140 ÷ 2 = 70. (Divide by the smallest prime, 2.)
⇒ 70 ÷ 2 = 35. (Divide by 2 again.)
⇒ 35 ÷ 5 = 7. (Divide by the next prime factor, 5.)
7 is prime, so we stop.
Therefore, 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7.
Page 5 — Exercise 1.1
Q1(i)
Express 140 as a product of its prime factors.
MEMORY TRICK
Start with the smallest prime: 2, 3, 5, 7… Stop when the remaining number is prime.
Q1(ii)
Express 156 as a product of its prime factors.
156 ÷ 2 = 78. (2 is the smallest prime factor.)
⇒ 78 ÷ 2 = 39. (Divide by 2 again.)
⇒ 39 ÷ 3 = 13. (3 divides 39.)
13 is prime, so we stop.
Therefore, 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13.
MEMORY TRICK
Keep dividing by the smallest possible prime. When the last number is prime, stop.
Q1(iii)
Express 3825 as a product of its prime factors.
3825 ÷ 3 = 1275. (Digit sum = 3+8+2+5 = 18, so 3 divides it.)
⇒ 1275 ÷ 3 = 425. (Digit sum = 15, so 3 divides it.)
⇒ 425 ÷ 5 = 85. (The number ends in 5.)
⇒ 85 ÷ 5 = 17. (The number ends in 5.)
17 is prime, so we stop.
Therefore, 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17.
MEMORY TRICK
For a number ending in 5, try 5. For a digit sum divisible by 3, try 3.
Q1(iv)
Express 5005 as a product of its prime factors.
5005 ÷ 5 = 1001. (The number ends in 5.)
⇒ 1001 ÷ 7 = 143. (7 × 143 = 1001.)
⇒ 143 ÷ 11 = 13. (11 × 13 = 143.)
13 is prime, so we stop.
Therefore, 5005 = 5 × 7 × 11 × 13.
MEMORY TRICK
Check easy divisibility first: ending 5 → 5; then test small primes such as 7 and 11.
Q1(v)
Express 7429 as a product of its prime factors.
7429 ÷ 17 = 437. (17 × 437 = 7429.)
⇒ 437 ÷ 19 = 23. (19 × 23 = 437.)
23 is prime, so we stop.
Therefore, 7429 = 17 × 19 × 23.
MEMORY TRICK
If 2, 3 and 5 do not work, test small primes systematically. Stop when the quotient is prime.
Q2(i)
Find the LCM and HCF of 26 and 91 and verify that LCM × HCF = product of the two numbers.
26 = 2 × 13 and 91 = 7 × 13. (Prime factorisation.)
⇒ HCF = 13. (Take the smallest power of each common prime.)
⇒ LCM = 2 × 7 × 13 = 182. (Take the greatest power of every prime involved.)
⇒ LCM × HCF = 182 × 13 = 2366.
⇒ 26 × 91 = 2366. (Direct multiplication.)
Therefore, LCM × HCF = 26 × 91 = 2366, so the result is verified.
MEMORY TRICK
HCF takes common primes with the smallest powers; LCM takes all primes with the greatest powers.
Q2(ii)
Find the LCM and HCF of 510 and 92 and verify that LCM × HCF = product of the two numbers.
510 = 2 × 3 × 5 × 17. (Prime factorisation.)
⇒ 92 = 2² × 23. (Prime factorisation.)
⇒ HCF = 2. (2 is the only common prime, with smallest power 2¹.)
⇒ LCM = 2² × 3 × 5 × 17 × 23 = 23460. (Greatest powers of all primes.)
⇒ LCM × HCF = 23460 × 2 = 46920.
⇒ 510 × 92 = 46920. (Direct multiplication.)
Therefore, LCM × HCF = 510 × 92 = 46920, so the result is verified.
MEMORY TRICK
First factor both numbers. Circle common primes for HCF; collect every prime for LCM.
Q2(iii)
Find the LCM and HCF of 336 and 54 and verify that LCM × HCF = product of the two numbers.
