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Theorem (Euclid’s Division Lemma):

For a pair of given positive integers ‘a’ and ‘b’, there exist unique integers ‘q’ and ‘r’ such that
Real Number 1
Explanation:
Thus, for any pair of two positive integers a and b; the relation
Real Number 2
Example:-
(a) 20, 8
Let 20 = a and 8 = b
Real Number 3
(b) 17, 5
Let 17 = a and 5 = b
Real Number 4

Exercise 1.1 (NCERT Book)

Question - 1: Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Solution: (i) 135 and 225
In the given problem, let 225 = a, and 135 = b
Real Number 5
Since, r (remainder) is not equal to zero (0). Thus, by applying the Euclid’s division algorithm, by taking 135 = a, and 90 = b we get
Real Number 6
Since, in this step also, r is not equal to zero(0). Thus by continuing the Euclid’s division algorithm, by taking this time, 90 = a, and 45 = b we get
Real Number 7
Therefore, 45 is the HCF of given pair 225 and 135
Thus, Answer: 45
Solution: (ii) 196 and 38220
In the given pair, 38220 > 196, thus let 38220 = a and 196 = b
Now by applying the Euclid’s division algorithm, we get
Real Number 8
Since, in the above equation we get, r = 0, therefore, 196 is the HCF of the given pair 196 and 38220.
Answer: 196
Solution:- (iii) 867 and 255
Let a = 867 and b = 255, thus after applying the Euclid’s division algorithm we get
867 = 255 X 3 + 102 ( Where r = 102)
Since, r ≠ 0, therefore, by taking 255 and 102 as a and b respectively we get
255 = 102 X 2 + 51
Similarly, 102 = 51 X 2 + 0 (Where, r = 0)
Since, in this term r = 0, thus HCF of the given pair 867 and 255 is equal to 51
Answer: 51
Question - 2 – Show that any positive odd integer is of the form 6q + 1 or, 6q +3 or, 6q + 5, where q is some integer.
Solution:
Let ‘a’ be any positive odd integer and ‘b = 6’.
Therefore,
Real Number 9
Now, by placing r = 0, we get, a = 6q + 0 = 6q
By placing r = 1, we get, a = 6q +1
By placing, r = 2, we get, a = 6q + 2
By placing, r = 3, we get, a = 6q + 3
By placing, r = 4, we get, a = 6q + 4
By placing, r = 5, we get, a = 6q +5
Thus, a = 6q or, 6q +1 or, 6q + 2 or, 6q + 3 or, 6q + 4 or, 6q +5
But here, 6q, 6q + 2, 6q +4 are the even integers
Therefore, 6q + 1 or, 6q + 3 or, 6q + 5 are the forms of any positive odd integers.
Alternate Method: Assume any value for q; like 1, 2, 3, ………….n
If q = 1;
Then;
Real Number 10
If q = 2; Then;
Real Number 11
Similarly, we can go on substitiuting different values for q and the result would always be a positive odd integer.
Another rationale for this can be as follows:
Since 6 is an even number and product of any number with an even number is always an even number. Moreover, if an odd number is added to any even number, we get an odd number as result.
Question: 3 – An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution: The required number of column is obtained by the HCF of 616 and 32
Let, a = 616 and b = 32, therefore, by applying Euclid’s division algorithm, we get
616 = 32 X 19 + 8, since, here r = 8 and ≠ 0, thus, by continuing the process, we get
32 = 8 X 2 + 0, here r = 0
Thus, the HCF of 616 and 32 is equal to 8
Thus, the required maximum number of column = 8
Question: 4 – Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m +1 for some integer m.
[Hint: Let x be any positive integer then it is of the form 3q, 3q +1 or 3q + 2. Now square each of these and show that they can rewritten in the form 3m or 3m +1]
Solution: Let ‘n’ is any positive integer, then it is of the form of 3q or, 3q + 1 or, 3q + 2
Real Number 12
Therefore, square of any positive integer is either of the form of 3m or 3m + 1
Alternate method:
Let us start with the smallest square number, i.e. 4
Real Number 13
Let us take the next square number, i.e. 9
Real Number 14
Let us take the next square number, i.e. 16
Real Number 15
Therefore, square of any positive integer is either of the form of 3m or 3m + 1
Question: 5 – Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let ‘a’ is a positive integer and b = 4
Therefore, according Euclid’s division lemma
Real Number 16 
Real Number 17
Thus, cube of any positive integer is in the form of 9m, 9m +1 or 9m +8
Alternate method:
Let us start with the smallest cube number, i.e. 8
Real Number 18
Let us take the next cube number, i.e. 27
Real Number 19
Let us take the next cube number, i.e. 64
Real Number 20
Thus, cube of any positive integer is in the form of 9m, 9m +1 or 9m +8

