Selina Concise Solutions for ICSE Class 6 Mathematics Chapter 8 HCF and LCM

IMPORTANT POINTS
Factor of a given number is that number by which the given number can be divided completely.
1. Prime Numbers :
A Natural number, which is divisible by 1 (one) and itself only is called a prime number.
2. Highest Common Factor :
H.C.F. stands for Highest Common Factor and H.C.F. of two or more given numbers is the greatest number (factor) which divides each given number completely.
3. Lowest Common Factor :
L.C.M. stands for Lowest Common Multiple. The L.C.M. of two or more given numbers is the lowest (smallest) number which is exactly divisible by each of the given numbers.

EXERCISE 8(A)

Question 1: Write all the factors of :
(i) 15
(ii) 55
(iii) 48
(iv) 36
(v) 84

Answer:
(i) Factors of 15 = F₁₅ = 1, 3, 5 and 15
(ii) Factors of 55 = F₅₅ = 1, 5, 11 and 55
(iii) Factors of 48 = F₄₈ = 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48
(iv) Factors of 36 = F₅₆ = 1, 2, 3, 4, 6, 9, 12, 18 and 36.
(v) Factors of 84 = F₈₄ = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.
Factors are the whole numbers that can divide a specific number without leaving any remainder behind. Every number will always have at least two factors: 1 and the number itself.
Teacher's Tip: Factors always come in pairs (like 3 × 5 = 15), so if you find one, you've usually found another!
Exam Tip: To ensure full marks, always list the factors in ascending order and don't forget the number itself.

Question 2: Write all prime numbers :
(i) less than 25
(ii) between 15 and 35
(iii) between 8 and 76

Answer:
(i) 2, 3, 5, 7, 11, 13, 17, 19 and 23
(ii) 17, 19, 23, 29 and 31
(iii) 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71 and 73.
Prime numbers are special numbers that only have two factors, which are 1 and the number itself. They cannot be divided evenly by any other numbers, making them the building blocks of mathematics.
Teacher's Tip: Remember that 2 is the only even prime number; every other prime number is odd!
Exam Tip: Check your list carefully to make sure you didn't accidentally include 1, as 1 is neither prime nor composite.

Question 3: Write the prime-numbers from :
(i) 5 to 45
(ii) 2 to 32
(iii) 8 to 48
(iv) 9 to 59

Answer:
(i) 7, 11, 13, 17, 19, 23, 29, 31, 37, 41 and 43.
(ii) 3, 5, 7, 11, 13, 17, 19, 23 29 and 31.
(iii) 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.
(iv) 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 and 53.
Finding prime numbers in a range involves checking each number to see if it can be divided by anything other than 1 and itself. You can use divisibility rules for 2, 3, and 5 to quickly eliminate most non-prime numbers.
Teacher's Tip: To find primes, ignore all even numbers except 2 and all numbers ending in 5 except 5.
Exam Tip: Read the question carefully to see if "from" means you should include the start and end numbers if they are prime.

Question 4: Write the prime factors of:
(i) 16
(ii) 27
(iii) 35
(iv) 49

Answer:

(i) Prime factors of 16 = 2
2 | 16
2 | 8
2 | 4
2 | 2
| 1

(ii) Prime factors of 27 = 3
3 | 27
3 | 9
3 | 3
| 1

(iii) Prime factors of 35 = 5, 7
5 | 35
7 | 7
| 1

(iv) Prime factors of 49 = 7
7 | 49
7 | 7
| 1
Prime factorization is the process of breaking down a composite number into a product of only prime numbers. We use the "ladder" or "division" method, dividing by the smallest possible prime until we reach 1.
Teacher's Tip: Always start dividing by the smallest prime (2, then 3, then 5) to keep your work organized.
Exam Tip: Show the full division ladder in your rough work to prove how you arrived at your prime factors.

