Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
It uses one big idea, the Fundamental Theorem of Arithmetic, for two jobs: finding the HCF and LCM of numbers, and proving that numbers such as √2, √3 and √5 are irrational.
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.
So 2 × 3 × 5 × 7 and 7 × 5 × 3 × 2 count as the same factorisation. We usually write the primes in ascending order and group equal primes as powers.
Split the number again and again until every branch ends in a prime.
Because the factorisation is unique, a number contains only the primes in its factorisation. This settles questions like:
For any two positive integers a and b:
So if the HCF is known, LCM = a × bHCF(a, b). Example: HCF(96, 404) = 4, so LCM = 96 × 4044 = 9696.
The product rule does not hold for three numbers. For 6, 72 and 120: HCF = 6 and LCM = 360, but 6 × 360 = 2160 while 6 × 72 × 120 = 51840.
For three numbers p, q, r the NCERT note gives: LCM(p, q, r) = p · q · r · HCF(p, q, r)HCF(p, q) · HCF(q, r) · HCF(p, r).
Theorem 1.2: if a prime p divides a², then p divides a (a a positive integer).
Reason: the primes in a² are exactly the primes in a, each used twice. So if p is a prime factor of a², it is already a prime factor of a.
It needs p to be prime: 4 divides 6² = 36, but 4 does not divide 6.
We use proof by contradiction: assume the opposite and reach something impossible.
The proof works for √3, √5, √7 and, in general, √p for any prime p: replace 2 by p throughout.
To show 5 − √3 is irrational, assume 5 − √3 = ab (a, b coprime, b ≠ 0). Then
The right side is rational, so √3 would be rational. That is a contradiction, so 5 − √3 is irrational.
For 3√2: assume 3√2 = ab, then √2 = a3b, which is rational. Contradiction again.
| Result | Statement |
|---|---|
| Fundamental Theorem of Arithmetic | Every composite number is a product of primes in one way only, apart from order |
| Theorem 1.2 | If prime p divides a², then p divides a |
| HCF | Smallest power of each common prime |
| LCM | Greatest power of every prime involved |
| Two numbers | HCF × LCM = product of the numbers |
| Irrational | √p for prime p; rational ± irrational; non-zero rational × irrational |
Try each problem first, then open the solution to check your working.
Express 2520 as a product of its prime factors.
Given: n = 2520
Use: Divide by the smallest prime repeatedly
2520 ÷ 2 = 1260, ÷ 2 = 630, ÷ 2 = 315
315 ÷ 3 = 105, ÷ 3 = 35
35 ÷ 5 = 7, and 7 is prime
Answer: 2520 = 2³ × 3² × 5 × 7
Express 9009 as a product of its prime factors.
Given: n = 9009
Use: Test 3, 7, 11, 13, …
Digit sum 9 + 0 + 0 + 9 = 18, so 9 divides it: 9009 ÷ 9 = 1001
1001 = 7 × 143 = 7 × 11 × 13
Answer: 9009 = 3² × 7 × 11 × 13
A number N = 2³ × 3² × 5. Without multiplying it out, decide which of 12, 25, 45 and 90 divide N.
Given: N = 2³ × 3² × 5
Use: d divides N only if every prime power in d appears in N
12 = 2² × 3: yes
25 = 5²: no, N has only one 5
45 = 3² × 5: yes
90 = 2 × 3² × 5: yes
Answer: 12, 45 and 90 divide N; 25 does not
Check whether (a) 12ⁿ and (b) 15ⁿ can end with the digit 0 for any natural number n.
Use: A number ending in 0 must have both 2 and 5 as prime factors
(a) 12ⁿ = 2²ⁿ × 3ⁿ has no factor 5.
(b) 15ⁿ = 3ⁿ × 5ⁿ has no factor 2.
By uniqueness of prime factorisation no other primes can appear.
Answer: Neither can end with 0 for any n
Show that 3 × 5 × 7 + 7 is composite.
Use: Take out the common factor
3 × 5 × 7 + 7 = 7 × (15 + 1) = 7 × 16 = 112
It has the factor 7 besides 1 and itself.
Answer: It is composite (112 = 7 × 16)
Is 2 × 3 × 5 × 7 × 11 + 1 composite?
Given: N = 2310 + 1 = 2311
Use: Test primes up to √2311 ≈ 48
Here there is no common factor to take out: 2, 3, 5, 7, 11 each leave remainder 1.
Testing 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47 also gives a remainder each time.
Answer: No; 2311 is prime. Only a sum with a common factor, like 7 × 11 × 13 + 13, is sure to be composite
Find the HCF and LCM of 84 and 120 and verify that HCF × LCM = product of the numbers.
Given: 84 = 2² × 3 × 7; 120 = 2³ × 3 × 5
Use: HCF: smallest common powers; LCM: greatest powers
HCF = 2² × 3 = 12
LCM = 2³ × 3 × 5 × 7 = 840
HCF × LCM = 12 × 840 = 10080 and 84 × 120 = 10080
Answer: HCF = 12, LCM = 840; verified
Find the HCF and LCM of 24, 36 and 60. Is HCF × LCM equal to the product of the three numbers?