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.
This theorem also says that the prime factorisation of a natural number is unique, except for the order of its factors.
For example 20 can be expressed as 2 × 2 × 5
Using this theorem the LCM and HCF of the given pair of positive integers can be calculated.

LCM = Product of the greatest power of each prime factor, involved in the numbers.

HCF = Product of the smallest power of each common prime factor in the numbers.

Exercise 1.2 (NCERT Book)
Question: 1 - Express each number as a product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Solution:
realnumbe 121
Question: 2 - 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
Solution:-
(i) 26 and 91
The prime factors of 26 = 2 × 13
The prime factors of 91 = 7 × 13
Therefore, LCM = 2 × 7 × 13 = 182
And, HCF = 13
Now, LCM × HCF = 182 × 13 = 2366
Product of given numbers = 26 × 91 = 2366
Therefore, LCM × HCF = Product of the given two numbers
(ii) 510 and 92
The prime factors of 510 = 2 × 3 × 5 × 17
The prime factors of 92 = 2 × 2 × 23
Therefore, LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460
And, HCF = 2
Now,
LCM × HCF = 23460 × 2 = 46920
Product of given two Numbers = 510 × 92 = 46920
Therefore, LCM × HCF = Product of given two numbers
(iii) 336 and 54
The prime factors of 336 = 2 × 2 × 2 × 2 × 3 × 7 = 24 × 3 × 7
The prime factors of 54 = 2 × 3 × 3 × 3 = 2 × 33
Therefore, LCM of 336 and 54 = 24 × 33 × 7 = 3024
And, HCF = 2 × 3 = 6
Now, LCM × HCF = 3024 × 6 = 18144
And the product of given numbers = 336 × 54 = 18144
Therefore, LCM × HCF = Product of given numbers
Question: 3. Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
Solution:-
(i) 12, 15 and 21
Prime factors of 12 = 2 × 2 × 3 = 22 × 3
Prime factors of 15 = 3 × 5
Prime factors of 21 = 3 × 7
Therefore, LCM = 2 × 2 × 3 × 5 × 7 = 420
HCF = 3
(ii) 17, 23 and 29
Prime factors of 17 = 17 × 1
Prime factors of 23 = 23 × 1
Prime factors of 29 = 29 × 1
Therefore, LCM = 17 × 23 × 29 = 11339
And HCF = 1
(iii) 8, 9 and 25
Prime factors of 8 = 2 × 2 × 2 = 23
Prime factors of 9 = 3 × 3 = 32
Prime factors of 25 = 5 × 5= 52
Therefore, LCM = 23 × 32 × 52 = 8 × 9 × 25 = 1800
Since, there is no common factors among the prime factors of given three numbers,
Thus HCF = 1
Question: 4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:-
realnumbe 122
Question: 5. Check whether 6n can end with the digit 0 for any natural number n.
Solution:-
Numbers that ends with zero are divisible by 5 and 10. Simultaneously they are divisible by 2 and 5 both.
If the number 6n will be divisible by 2 and 5, then, it will end with the digit 0 otherwise not.
Prime factor of 6n = (2 × 3) n
Since, 5 is not a prime factor of 6n
Therefore, for any value of n, 6n will not be divisible by 5.
Therefore, 6n cannot end with the digit 0 for any natural number n.
Note: 6n always has 6 at unit’s place. Following examples illustrate this:
61 = 6
62 = 36
63 = 216
64 = 1296
Hence, 6n can never end with the digit zero for any natural number n.
Question: 6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Numbers which have at least one factor other than 1 and number itself are called composite numbers.
Given, first expression:
7 × 11 × 13 + 13
= 13 × (7 × 11 + 1)
= 13 × (77 + 1)
= 13 × 78
Since, given expression 7 × 11 × 13 + 13 has two prime factors other than 1, thus it is a composite number.
Second expression:
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 ( 7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 (1008 + 1)
= 5 × 1009
Since, the given expression has two factors other than 1, thus it is a composite number.
Question: 7. 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?
Solution:
Since, they will take different times to complete one round of the sports field. Thus, time after which they will meet again at the starting point will be given by the LCM of time taken to complete one round for each of them.
Time taken by Sonia to complete one round = 18 minute
Time taken by Ravi to complete one round = 12 minute
Prime factors of 18 = 2 × 3 × 3 = 2 × 32
Prime factors of 12 = 2 × 2 × 3 = 22 × 3
Therefore, LCM = 22 × 32 = 4 × 9 = 36
Thus, after starting simultaneously Ravi and Sonia will meet at starting point after 36 minute.