Question 5: If P_n means prime factors of n, find:
(i) p₆
(ii) P₂₄
(iii) p₅₀
(iv) P₄₂

Answer:
(i) F₆ = 1, 2, 3, 6
P.F₆ (Prime factor of 6) = 2 and 3.
(ii) F₂₄ = 1, 2, 3, 4, 6, 8, 12, 24
P.F₂₄ = 2 and 3.
(iii) F₅₀ = 2, 5, 5
P.F₅₀ = 2 and 5.
(iv) F₄₂ = 1, 2, 3, 6, 7, 14, 21, 42
P.F₄₂ = 2, 3 and 7.
This notation asks you to identify the distinct prime numbers that divide the given number. Even if a prime factor repeats (like 2 in 24), we only list the unique prime numbers.
Teacher's Tip: P_n just wants the names of the prime "ingredients" used to make the number.
Exam Tip: Differentiate between "factors" (all divisors) and "prime factors" (only prime divisors) to avoid losing marks.

EXERCISE 8(B)

Question 1: Using the common factor method, find the H.C.F. of :
(i) 16 and 35
(ii) 25 and 20
(iii) 27 and 75
(iv) 8, 12 and 18
(v) 24, 36, 45 and 60

Answer:
(i) F₁₆ = 1, 2, 4, 8, 16
F₃₅ = 1, 5, 7, 35
Common factors between 16 and 35 = 1
H.C.F. of 16 and 35 = 1

(ii) F₂₅ = 1, 5, 25
F₂₀ = 1, 2, 4, 5, 10, 20
Common factors between 25 and 20 = 1, 5
H.C.F. of 25 and 20 = 5

(iii) F₂₇ = 1, 3, 9, 27
F₇₅ = 1, 3, 5, 15, 25, 75
Common factors between 27 and 75 = 1, 3
H.C.F. of 27 and 75 = 3

(iv) F₈ = 1, 2, 4, 8
F₁₂ = 1, 2, 3, 4, 6, 12
F₁₈ = 1, 2, 3, 6, 9, 18
Common factors between 8, 12 and 18 = 1, 2
H.C.F. of 8, 12 and 18 = 2

(v) F₂₄ = 1, 2, 3, 4, 6, 8, 12, 24
F₃₆ = 1, 2, 3, 4, 6, 12, 18, 36
F₄ = 1, 3, 5, 9, 15, 45
F₆₀ = 1, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Common factor between 24, 36, 45 and 60 = 1, 3
H.C.F. of 24, 36, 45 and 60 = 3
The common factor method involves listing every factor for each number and then circling the ones they all share. The highest number in that circle of common factors is your H.C.F.
Teacher's Tip: H.C.F. is the largest "shared" factor; it can never be larger than the smallest number in the group.
Exam Tip: Clearly underline or circle the common factors before stating the final H.C.F. to show your logical steps.

Question 2: Using the prime factor method, find the H.C.F. of:
(i) 5 and 8
(ii) 24 and 49
(iii) 40, 60 and 80
(iv) 48, 84 and 88
(v) 12, 16 and 28

Answer:

(i) Prime factor of 5 = 5
Prime factor of 8 = 2 × 2 × 2
No common prime factor
H.C.F. of 5 and 8 = 1
(as both the number are co-prime)

(ii) Prime factor of 24 = 2 x 2 x 2 x 3
Prime factor of 49 = 7 x 7
No common prime factor, number are co-prime.
H.C.F. of 24 and 49 = 1.

(iii) Prime factor of 40 = 2 x 2 x 2 x 5
Prime factor of 60 = 2 x 2 x 3 x 5
Prime factor of 80 = 2 x 2 x 2 x 2 x 5
Common prime factor = 2 x 2 x 5
H.C.F. of 40, 60 and 80 = 2 x 2 x 5 = 20

(iv) Prime factor of 48 = 2 x 2 x 2 x 2 x 3
Prime factor of 84 = 2 x 2 x 3 x 7
Prime factor of 88 = 2 x 2 x 2 x 11
Common prime factor of 48, 84 and 88 = 2 x 2
H.C.F. of 48, 84 and 88 = 2 x 2 = 4

(v) Prime factor of 12 = 2 x 2 x 3
Prime factor of 16 = 2 x 2 x 2 x 2
Prime factor of 28 = 2 x 2 x 7
Common prime factor between 12, 16 and 28 = 2 x 2
H.C.F. of 12, 16 and 28 = 2 × 2 = 4
In the prime factor method, you express each number as a product of primes and then pick the lowest power of each common prime. Multiplying these shared prime parts together gives you the H.C.F.
Teacher's Tip: If there are no common prime factors at all, the H.C.F. is always 1.
Exam Tip: Write out each number's full prime expansion on a new line so you can easily vertically align common factors.

