiction has arisen because of our incorrect assumption that \(5 - \sqrt{3}\) is rational. So we conclude that \(5 - \sqrt{3}\) is irrational. ∎
Show that \(3\sqrt{2}\) is irrational.
Proof by contradiction: Let us assume, to the contrary, that \(3\sqrt{2}\) is rational. That is, we can find co-prime a and b (\(b \ne 0\)) such that:
Rearranging:
Since 3, a and b are integers, \(\dfrac{a}{3b}\) is rational, and so \(\sqrt{2}\) would be rational. But this contradicts the fact that \(\sqrt{2}\) is irrational.
So we conclude that \(3\sqrt{2}\) is irrational. ∎
Prove that \(\sqrt{5}\) is irrational.
Proof by contradiction: Assume, to the contrary, that \(\sqrt{5}\) is rational. Then we can write:
where a and b are co-prime integers (no common factor other than 1) and b ≠ 0.
Rearranging: \(b\sqrt{5} = a\). Squaring both sides:
So 5 divides a². Since 5 is prime, by the theorem “if a prime p divides a², then p divides a” — 5 divides a. Write a = 5c for some integer c. Substituting:
So 5 divides b², and hence 5 divides b.
Therefore a and b have 5 as a common factor — this contradicts the fact that a and b are co-prime.
This contradiction has arisen because of our wrong assumption that \(\sqrt{5}\) is rational. Hence \(\sqrt{5}\) is irrational. ∎
Prove that \(3 + 2\sqrt{5}\) is irrational.
Proof by contradiction: Assume, to the contrary, that \(3 + 2\sqrt{5}\) is rational. Then:
where a and b are co-prime integers and b ≠ 0. Rearranging:
Since a and b are integers, \(\dfrac{a - 3b}{2b}\) is a rational number. This would mean \(\sqrt{5}\) is rational.
But this contradicts the established fact that \(\sqrt{5}\) is irrational (proved in Q8).
Therefore our assumption is false, and \(3 + 2\sqrt{5}\) is irrational. ∎
Prove that the following are irrationals:
(i) \(\dfrac{1}{\sqrt{2}}\) (ii) \(7\sqrt{5}\) (iii) \(6 + \sqrt{2}\)
In each part we assume the number is rational and derive a contradiction.
(i) Assume \(\dfrac{1}{\sqrt{2}} = \dfrac{a}{b}\) with a, b co-prime integers, b ≠ 0, a ≠ 0. Then:
Since a and b are integers, \(\dfrac{b}{a}\) is rational — so \(\sqrt{2}\) would be rational, contradicting the fact that \(\sqrt{2}\) is irrational. Hence \(\dfrac{1}{\sqrt{2}}\) is irrational.
(ii) Assume \(7\sqrt{5} = \dfrac{a}{b}\) with a, b co-prime integers, b ≠ 0. Then:
Since a and b are integers, \(\dfrac{a}{7b}\) is rational — so \(\sqrt{5}\) would be rational, a contradiction. Hence \(7\sqrt{5}\) is irrational.
(iii) Assume \(6 + \sqrt{2} = \dfrac{a}{b}\) with a, b co-prime integers, b ≠ 0. Then:
Since a and b are integers, \(\dfrac{a - 6b}{b}\) is rational — so \(\sqrt{2}\) would be rational, a contradiction. Hence \(6 + \sqrt{2}\) is irrational. ∎
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