336 = 2⁴ × 3 × 7. (Prime factorisation.)
⇒ 54 = 2 × 3³. (Prime factorisation.)
⇒ HCF = 2 × 3 = 6. (Smallest powers of common primes.)
⇒ LCM = 2⁴ × 3³ × 7 = 3024. (Greatest powers of all primes.)
⇒ LCM × HCF = 3024 × 6 = 18144.
⇒ 336 × 54 = 18144. (Direct multiplication.)
Therefore, LCM × HCF = 336 × 54 = 18144, so the result is verified.
MEMORY TRICK
HCF = common + smallest powers; LCM = all + greatest powers.
Q3(i)
Find the LCM and HCF of 12, 15 and 21 by the prime factorisation method.
12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7. (Prime factorisation.)
⇒ HCF = 3. (3 is the only prime common to all three numbers.)
⇒ LCM = 2² × 3 × 5 × 7 = 420. (Take the greatest power of every prime.)
Therefore, HCF(12, 15, 21) = 3 and LCM(12, 15, 21) = 420.
MEMORY TRICK
For 3 or more numbers, HCF must be common to ALL; LCM must contain every prime needed.
Q3(ii)
Find the LCM and HCF of 17, 23 and 29 by the prime factorisation method.
17, 23 and 29 are all prime. (Each has only 1 and itself as factors.)
⇒ HCF = 1. (No prime factor is common to all three.)
⇒ LCM = 17 × 23 × 29 = 11339. (All three primes are needed.)
Therefore, HCF(17, 23, 29) = 1 and LCM(17, 23, 29) = 11339.
MEMORY TRICK
Different prime numbers have HCF 1, and their LCM is their product.
Q3(iii)
Find the LCM and HCF of 8, 9 and 25 by the prime factorisation method.
8 = 2³, 9 = 3², 25 = 5². (Prime factorisation.)
⇒ HCF = 1. (There is no prime factor common to all three.)
⇒ LCM = 2³ × 3² × 5² = 1800. (Take every greatest prime power.)
Therefore, HCF(8, 9, 25) = 1 and LCM(8, 9, 25) = 1800.
MEMORY TRICK
Different prime bases mean HCF = 1; multiply the prime powers for the LCM.
Q4
Given that HCF(306, 657) = 9, find LCM(306, 657).
HCF × LCM = 306 × 657. (HCF–LCM relation for two positive integers.)
⇒ 9 × LCM = 306 × 657. (Given HCF = 9.)
⇒ LCM = (306 × 657) ÷ 9.
⇒ LCM = 34 × 657. (306 ÷ 9 = 34.)
⇒ LCM = 22338.
Therefore, LCM(306, 657) = 22338.
MEMORY TRICK
When HCF is given, use: LCM = (first number × second number) ÷ HCF.
Q5
Check whether 6ⁿ can end with the digit 0 for any natural number n.
6ⁿ = (2 × 3)ⁿ. (Prime factorisation of 6.)
⇒ 6ⁿ = 2ⁿ × 3ⁿ.
⇒ The prime factors of 6ⁿ are only 2 and 3.
⇒ A number ending in 0 must be divisible by 10 = 2 × 5. (Divisibility by 10.)
⇒ 6ⁿ has no factor 5, so it cannot be divisible by 10.
Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.
MEMORY TRICK
To end in 0, you need factors 2 and 5. If there is no 5, the answer is NO.
Q6(i)
Explain why 7 × 11 × 13 + 13 is a composite number.
7 × 11 × 13 + 13 = 13(7 × 11 + 1). (Take 13 common.)
⇒ 13(77 + 1) = 13 × 78.
⇒ 13 × 78 is a product of two integers greater than 1.
Therefore, 7 × 11 × 13 + 13 is composite.
MEMORY TRICK
Look for a common factor. Here, 13 is common in both terms.
Q6(ii)
Explain why 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5(7 × 6 × 4 × 3 × 2 × 1 + 1). (Take 5 common.)