Question 3: Using the division method, find the H.C.F. of the following :
(i) 16 and 24
(ii) 18 and 30
(iii) 7, 14 and 24
(iv) 70, 80, 120 and 150
(v) 32, 56 and 46

Answer:
(i) 16 and 24
16 ) 24 ( 1
16
--
8 ) 16 ( 2
16
--
x
Since last division is 8, therefore H.C.F. of 16 and 24 = 8

(ii) 18 and 30
18 ) 30 ( 1
18
--
12 ) 18 ( 1
12
--
6 ) 12 ( 2
12
--
x
Since last division is 6, therefore H.C.F. of 18 and 30 = 6

(iii) 7, 14 and 24
7 ) 14 ( 2
14
--
x
7 ) 24 ( 3
21
--
3 ) 7 ( 2
6
--
1 ) 3 ( 3
3
--
x
Since the last division is 1, therefore H.C.F. of 7, 14 and 24 = 1

(iv) 70, 80, 120 and 150
70 ) 80 ( 1
70
--
10 ) 70 ( 7
70
--
x
10 ) 120 ( 12
120
---
x
10 ) 150 ( 15
150
---
x
Since the last division = 10, therefore H.C.F. of 70, 80, 120 and 150 = 10

(v) 32, 56 and 46
32 ) 56 ( 1
32
--
24 ) 32 ( 1
24
--
8 ) 24 ( 3
24
--
x
8 ) 46 ( 5
40
--
6 ) 8 ( 1
6
--
2 ) 6 ( 3
6
--
x
Since last division = 2, therefore H.C.F. of 32, 56 and 46 = 2
The long division method for H.C.F. is great for large numbers where factors are hard to find. You divide the larger number by the smaller, and then use the remainder as your new divisor until there is no remainder left.
Teacher's Tip: The H.C.F. is the *last* number that divides perfectly, not the quotient (answer at the side)!
Exam Tip: When dealing with more than two numbers, find the H.C.F. of the first two, then find the H.C.F. of that result and the third number.

Question 4: Use a method of your own choice to find the H.C.F. of :
(i) 45, 75 and 135
(ii) 48, 36 and 96
(iii) 66, 33 and 132
(iv) 24, 36, 60 and 132
(v) 30, 60, 90 and 105

Answer:
(i) Factor of 45 = F₄₅ = 3 × 3 × 5
Factor of 75 = F₇₅ = 3 × 5 × 5
and Factor of 135 = F₁₃₅ = 3 × 3 × 3 × 5
Now the common factors of 45, 75 and 135 = 3 and 5
H.C.F. = 3 × 5 = 15

(ii) Factor of 48 = F₄₈ = 2 × 2 × 2 × 2× 3
Factor of 36 = F₃₆ = 2 × 2 × 3 × 3
and factor of 96 = 2 × 2 × 2 × 2 × 2 × 3
Now the common factor of 48, 36 and 96 = 2, 2 and 3
H.C.F. = 2 × 2 × 3 = 12

(iii) Factor of 66 = F₆₆ = 2 × 3 × 11
Factor of 33 = F₃₃ = 3 × 11
and factor of 132 = F₁₃₂ = 2 × 2 × 3 × 11
Now the common factor of 66, 33 and 132 = 3 and 11
H.C.F. = 3 × 11 = 33

(iv) Factor of 24 = F₂₄ = 2 × 2 × 2 × 3
Factor of 36 = F₃₆ = 2 × 2 × 3 × 3
Factor of 60 = F₆₀ = 2 × 2 × 3 × 5
and Factor of 132 = F₁₃₂ = 2 × 2 × 3 × 11
Now the common factors of 24, 36, 60 and 132 = 2, 2 and 3
H.C.F. = 2 × 2 × 3 = 12

(v) Factor of 30 = F₃₀ = 2 × 3 × 5
Factor of 60 = F₆₀ = 2 × 2 × 3 × 5
Factor of 90 = F₉₀ = 2 × 3 × 3 × 5
and factor of 105 = F₁₀₅ = 3 × 5 × 7
Now the common factor of 30, 60, 90 and 105 = 3 and 5
H.C.F. = 3 × 5 = 15
Choosing the right method depends on the numbers; smaller numbers are often easier with prime factorization, while larger numbers might be faster with the division method. Regardless of the method, the H.C.F. represents the largest possible measure that can divide into all these quantities evenly.
Teacher's Tip: Use the "method of your choice" to your advantage by picking the one you find easiest and quickest.
Exam Tip: Always double-check your multiplication at the final step to ensure your H.C.F. is correct.

