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NCERT Solutions: Real Numbers (Exercise 1.1) Notes | Study Mathematics (Maths) Class 10 - Class 10

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Q1. Use Euclid’s division algorithm to find the HCF of
(i) 135 and 225 
(ii) 196 and 38220 
(iii) 867 and 255
Ans: 

(i) HCF of 135 and 225
Step 1: Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
⇒ 225 = 135 × 1 + 90
Step 2: Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
 135 = 90 × 1 + 45
Step 3: We consider the new divisor 90 and new remainder 45 and apply the division lemma to obtain: 90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.

(ii) HCF of 196 and 38220
Step 1: Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
⇒ 38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.

(iii) HCF of 867 and 255
Step 1: Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
⇒ 867 = 255 × 3 + 102
Step 2: Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
⇒ 255 = 102 × 2 + 51
Step 3: We consider the new divisor 102 and new remainder 51 and apply the division lemma to obtain
⇒ 102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51.


Q2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Ans: Let us consider a positive odd integer as ‘a’.
On dividing ‘a’ by 6, Let ‘a’ be the quotient and ‘r’ be the remainder.
∴ Using Euclid’s lemma, we have:
a = 6q + r where 0 ≤ r < 6 i.e., r = 0, 1, 2, 3, 4, or 5

Now substituting the value of r, we get:
If r = 0, then a = 6q
Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.
If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2.
A positive integer can be either even or odd.
Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.


Q3. 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?
Ans: Given,
Number of army contingent members = 616
Number of army band members = 32
If the two groups have to march in the same column, we have to find out the highest common factor between the two groups.
HCF(616, 32) gives the maximum number of columns in which they can march.

By Using Euclid’s algorithm to find their HCF, we get,
Since, 616 > 32
∴ 616 = 32 × 19 + 8
Since, 8 ≠ 0, therefore, taking 32 as the new divisor, we have,
32 = 8 × 4 + 0
Now we have got remainder as 0, therefore, HCF (616, 32) = 8.
Hence, the maximum number of columns in which they can march is 8.


Q4. Use Euclid’s division lemma to show that the square of any positive integer is either of 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 be rewritten in form 3m or 3m + 1.]

Ans: Let us consider an arbitrary positive integer as ‘x’ such that it is of the form 3q, (3q + 1) or (3q + 2).

Note:
For any positive integer x and b = 3,
x = 3q + r, where q is quotient and r is remainder such that 0 < r < 3.
If r = 0 then x = 3q
If r = 1 then x = 3q + 1
If r = 2 then x = 3q + 2

For x = 3q, we have
x2 = (3q)2
⇒ x2 = 9q2
= 3 (3q2)
= 3m ...(1) (Putting 3q2 = m where m is an integer)

For x = 3q + 1
x2 = (3q + 1)2
= 9q2 + 6q + 1
= 3 (3q2 + 2q) + 1
= 3m + 1 ...(2) (Puting 3q2 + 2q = m, where m is an integer)

For x = 3q + 2
x2 = (3r + 2)2
= 9q2 + 12q + 4 

= (9q2 + 12q + 3) + 1
= 3 (3q2 + 4q + 1) + 1
= 3m + 1 ...(3) (Putting 3q2 + 4q + 1 = m, where m is an integer)
From (1), (2) and (3)
x2 = 3m or 3m + 1
Thus, the square of any positive integer is either of form 3m or 3m + 1 for some integer m.


Q5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Ans: Let us consider an arbitrary positive integer x such that it is in the form of 3q, (3q + 1) or (3q + 2).

Note:
For any positive integer x and b = 3,
x = 3q + r, where q is quotient and r is remainder such that 0 < r < 3.
If r = 0 then x = 3q
If r = 1 then x = 3q + 1
If r = 2 then x = 3q + 2

∴ x is of the form 3q, (3q + 1) or (3q + 2).

For x = 3q
x3 = (3q)3
= 27q3
= 9 (3q3)
= 9m ...(1) (Putting 3q3 = m, where m is an integer)

For x = 3q + 1
x3 = (3q + 1)3
= 27q3 + 27q2 + 9q + 1
= 9 (3q3 + 3q2 + q) + 1
= 9m + 1 ...(2) (Putting (3q3 + 3q2 + 1) = m, where m is an integer)

For x = 3q + 2
x3 = (3q + 2)3
= 27q3 + 54q2 + 36q + 8
= 9 (3q3 + 6q2 + 4q) + 8
= 9m + 8 ...(3) (Putting (3q3 + 6q2 + 4q) = m, where m is an integer)
From (1), (2), (3) we have:
x3 = 9m, (9m + 1) or (9m + 8)
Thus, x3 of any positive integer can be in the form 9m, (9m + 1) or (9m + 8).


Check out the NCERT Solutions of all the exercises of Real Numbers:

Exercise 1.2 NCERT Solutions: Real Numbers

Exercise 1.3 NCERT Solutions: Real Numbers

Exercise 1.4 NCERT Solutions: Real Numbers

The document NCERT Solutions: Real Numbers (Exercise 1.1) Notes | Study Mathematics (Maths) Class 10 - Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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