Kerala Syllabus Class 7 Mathematics - Unit 9 Number Relations - Questions and Answers | Teaching Manual 

Questions and Answers for Class 7 Mathematics - Unit 9 Number Relations - Study Notes | Text Books Solution STD 7 - Maths: Unit 9 Number Relations - Questions and Answers | ഗണിതം - സംഖ്യാബന്ധങ്ങൾ - ചോദ്യോത്തരങ്ങൾ 

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Kerala Syllabus STD 7 Maths: Unit 9 Number Relations - Textbook Solutions

♦ Textbook Activities (Textbook Page No: 131)
♦ Try these problems:
1. Find the number of factors of each number below:
(i)  40  (ii)  54  (iii) 60  (iv) 100  (v)  210
Answer:
(i)  40  
40 = 2³ x 5
Number of factors of 40 = (3 + 1) x (1 + 1)
                                     = 4 x 2 = 8

(ii)  54
54 = 2 x 3³
Number of factors of 54 = (1 + 1) x (3 + 1)
                                     = 2 x 4 = 8

(iii) 60  
60 = 2² x 3 x 5
Number of factors of 60 = (2 + 1) x (1 + 1) x (1 + 1)
                                     = 3 x 2 x 2 = 12

(iv) 100  
100 = 2² x 5²
Number of factors of 100 = (2 + 1) x (2 + 1)
                                       = 3 x 3 = 9

(v) 210
210 = 2 x 3 x 5 x 7
Number of factors of 210 = (1 + 1) x (1 + 1) x (1 + 1) x (1 + 1)
                                       = 2 x 2 x 2 x 2 = 16

(2) From the number of factors of a number, we can deduce some peculiarities of the number. The table below lists these for number of factors up to 5. Extend it to number of factors up to 10.
Answer:
Let's look at an example to expand this table. Let's look at the line: 'Number of factors equals four'. We have learned that, if p and q are two different primes and m, n are any two natural numbers, then the number of factors of pᵐ qⁿ is (m + 1) (n + 1). Since the number of factors is 4, we can write 4 = (m + 1) (n + 1).
If we split 4 into its factors, we can write 4 = 4 x 1 and 4 = 2 x 2.
Then we get (m + 1) (n + 1) = 4 x 1 --------- (1), and (m +1) (n + 1) = 2 x 2 --------- (2)
From (1), we obtain
m + 1 = 4 and n + 1 = 1 ie, m = 3 and n = 0
∴ pᵐ qⁿ = p³ q⁰ = p³. That is, third power of a prime.
From (2), we get m + 1 = 2 and n + 1 = 2 ie m = 1 and n = 1
∴ pᵐ qⁿ = p¹ q¹ = pq. That product of two primes. See how the table is expanded by writing out each factor:

♦ Textbook Activities (Textbook Page No: 134)
♦ Try these problems:
(1)  For each pair of numbers given below, find the largest common factor and all the other common factors:
(i) 45, 75 (ii) 225, 275 (iii) 360, 300 (iv) 210, 504 (v) 336, 588
Answer:
(i) 45, 75 
45 = 3² x 5
75 = 3 x 5²
Prime numbers common to both : 3, 5
smallest powers of these primes : 3, 5
That is, the common factors are the factors of 3 x 5
Common factors 1, 3, 5, 15
Largest common factor = 15

(ii) 225, 275
225 = 3² x 5²
275 = 5² x 11
That is the common factors are the factors of 5²
Common factors 1, 5, 25
Largest common factor = 25

(iii) 360, 300 
360 = 2³ x 3² x 5
300 = 2² x 3 x 5²
Common factors are the factors of 2² x 3 x 5
Common factors 1, 2, 4, 3, 6, 12, 5, 10, 20, 15, 30, 60
Largest common factor = 60

(iv) 210, 504 
210 = 2 x 3 x 5 x 7
504 = 2³ x 3² x 7
Common factors are the factors of 2 x 3 x 7
Common factors 1, 2, 3, 7, 6, 14, 21, 42
Largest common factor = 42

(v) 336, 588
336 = 2⁴ x 3 x 7
588 = 2² x 3 x 7² 
Common factors are the factors of 2² x 3 x 7
Common factors 1, 2, 4, 3, 6, 12, 7, 14, 28, 21, 42, 147
Largest common factor = 147

♦ Textbook Activities (Textbook Page No: 134)
Project
Just as common factors of two numbers, we can also consider common multiples of two numbers. For example, look at the multiples of 2 and 3:
Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ...
Multiples of 3 : 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ...
Common multiples of 2 and 3: 6, 12, 18, 24, 30, ....

Let's examine these common multiples: 6, 12, 18, ...
Why are all common multiples of 2 and 3 multiples of 6?
On the other hand, are all multiples of 6, also common multiples of 2 and 3?
6 = 2 x 3
12 = (2 x 3) x 2
18 = (2 x 3) x 3
24 = (2 x 3) x 4
---------------------
----------------------
These are all multiples of 2 x 3 = 6, which is the smaallest common multiple of both 2 and 3.
The smallest common multiple of two numbers can be found using their prime factors.
From the above series, we can also see that the multiples of 6 are all the common multiples of both 2 and 3.
This is an important principle in arithmetic.
All common multiples of two numbers are multiples of their smallest common multiple. To find the smallest common multiple.
i) First, write each number as the product of powers of different primes.
ii) Take the largest powers of each prime number seen among these factors.
iii) Their product will be the smallest common multiple.
Think of the following about common multiples:

(1) Are all common multiples of 3 and 4, multiples of a single number?
Yes, They willbe the multiples of the smallest common multiple of 3 and 4. To find that, first let's write 3 and 4 as the product of different primes.
4 = 2 x 2 = 2², 3 = 3¹ = 3
The product of the largest powers of each prime number among these factors = 2² x 3¹ = 4 x 3 = 12
That is the common multiples of both 3 and 4 are all multiples of 12.

(2)  Are all common multiples of 4 and 6, multiples of 24? If not, are they all multiples of some other number?
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40,.......
Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60,......
Common multiples: 12, 24, 36, 48, ............
Among these 12, 36, 60, ............. are not multiples of 24. But these are all multiples of the smallest common multiple of both 4 and 6. To find the smallest common multiple, let's write:
4 = 2²,       6 = 2 x 3
The product of the largest powers of each prime number among these factors = 2² x 3 = 4 x 3 = 12 
That is the common multiples are all multiples of 12.

(3)  If two numbers are written as the product of powers of different primes, can we find the smallest of their common multiples?
Yes, we can. This is already explained.

(4) Are all common multiples of two numbers multiples of the smallest common multiple?
Yes, this is also explained at the beginning of the project