Question 5: Find the greatest number that divides each of 180, 225 and 315 completely.

Answer:
The greatest number that divides 180, 225 and 315 will be HCF of 180, 225, 315
Let us first find HCF of 180 and 225
180 ) 225 ( 1
180
---
45 ) 180 ( 4
180
---
x
Since third number is 315, and HCF obtained above is 45, find the HCF of 315 and 45.
45 ) 315 ( 7
315
---
x
therefore HCF of given number 180, 225 and 315 = 45
When a question asks for the "greatest number that divides," it is a direct clue to find the Highest Common Factor. This number is the largest common divisor shared by all three values.
Teacher's Tip: "Greatest" is a keyword for H.C.F. while "Smallest" is often a keyword for L.C.M.
Exam Tip: State clearly that the "Greatest number = H.C.F." at the beginning of your solution to show your understanding of the problem.

Question 6: Show that 45 and 56 are co-prime numbers.

Answer:
The HCF of two co-prime numbers is always 1.
HCF of 45 and 56
45 ) 56 ( 1
45
--
11 ) 45 ( 4
44
--
1 ) 11 ( 11
11
--
x
From above it is proved that HCF of 45 and 56 is 1
Hence 45 and 56 are co-prime numbers.
Co-prime numbers are any two numbers whose only common factor is 1. This doesn't mean the numbers themselves have to be prime; it just means they don't share any divisors besides 1.
Teacher's Tip: To prove co-primes, just find their H.C.F. and see if it equals 1!
Exam Tip: Explicitly write "Since H.C.F. is 1, they are co-prime" to complete your proof.

Question 7: Out of 15, 16, 21 and 28, find out all the pairs of co-prime numbers.

Answer:
The pair will be 15 - 16, 16 - 21, 21 - 28, 15 - 28 and 16 - 28.

The HCF of 15 and 16
15 ) 16 ( 1
15
--
1 ) 15 ( 15
15
--
x
and HCF of 21 and 28

HCF of 16 and 21
16 ) 21 ( 1
16
--
5 ) 16 ( 3
15
--
1 ) 5 ( 5
5
--
x
15 ) 28 ( 1
15
--
13 ) 15 ( 1
13
--
2 ) 13 ( 6
12
--
1 ) 2 ( 1
2
--
x

HCF of 16, 28
16 ) 28 ( 1
16
--
12 ) 16 ( 1
12
--
4 ) 12 ( 3
12
--
x
From above it is clear that 15 and 16 are co-prime because common factor is 1
Hence pairs 15 and 16, 16 and 21, 15 and 28 are co-prime number.
To find all co-prime pairs, you must test every possible combination of two numbers from the set. If their Highest Common Factor is exactly 1, they are part of your final list.
Teacher's Tip: Numbers that are consecutive (like 15 and 16) are *always* co-prime!
Exam Tip: List the final pairs clearly at the end of your answer after showing the H.C.F. calculations for each.

Question 8: Find the greatest no. that will divide 93, 111 and 129, leaving remainder 3 in each case.

Answer:
Since Remainder is 3 in each case numbers are
93 - 3 = 90
111 - 3 = 108
129 - 3 = 126
Required number will be HCF of 90, 108 and 126
HCF of 90 and 108
90 ) 108 ( 1
90
--
18 ) 90 ( 5
90
--
x
HCF of 18 and 126
18 ) 126 ( 7
126
---
x
therefore Greatest number will be = 18
If a number leaves a remainder, it means it doesn't divide perfectly. By subtracting that remainder first, we find the numbers that *can* be divided perfectly, and then we find their H.C.F.
Teacher's Tip: Always subtract the remainder from the numbers *before* starting the H.C.F. division.
Exam Tip: State the subtractions clearly so the examiner sees why you are finding the H.C.F. of 90, 108, and 126 instead of the original numbers.