Given: 24 = 2³ × 3; 36 = 2² × 3²; 60 = 2² × 3 × 5
HCF = 2² × 3 = 12
LCM = 2³ × 3² × 5 = 360
HCF × LCM = 4320, but 24 × 36 × 60 = 51840
Answer: HCF = 12, LCM = 360; the product rule fails for three numbers
The HCF of two numbers is 18 and their product is 12960. Find their LCM.
Given: HCF = 18; a × b = 12960
Use: LCM = a × b ÷ HCF
LCM = 12960 ÷ 18 = 720
Answer: 720
The HCF and LCM of two numbers are 12 and 2520. One number is 180. Find the other.
Given: HCF = 12; LCM = 2520; a = 180
Use: b = HCF × LCM ÷ a
b = 12 × 2520 ÷ 180 = 30240 ÷ 180 = 168
Check: 180 = 2² × 3² × 5 and 168 = 2³ × 3 × 7 give HCF 2² × 3 = 12 and LCM 2³ × 3² × 5 × 7 = 2520
Answer: 168
Can two numbers have HCF 16 and LCM 380?
Use: The HCF always divides the LCM
380 ÷ 16 = 23.75, not a whole number
Answer: No, because 16 does not divide 380
Bus A leaves a depot every 15 minutes and bus B every 25 minutes. Both leave together at 6:00 am. When do they next leave together?
Given: intervals 15 min and 25 min
Use: Next meeting = LCM
15 = 3 × 5; 25 = 5²
LCM = 3 × 5² = 75 minutes
Answer: After 75 minutes, at 7:15 am
Three traffic lights change every 48 s, 72 s and 108 s. They change together at 7:00:00 am. When will they next change together?
Given: 48 = 2⁴ × 3; 72 = 2³ × 3²; 108 = 2² × 3³
Use: LCM
LCM = 2⁴ × 3³ = 432 s
432 s = 7 min 12 s
Answer: At 7:07:12 am
A floor is 4.5 m long and 3.6 m wide. Find the side of the largest square tile that covers it exactly, and the number of tiles.
Given: 450 cm and 360 cm
Use: Largest equal side = HCF
450 = 2 × 3² × 5²; 360 = 2³ × 3² × 5
HCF = 2 × 3² × 5 = 90 cm
Tiles = (450 ÷ 90) × (360 ÷ 90) = 5 × 4 = 20
Answer: 90 cm tiles; 20 tiles
A shop has 144 cartons of mango juice and 90 cartons of orange juice. They are stacked so that each stack has one kind only and all stacks have the same number of cartons. Find the greatest number per stack and the number of stacks.
Given: 144 = 2⁴ × 3²; 90 = 2 × 3² × 5
Use: Greatest equal stack = HCF
HCF = 2 × 3² = 18
Stacks = 144 ÷ 18 + 90 ÷ 18 = 8 + 5 = 13
Answer: 18 cartons per stack; 13 stacks
Find the smallest number which leaves remainder 5 when divided by each of 12, 15 and 20.
Given: 12 = 2² × 3; 15 = 3 × 5; 20 = 2² × 5
Use: Required number = LCM + remainder
LCM = 2² × 3 × 5 = 60
60 + 5 = 65
Answer: 65
Find the largest number that divides 245 and 1029 leaving remainder 5 in each case.
Given: 245 − 5 = 240; 1029 − 5 = 1024
Use: Required number = HCF of (number − remainder)
240 = 2⁴ × 3 × 5; 1024 = 2¹⁰
HCF = 2⁴ = 16 (and 16 > 5, so remainder 5 is possible)
Answer: 16
Prove that √7 is irrational.
Use: Theorem 1.2 with p = 7
Assume √7 = ab, a and b coprime, b ≠ 0. Then 7b² = a².
So 7 divides a², hence 7 divides a. Write a = 7c.
Then 7b² = 49c², so b² = 7c². So 7 divides b², hence 7 divides b.
Now 7 divides both a and b, which contradicts that they are coprime.
Answer: √7 is irrational
Given that √3 is irrational, prove that 2 + 5√3 is irrational.
Use: Isolate the root
Assume 2 + 5√3 = ab (a, b integers, b ≠ 0).
The right side is rational, so √3 would be rational. Contradiction.
Answer: 2 + 5√3 is irrational
Show that 4√3 is irrational.
Use: Rationalise, then use non-zero rational × irrational
43 is a non-zero rational and √3 is irrational, so the product is irrational.
Or directly: if 4√3 = ab, then √3 = 4ba, which is rational. Contradiction.
Answer: It is irrational
Find √2 × √8 and (√5 + 2)(√5 − 2). What does this show?
