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- Real Numbers : Theorem (Euclid's Division Lemma)

For a pair of given positive integers 'a' and 'b', there exist unique integers 'q' and 'r' such that

Explanation:

Thus, for any pair of two positive integers a and b; the relation

Example:-

(a) 20, 8

Let 20 = a and 8 = b

(b) 17, 5

Let 17 = a and 5 = b

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

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

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

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

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 â‰ 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,

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;

If q = 2; Then;

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 â‰ 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

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

Let us take the next square number, i.e. 9

Let us take the next square number, i.e. 16

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

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

Let us take the next cube number, i.e. 27

Let us take the next cube number, i.e. 64

Thus, cube of any positive integer is in the form of 9m, 9m +1 or 9m +8

- 10th Maths Paper Solutions Set 2 : CBSE Delhi Previous Year 2015
- 10th Maths Paper Solutions Set 3 : CBSE Delhi Previous Year 2015
- 10th Maths Paper Solutions Set 1 : CBSE Delhi Previous Year 2015
- 10th Maths Paper Solutions Set 1 : CBSE Abroad Previous Year 2015
- 10th Maths Paper Solutions Set 1 : CBSE All India Previous Year 2015

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Introduction
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NCERT Solution - Exercise 3.1
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NCERT Solution - Exercise 3.2 Part-1
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NCERT Solution - Exercise 3.2 Part-2
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NCERT Solution - Exercise 3.3 Part-1
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NCERT Solution - Exercise 3.3 Part-2
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NCERT Solution - Exercise 3.4 Part-1
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NCERT Solution - Exercise 3.4 Part-2
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NCERT Solution - Exercise 3.5 Part-1
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NCERT Solution - Exercise 3.5 Part-2
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NCERT Solution - Exercise 3.6 Part-1
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NCERT Solution - Exercise 3.6 Part-2
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NCERT Solution - Exercise 4.1 Part-1
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NCERT Solution - Exercise 4.1 Part-2
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NCERT Solution - Exercise 4.2 Part-1
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NCERT Solution - Exercise 4.2 Part-2
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NCERT Solution - Exercise 4.2 Part-3
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NCERT Solution - Exercise 4.3 Part-1
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NCERT Solution - Exercise 4.3 Part-2
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NCERT Solution - Exercise 4.3 Part-3
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NCERT Solution - Exercise 4.3 Part-4
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NCERT Solution - Exercise 4.3 Part-5
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NCERT Solution - Exercise 4.4 Part-1
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NCERT Solution - Exercise 4.4 Part-2
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Solution of Exercise 5.1 (NCERT)-Part-1
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Solution of Exercise 5.1 (NCERT)-Part-2
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Solution of Exercise 5.1 (NCERT)-Part-3.1
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Solution of Exercise 5.1 (NCERT)-Part-3.2
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Solution of Exercise 5.2 (NCERT) Part -1
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Solution of Exercise 5.2 (NCERT) Part -2
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Solution of Exercise 5.2 (NCERT) Part -3
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Solution of Exercise 5.2 (NCERT) Part -4
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Solution of Exercise 5.2 (NCERT) Part -5
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Solution of Exercise 5.2 (NCERT) Part -6
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Solution of Exercise 5.3 (NCERT)Part -1
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Solution of Exercise 5.3 (NCERT)Part - 2
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Solution of Exercise 5.3 (NCERT)Part - 3
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Solution of Exercise 5.3 (NCERT)Part - 4
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Solution of Exercise 5.3 (NCERT)Part - 5
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Solution of Exercise 5.3 (NCERT)Part - 6
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Solution of Exercise 5.3 (NCERT)Part - 7
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Solution of Exercise 5.3 (NCERT)Part - 8
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Solution of NCERT Exercise 5.4 (Optional)
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Introduction : Theorems
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NCERT Solution of Exercise 6.1
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NCERT Solution Exercise 6.2 Part - I
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NCERT Solution Exercise 6.2 Part - II
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NCERT Solution Exercise 6.2 Part - III
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NCERT Solution of Exercise 6.3 Part - III
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NCERT Solution of Exercise 6.4
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NCERT Solution of Exercise 6.5
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NCERT Solution of Exercise 6.5 Part - 2
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NCERT Solution of Exercise 6.6 - Part - 1
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NCERT Solution of Exercise 6.6 - Part - 2
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THEOREM 1:
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Solution of NCERT Exercise 10.1
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Solution of NCERT Exercise 10.2
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Solution of NCERT Exercise 10.2 - Part 2
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Solution of NCERT Exercise 10.2 - part - 3
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Solution of NCERT Exercise 10.2 - part - 4
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Solution of NCERT Exercise 10.2 - part - 5
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Solution of NCERT Exercise 10.2 - part - 6
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