EXERCISE 8(C)

Question 1: Using the common multiple method, find the L.C.M. of the following :
(i) 8, 12 and 24
(ii) 10, 15 and 20
(iii) 3, 6, 9 and 12

Answer:
(i) 8, 12 and 24
4 | 8, 12, 24
3 | 2, 3, 6
2 | 2, 1, 2
| 1, 1, 1
therefore L.C.M. = 4 × 3 × 2 = 24

(ii) 10, 15 and 20
2 | 10, 15, 20
2 | 5, 15, 10
5 | 5, 15, 5
| 1, 3, 1
therefore L.C.M. = 2 × 2 × 5 × 3 = 60

(iii) 3, 6, 9 and 12
3 | 3, 6, 9, 12
2 | 1, 2, 3, 4
| 1, 1, 3, 2
therefore L.C.M. = 3 × 2 × 3 × 2 = 36
L.C.M. is the smallest number that is a multiple of every number in your set. We use common division, dividing by factors shared by at least two numbers until all rows end in 1.
Teacher's Tip: Multiples are "skip counting"; L.C.M. is the first time all your skip counting hits the same number!
Exam Tip: In L.C.M. common division, if a number isn't divisible by the divisor, just carry it down to the next line as it is.

Question 2: Find the L.C.M. of each the following groups of numbers, using
(i) the prime factor method and
(ii) the common division method :
(i) 18, 24 and 96
(ii) 100, 150 and 200
(iii) 14, 21 and 98
(iv) 22, 121 and 33
(v) 34, 85 and 51

Answer:
(i) L.C.M. of 18, 24 and 96
(i) By prime factors
Prime factors of 18 = 2 × 3 × 3
Prime factors of 24 = 2 × 2 × 2 × 3
Prime factors of 96 = 2 × 2 × 2 × 2 × 2 × 3
L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288
By common division method
L.C.M. of 18, 24 and 96 = 2 × 2 ×s 2 × 3× 3 × 4 = 288
2 | 18, 24, 96
2 | 9, 12, 48
2 | 9, 6, 24
3 | 9, 3, 12
| 3, 1, 4

(ii) 100, 150 and 200
Factor of 100 = 2 × 2 × 5 × 5 = 2² × 5²
Factor of 150 = 2 × 3 × 5 × 5 = 2¹ × 3¹ × 5²
Factor of 200 = 2 × 2 × 2 × 5 × 5 = 2³ × 5²
therefore L.C.M. of 100, 150 and 200 = 2³ × 3¹ × 5² = 600
Common Division Method :
2 | 100, 150, 200
2 | 50, 75, 100
5 | 25, 75, 50
5 | 5, 15, 10
| 1, 3, 2
therefore L.C.M. of 100, 150 and 200 = 2 × 2 × 5 × 5 × 3 × 2 = 600

(iii) 14, 21, 98
Factor of 14 = 2 × 7 = 2¹ × 7¹
Factor of 21 = 3 × 7 = 3¹ × 7¹
Factor of 98 = 2 × 7 × 7 = 2¹ × 7²
therefore L.C.M. of 14, 21 and 98 = 2¹ × 3¹ × 7² = 294
Common Division Method :
2 | 14, 21, 98
7 | 7, 21, 49
| 1, 3, 7
therefore L.C.M. of 14, 21, 98 = 2 × 7 × 3 × 7 = 294

(iv) 22, 121 and 33
Factor of 22 = 2 × 11 = 2¹ × 11¹
Factor of 121 = 11 × 11 = 11²
Factor of 33 = 3 ×11 = 3¹ × 11¹
therefore L.C.M. of 22, 121 and 33 = 2¹ × 3¹ × 11² = 726
Common Division Method :
2 | 22, 121, 33
11| 11, 121, 33
| 1, 11, 3
L.C.M. = 2 × 11 × 11 × 3 = 726

(v) 34, 85 and 51
Factor of 34 = 2 × 17 = 2¹ × 17¹
Factor of 85 = 5 × 17 = 5¹ × 17¹
Factor of 51 = 3 × 17 = 3¹ × 17¹
therefore L.C.M. of 34, 85 and 51 = 2¹ × 5¹ × 3¹ × 17 = 510
Common Division Method :
2 | 34, 85, 51
17| 17, 85, 51
| 1, 5, 3
L.C.M. of 34, 85 and 51 = 2 × 17 × 5 × 3 = 510
Prime factorization for L.C.M. requires taking the *highest* power of every prime factor present in any of the numbers. Common division is often faster for three or more numbers as you process them simultaneously.
Teacher's Tip: For prime factor L.C.M., collect all prime bases and use their "winner" (highest) exponents.
Exam Tip: Double-verify your final product; one small multiplication error in the string of prime factors will lead to the wrong L.C.M.