NCERT Solution - Exercise - 1.3

Question – 1 - Prove that Class 10 math NCERT Solution 1.3_1 is irrational.
Answer: Let us assume the contrary, i.e. Class 10 math NCERT Solution 1.3_2 is irrational.
Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;
Class 10 math NCERT Solution 1.3_3
Squaring on both sides, we get;
Class 10 math NCERT Solution 1.3_4
This means that a2 is divisible by 5 and hence a is also divisible by 5.
This contradicts our earlier assumption that a and b are coprime, because we have found 5 as at least one common factor of a and b.
This also contradicts our earlier assumption that Class 10 math NCERT Solution 1.3_2 is irrational.
Class 10 math NCERT Solution 1.3_5
Question – 2 - Prove that Class 10 math NCERT Solution 1.3_6 is irrational.
Answer: Let us assume to the contrary, i.e. Class 10 math NCERT Solution 1.3_7 is irrational.
Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;
Class 10 math NCERT Solution 1.3_8
Class 10 math NCERT Solution 1.3_9
Since a and b are rational, so  Class 10 math NCERT Solution 1.3_10 is rational and hence, Class 10 math NCERT Solution 1.3_11 is rational.
But this contradicts the fact that Class 10 math NCERT Solution 1.3_11 is irrational.
This happened because of our faulty assumption.
Class 10 math NCERT Solution 1.3_12
Question – 3 - Prove that following are irrationals:
Class 10 math NCERT Solution 1.3_13
Answer: Let us assume to the contrary, i.e. Class 10 math NCERT Solution 1.3_14 is rational.
Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;
Class 10 math NCERT Solution 1.3_15
Squaring on both sides, we get;
Class 10 math NCERT Solution 1.3_16
This means that b2 is divisible by 2 and hence a is also divisible by 2.
This contradicts our earlier assumption that a and b are co-prime, because 2 is at least one common factor of a and b.
This also contradicts our earlier assumption that Class 10 math NCERT Solution 1.3_14 is rational.
Hence, Class 10 math NCERT Solution 1.3_14 is irrational proved.

Class 10 math NCERT Solution 1.3_17
Answer: Let us assume to the contrary, i.e. Class 10 math NCERT Solution 1.3_18 is rational.
There can be two integers a and b (b≠0) and a and b are coprime, so that;
Class 10 math NCERT Solution 1.3_19
Squaring on both sides, we get;
Class 10 math NCERT Solution 1.3_20
This means that a2 is divisible by 245; which means that a is also divisible by 245.
This contradicts our earlier assumption that a and b are coprime, because 245 is at least one common factor of a and b.
This happened because of our faulty assumption and hence, Class 10 math NCERT Solution 1.3_18 is irrational proved.
Class 10 math NCERT Solution 1.3_21
Answer: Let us assume to the contrary, i.e. Class 10 math NCERT Solution 1.3_22 is rational.
Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;
Class 10 math NCERT Solution 1.3_23
Since a and b are rational, so Class 10 math NCERT Solution 1.3_24 is rational and hence, Class 10 math NCERT Solution 1.3_25 is rational.
But this contradicts the fact that Class 10 math NCERT Solution 1.3_25 is irrational.
This happened because of our faulty assumption.
Hence, Class 10 math NCERT Solution 1.3_26 is irrational proved.

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