Use: (x + y)(x − y) = x² − y²
√2 × √8 = √16 = 4
(√5 + 2)(√5 − 2) = 5 − 4 = 1
Answer: Both are rational, so the product of two irrational numbers need not be irrational
Pick a level and answer 10 questions with instant feedback. Each level has a bank of 20 questions, and New 10 questions draws a fresh set.
Read the Assertion (A) and the Reason (R), then pick one of the four standard options. You get 5 questions per set from a bank of 20 at each level, and New 5 questions draws a fresh set.
Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
(i) 140 = 2 × 70 = 2 × 2 × 35 = 2² × 5 × 7
(ii) 156 = 2 × 78 = 2 × 2 × 39 = 2² × 3 × 13
(iii) 3825 = 5 × 765 = 5 × 5 × 153 = 5² × 9 × 17 = 3² × 5² × 17
(iv) 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 437 = 17 × 19 × 23
Answer: 2² × 5 × 7; 2² × 3 × 13; 3² × 5² × 17; 5 × 7 × 11 × 13; 17 × 19 × 23
(i) 26 = 2 × 13, 91 = 7 × 13. HCF = 13; LCM = 2 × 7 × 13 = 182.
Check: 182 × 13 = 2366 and 26 × 91 = 2366. ✓
(ii) 510 = 2 × 3 × 5 × 17, 92 = 2² × 23. HCF = 2; LCM = 2² × 3 × 5 × 17 × 23 = 23460.
Check: 23460 × 2 = 46920 and 510 × 92 = 46920. ✓
(iii) 336 = 2⁴ × 3 × 7, 54 = 2 × 3³. HCF = 2 × 3 = 6; LCM = 2⁴ × 3³ × 7 = 3024.
Check: 3024 × 6 = 18144 and 336 × 54 = 18144. ✓
Answer: (i) HCF 13, LCM 182 (ii) HCF 2, LCM 23460 (iii) HCF 6, LCM 3024
(i) 12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7. HCF = 3; LCM = 2² × 3 × 5 × 7 = 420.
(ii) 17, 23 and 29 are all prime. HCF = 1; LCM = 17 × 23 × 29 = 11339.
(iii) 8 = 2³, 9 = 3², 25 = 5². No common prime, so HCF = 1; LCM = 2³ × 3² × 5² = 1800.
Answer: (i) HCF 3, LCM 420 (ii) HCF 1, LCM 11339 (iii) HCF 1, LCM 1800
For two numbers, HCF × LCM = product of the numbers.
LCM = 306 × 6579 = 34 × 657 = 22338
Answer: LCM (306, 657) = 22338
If 6ⁿ ended with 0, it would be divisible by 10 = 2 × 5, so its prime factorisation would contain 5.
But 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. By the uniqueness in the Fundamental Theorem of Arithmetic, its only primes are 2 and 3.
Answer: No, 6ⁿ cannot end with 0 for any natural number n
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × 78 = 1014. It has factors other than 1 and itself (for example 13 and 78).
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × 1009 = 5045. It has the factor 5 besides 1 and itself.
Answer: Both can be written as a product of two numbers greater than 1, so both are composite
Sonia is at the start after 18, 36, 54, … minutes; Ravi after 12, 24, 36, … minutes.
They are together at the start first at the LCM of 18 and 12.
18 = 2 × 3², 12 = 2² × 3, so LCM = 2² × 3² = 36.
Answer: After 36 minutes
Assume, to the contrary, that √5 is rational. Then √5 = ab, where a and b are coprime integers and b ≠ 0.
So b√5 = a. Squaring, 5b² = a². Hence 5 divides a², and since 5 is prime, 5 divides a (Theorem 1.2).
Write a = 5c for some integer c. Then 5b² = 25c², that is, b² = 5c².
So 5 divides b², and therefore 5 divides b.
Now 5 is a common factor of a and b, which contradicts the fact that a and b are coprime.
This contradiction arose from assuming √5 is rational.
Answer: √5 is irrational
Assume, to the contrary, that 3 + 2√5 is rational. Then 3 + 2√5 = ab, with a, b coprime integers and b ≠ 0.
Since a and b are integers, a − 3b2b is rational, so √5 would be rational.
This contradicts the fact that √5 is irrational (Question 1).
Answer: 3 + 2√5 is irrational
(i) Assume 1√2 = ab, with a, b coprime integers, a ≠ 0, b ≠ 0. Then √2 = ba, which is rational. This contradicts the fact that √2 is irrational. So 1√2 is irrational.
(ii) Assume 7√5 = ab, with a, b coprime integers, b ≠ 0. Then √5 = a7b, which is rational. This contradicts the fact that √5 is irrational. So 7√5 is irrational.
(iii) Assume 6 + √2 = ab, with a, b coprime integers, b ≠ 0. Then √2 = ab − 6 = a − 6bb, which is rational. This contradicts the fact that √2 is irrational. So 6 + √2 is irrational.
Answer: All three are irrational