Question 3: The H.C.F. and the L.C.M. of two numbers are 50 and 300 respectively. If one of the numbers is 150, find the other one.

Answer:
H.C.F. = 50
L.C.M. = 300
Product of L.C.M. and H.C.F. = 300 × 50 = 15000
One number = 150
The other number = Product of L.C.M. and H.C.F.} / {One number = {15000}/{150} = 100
There is a powerful relationship where the product of any two numbers is always equal to the product of their H.C.F. and L.C.M. This formula allows us to find a missing value if the other three are known.
Teacher's Tip: Use the formula: Number 1 × Number 2 = HCF × LCM.
Exam Tip: Always write the formula first before plugging in the numbers to ensure you get marks for the correct approach.

Question 4: The product of two numbers is 432 and their L.C.M. is 72. Find their H.C.F.

Answer:
Product of two numbers = Product of their L.C.M. and H.C.F.
Here, product of two number = 432
L.C.M. = 72
H.C.F. = 432/72 = 6
This question uses the same mathematical rule that connects products, H.C.F., and L.C.M. Since the product of the numbers is already given, we simply divide it by the L.C.M. to isolate the H.C.F.
Teacher's Tip: Think of this like a balance scale; the product of numbers on one side must equal HCF × LCM on the other.
Exam Tip: Check your division by multiplying 72 × 6 to see if you get 432 back.

Question 5: The product of two numbers is 19,200 and their H.C.F. is 40. Find their L.C.M.

Answer:
L.C.M. = Product of number/H.C.F.
Product of number = 19,200
H.C.F. = 40
therefore L.C.M. = 19,200/40 = 480
By rearranging our core formula, we find that the L.C.M. is the result of dividing the total product of the numbers by their Highest Common Factor. This is a common shortcut for these types of word problems.
Teacher's Tip: When dividing by 40, you can "cancel" a zero from both 19,200 and 40 to make the math easier: 1920 \div 4.
Exam Tip: Ensure you specify which value you are solving for (L.C.M. or H.C.F.) at the start of your calculation.

Question 6: Find the smallest number which, when divided by 12, 15, 18, 24 and 36 leaves no remainder

Answer:
The least number which is exactly divisible by each given number is their L.C.M.
Required number L.C.M. of 12, 15, 18, 24 and 36.
2 | 12, 15, 18, 24, 36
2 | 6, 15, 9, 12, 18
3 | 3, 15, 9, 6, 9
3 | 1, 5, 3, 2, 3
| 1, 5, 1, 2, 1
therefore L.C.M. = least required number = 2 × 2 × 3 × 3 × 5 × 2 = 360
Hence, the least required number = 360
This problem asks for the Lowest Common Multiple because we need a number that "belongs" to the multiplication table of every number in the list. 360 is the first number that all these divisors can fit into without leaving any "leftovers" or remainders.
Teacher's Tip: "Smallest number divisible by..." is the textbook definition of L.C.M.
Exam Tip: If you have a long list of numbers, be very neat with your common division columns so you don't miss any values.

Question 7: Find the smallest number which, when increased by one is exactly divisible by 12, 18, 24, 32 and 40

Answer:
L.C.M. of given numbers
2 | 12, 18, 24, 32, 40
2 | 6, 9, 12, 16, 20
2 | 3, 9, 6, 8, 10
3 | 3, 9, 3, 4, 5
| 1, 3, 1, 4, 5
therefore L.C.M. = 2 × 2 × 2 × 3 × 3 × 4 × 5 = 1440 = One increasing
therefore The required number = 1440 - 1 = 1439
In this scenario, we find the perfect target (the L.C.M.) first. Since the question says our number becomes perfect *after* being increased by 1, we must subtract 1 from the L.C.M. to find our starting point.
Teacher's Tip: For "Increased by x" problems, find the LCM and then do LCM - x.
Exam Tip: Don't forget the final subtraction! 1440 is the multiple, but 1439 is the specific answer the question is asking for.