⇒ 5(1008 + 1) = 5 × 1009.
⇒ 5 × 1009 is a product of two integers greater than 1.
Therefore, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is composite.
MEMORY TRICK
If every term contains 5, take 5 common. A product of two numbers greater than 1 is composite.
Q7
Sonia takes 18 minutes to drive one round of a circular path, while Ravi takes 12 minutes for one round. They start together at the same point and move in the same direction. After how many minutes will they meet again at the starting point?
They will meet at the starting point after a time that is a common multiple of 18 and 12. (Both must complete whole rounds.)
⇒ 18 = 2 × 3² and 12 = 2² × 3. (Prime factorisation.)
⇒ LCM(18, 12) = 2² × 3². (Take greatest powers.)
⇒ LCM = 4 × 9 = 36 minutes.
Therefore, They will meet again at the starting point after 36 minutes.
MEMORY TRICK
When people repeat rounds and must meet together again, think LCM.
Page 9 — Exercise 1.2
Q1
Prove that √5 is irrational.
Assume that √5 is rational. (Proof by contradiction.)
⇒ √5 = a/b, where a and b are coprime integers and b ≠ 0. (Write a rational number in lowest form.)
⇒ 5 = a²/b².
⇒ 5b² = a².
⇒ 5 divides a². (5 divides the right side.)
⇒ 5 divides a. (Theorem 1.2: if prime p divides a², then p divides a.)
⇒ a = 5c, for some integer c.
⇒ 5b² = 25c².
⇒ b² = 5c².
⇒ 5 divides b².
⇒ 5 divides b. (Theorem 1.2.)
⇒ 5 is a common factor of a and b.
But this contradicts that a and b are coprime.
Therefore, our assumption is false; therefore √5 is irrational.
MEMORY TRICK
For √p, assume a/b, square it, show p divides a, then p divides b. That contradicts coprime a and b.
Q2
Prove that 3 + 2√5 is irrational.
Assume that 3 + 2√5 is rational. (Proof by contradiction.)
⇒ 2√5 = (3 + 2√5) − 3. (Rearrange.)
⇒ 2√5 is rational. (Difference of two rational numbers.)
⇒ √5 = (2√5)/2 is rational. (Quotient of rational numbers, denominator ≠ 0.)
But √5 is irrational by Q1.
This is a contradiction.
Therefore, 3 + 2√5 is irrational.
MEMORY TRICK
Move the rational part to the other side, then divide by the non-zero rational coefficient.
Q3(i)
Prove that 1/√2 is irrational.
Assume that 1/√2 is rational. (Proof by contradiction.)
⇒ √2 = 1 ÷ (1/√2). (Rearrange.)
⇒ √2 is rational. (The reciprocal of a non-zero rational number is rational.)
But √2 is irrational. (Known result proved in the chapter.)
This is a contradiction.
Therefore, 1/√2 is irrational.
MEMORY TRICK
If 1/√p were rational, its reciprocal √p would also be rational. That gives the contradiction.
Q3(ii)
Prove that 7√5 is irrational.
Assume that 7√5 is rational. (Proof by contradiction.)
⇒ √5 = (7√5)/7. (Divide by 7.)
⇒ √5 is rational. (Quotient of rational numbers.)
But √5 is irrational.
This is a contradiction.
Therefore, 7√5 is irrational.
MEMORY TRICK
A non-zero rational number × an irrational number stays irrational. To prove it, divide by the rational number.
Q3(iii)
Prove that 6 + √2 is irrational.
Assume that 6 + √2 is rational. (Proof by contradiction.)
⇒ √2 = (6 + √2) − 6. (Rearrange.)
⇒ √2 is rational. (Difference of rational numbers.)
But √2 is irrational.
This is a contradiction.
Therefore, 6 + √2 is irrational.
MEMORY TRICK
Rational + irrational = irrational. For the proof, subtract the rational number and expose the irrational part.