Question 8: Find the smallest number which, on being decreased by 3, is completely divisible by 18, 36, 32 and 27.

Answer:
LCM of 18, 36, 32 and 27
2 | 18, 36, 32, 27
2 | 9, 18, 16, 27
3 | 9, 9, 8, 27
3 | 3, 3, 8, 9
| 1, 1, 8, 3
= 2 × 2 × 3 × 3 × 3 × 8 = 864
therefore Required number = 864 + 3 = 867
Since the number is perfect after losing 3, it must have originally been 3 units larger than the L.C.M. We calculate the L.C.M. to find the "perfect" multiple and then add the 3 back on.
Teacher's Tip: For "Decreased by x" problems, find the LCM and then do LCM + x.
Exam Tip: Read the question twice to make sure you know whether to add or subtract at the end; it's the most common mistake in HCF/LCM problems.

REVISION EXERCISE

Question 1: Find the H.C.F. of :
(i) 108, 288 and 420
(ii) 36, 54 and 138

Answer:

(i) H.C.F. of 108, 288, 420 = 12
108 ) 288 ( 2
216
---
72 ) 108 ( 1
72
---
36 ) 72 ( 2
72
--
x
36 ) 420 ( 11
396
---
24 ) 36 ( 1
24
--
12 ) 24 ( 2
24
--
x

(ii) H.C.F. of 36, 54 and 138 = 6
36 ) 54 ( 1
36
--
18 ) 36 ( 2
36
--
x
18 ) 138 ( 7
126
---
12 ) 18 ( 1
12
--
6 ) 12 ( 2
12
--
x
The H.C.F. is found by continuing the long division process until a divisor leaves zero as a remainder. For multiple numbers, you take the H.C.F. of the first pair and use that to divide the next number in the list.
Teacher's Tip: In long division H.C.F., the remainder becomes the new "outside" number and the old divisor becomes the new "inside" number.
Exam Tip: Keep your subtraction accurate! One error in the long division chain will make the whole H.C.F. incorrect.

Question 2: Find the L.C.M. of:
(i) 72, 80 and 252
(ii) 48, 66 and 120

Answer:
L.C.M. 72, 80, 252
2 | 72, 80, 252
2 | 36, 40, 126
2 | 18, 20, 63
3 | 9, 10, 63
3 | 3, 10, 21
| 1, 10, 7
= 2 × 2 × 2 × 3 × 3 × 10 × 7 = 5040

(ii) L.C.M. of 48, 66 and 120
2 | 48, 66, 120
2 | 24, 33, 60
2 | 12, 33, 30
3 | 6, 33, 15
| 2, 11, 5
= 2 × 2 × 2 × 2 × 3 × 5 × 11 = 2640
To calculate the L.C.M. of three numbers, we use common division to extract shared prime factors as long as they divide at least two numbers. We then multiply all the side divisors and the final bottom row to get our common multiple.
Teacher's Tip: If only one number is divisible by a prime, you can stop the ladder and just multiply that number into your final L.C.M. string.
Exam Tip: Write down the multiplication string (2 × 2 × 3...) before doing the actual math to show you have the right components.

Question 3: State true or false : Give an example.
(i) H.C.F. of two prime numbers is 1.
(ii) H.C.F. of two co-prime numbers is 1.
(iii) L.C.M. of two prime numbers is equal to their product.
(iv) L.C.M. of two co-prime numbers is equal to their product.

Answer:
(i) True : Because the prime numbers have no common factor except 1.
(ii) True : Because co-prime numbers have no common factor except 1.
(iii) True : Because the prime number have no common factor except 1.
(iv) True : Because co-prime numbers have no common factor except 1.
Since prime and co-prime numbers don't share any building blocks (factors), their H.C.F. is always the baseline 1. Consequently, their L.C.M. must be the two numbers multiplied together because there are no shared parts to simplify.
Teacher's Tip: For numbers like 7 and 11 (primes) or 8 and 9 (co-primes), H.C.F. is always 1 and L.C.M. is always the product!
Exam Tip: If an exam asks for an example, always provide specific numbers like "Ex: 3 and 5" to prove the rule.

Question 4: The product of two numbers is 12096 and their H.C.F. is 36. Find their L.C.M.

Answer:
We know that
Product of two numbers = Product of their H.C.F. and L.C.M.
12096 = 36 × L.C.M.
L.C.M. = 12096/36 = 336
This problem reinforces the formulaic link between products and common factors/multiples. By dividing the known product by the known H.C.F., we solve for the L.C.M. in one simple step.
Teacher's Tip: Think of "36" as the divisor that undoes the product to reveal the L.C.M.
Exam Tip: Show the long division for 12096 \div 36 in your rough column to avoid simple calculation errors.

Question 5: The product of the H.C.F. and the L.C.M. of two numbers is 1152. If one number is 48, find the other one.

Answer:
We know that:
Product of two numbers = Product of their H.C.F. and L.C.M.
1st number × 2nd number = Product of their H.C.F. and L.C.M.
48 × 2nd number = 1152
2nd number = 1152/48 = 24
The core theorem states that HCF × LCM is the same as multiplying the two original numbers. Knowing this product (1152) and one number (48) makes finding the second number a simple division task.
Teacher's Tip: H × L = A × B. If you have three, you can find the fourth!
Exam Tip: Clearly label your numbers as "1st number" and "2nd number" to keep your algebra organized.

Question 6: (i) Find the smallest number that is completely divisible by 28 and 42.
(ii) Find the largest number that can divide 28 and 42 completely.

Answer:
(i) We know that the least number which is divisible by 28 and 42 is their L.C.M.
2 | 28, 42
7 | 14, 21
| 2, 3
L.C.M. of 28 and 42 = 2 × 7 × 2 \times 3 = 84

(ii) We know that the largest number which can divide 28 and 42 completely will be their H.C.F.
28 ) 42 ( 1
28
--
14 ) 28 ( 2
28
--
x
H.C.F. of 28 and 42 = 14
Divisibility works two ways: finding a large target for small numbers (L.C.M.) or finding a small measure for large numbers (H.C.F.). 84 is the first number both 28 and 42 "reach" when counting, while 14 is the biggest "slice" that fits into both.
Teacher's Tip: "Smallest number divisible BY" = L.C.M. "Largest number that can DIVIDE" = H.C.F.
Exam Tip: This is a classic comparison question; make sure you answer both parts separately to get full credit.

Question 7: Find the L.C.M. of 140 and 168. Use the L.C.M. obtained to find the H.C.F. of the given numbers.

Answer:
Numbers are 140 and 168
L.C.M. of 140 and 168
2 | 140, 168
2 | 70, 84
7 | 35, 42
| 5, 6
= 2 × 2 × 7 × 5 × 6 = 840
H.C.F. = {1st number × 2nd number}/{L.C.M.} = {140 × 168}/840 = 28
This multi-step problem asks you to calculate L.C.M. first and then reverse-engineer the H.C.F. using the product-rule formula. It tests your ability to connect these two mathematical concepts through calculation.
Teacher's Tip: You can simplify the fraction {140 × 168}/840 by dividing both 140 and 840 by 140 first!
Exam Tip: If the question tells you to "Use the L.C.M. obtained," you MUST use the formula; don't find H.C.F. by division or factors, or you will lose "method marks."

Question 8: Find the H.C.F. of 108 and 450 and use the H.C.F. obtained to find the L.C.M. of the given numbers.

Answer:
Numbers are given : 108 and 450
H.C.F. of 108 and 450 = 18
108 ) 450 ( 4
432
---
18 ) 108 ( 6
108
---
x
therefore L.C.M. = {1st number × 2nd number}/{H.C.F.} = {108 × 450}/{18} = 2700
This is the inverse of the previous problem, starting with the Highest Common Factor and moving to the Lowest Common Multiple. 18 is the largest shared divisor, and 2700 is the smallest number that both 108 and 450 can divide into perfectly.
Teacher's Tip: Use division method for H.C.F. when numbers are over 100 to save time.
Exam Tip: Be careful when simplifying large fractions; it's often easier to divide one number by the H.C.F. first (like 108 / 18 = 6) and then multiply (6 × 450).