140-NCERT-10-1-Real Numbers - Ex-1.1

NCERT

10th Mathematics

Exercise 1.1

Topic: 1 Real Numbers

EXERCISE 1.1

 Q-1. Use Euclid’s division algorithm to find the HCF of :

        (i) 135 and 225          (ii) 196 and 38220           (iii) 867 and 255

The steps to use Euclid’s division algorithm are as follows.

To obtain the HCF of two positive integers, say p and q, with p > q, follow the steps below:  

Step 1: Apply Euclid’s division lemma, to p and q. So, we find whole numbers,

r and s such that p = qr + s, 0 ≤ s < q.      

Step 2: If s = 0, q is the HCF of p and q. If s ≠ 0, apply the division lemma to q and s

Step 3: Continue the process till the remainder is zero. The divisor at this stage

will be the required HCF.

Solution:

(i) 135 and 225

1) Step 1: Since 225 > 135, apply the division lemma to 135 and 225,

225 = (135 × 1) + 90

2) Step 2: Since the remainder is 90 ≠ 0, apply the division lemma to 135 and 90,

135 = (90 × 1) + 45

3) Step 3: Here the remainder of 45 ≠ 0, apply the division lemma to 90 and 45, 

90 = (45 × 2) + 0.

4) Here, the reminder s = 0, and 45 is the divisor, so 45 is the HCF of

135 and 225.

(ii) 196 and 38220

1) Step 1: Since 38220 > 196, apply the division lemma to 196 and 38220,

38220 = (196 × 195) + 0

2) Here, the reminder s = 0, and 195 is the divisor, so 195 is the HCF of 

196 and 38220.

(iii) 867 and 255

1) Step 1: Since 867 > 255, apply the division lemma to 867 and 255, to get

867 = (255 × 3) + 102

2) Step 2: Since the remainder is 102 ≠ 0, apply the division lemma to 255 and

102, to get

255 = (102 × 2) + 51

3) Step 3: Here the remainder of 51 ≠ 0, apply the division lemma to 102 and 51,

102 = (51 × 2) + 0.

4) Here, the reminder s = 0, and 51 is the divisor, we can say that 51 is the HCF

of 867 and 255.

Q-2. Show that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5,

where q is some integer.

Explanation:

1) As per Euclid’s division algorithm. 

2) Let, for any positive integer ‘p’ in the form of 6q + s, where q is some integer 

(as given in the problem), and 's' be a reminder. This means, 0 ≤ s < 6 i.e. 's' may be any integer between 0 and 6, i.e. s = 0 or 1 or 2 or 3 or 4 or 5 but it can’t be 6 because s < 6. 

3) So, by Euclid’s division lemma, possible values for p will be 6q, 6q+1, 6q+2,

6q+3, 6q+4, or 6q+5.

Solution:

1) Let ‘p’ be any positive integer. So according to Euclid’s algorithm, 

for a = 6q + s some integer q ≥ 0 and s = 0, 1, 2, 3, 4, 5 as 0 ≤ s < 6.

2)  ∴ p can be (6q +0), (6q +1), (6q + 2), (6q +3), (6q + 4), or (6q +5).

3) Since p is odd, it can't be (6q +0), (6q + 2), or (6q + 4), as they all are divisible

by 2.

4) We know that odd numbers are of the form (2n+1), for any integer n.

5) Here we can write (6q +1), (6q +3), (6q +5) in the form of (2n+1) as follows:

(6q+1) = 2(3q) + 1, taking 3q = n_1, so we have

   = 2( n₁) + 1 which is an odd number. 

(6q+3) = 2(3q) + 3, 

   = 2(3q) + 2 + 1

   = 2(3q + 1) + 1 taking (3q + 1) = n_2, so we have 

   = 2( n₂) + 1 which is an odd number.    

(6q+5) = 2(3q) + 5, 

   = 2(3q) + 4 + 1

   = 2(3q + 2) + 1 taking (3q + 2) = n_3, so we have 

   = 2( n₃) + 1 which is an odd number.

6) Therefore, any odd integer is of the form (6q +1), (6q +3), (6q +5).

Q-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?

Explanation:

1) We know that HCF means: HCF is the highest common factor, which can be

divided exactly into two or more numbers.

2) Here we have to find HCF of 616 and 32 to get the maximum number of columns

in which, they can march.

Solution:

1) Here we will find HCF of (616, 32)

2) Using Euclid’s algorithm,

Step 1: Since 616 > 32,

616 = (32 × 19) + 8

Step 2: Since the remainder, 8 ≠ 0,

32 = (8 × 4) + 0

Step 3: Here the remainder is 0, so HCF of 616 and 32 is 32.

3) The maximum number of columns they can march is 32.

Q-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 be rewritten in form 3m or 3m + 1.]

Explanation:

1) Let 'c' be a positive integer. Using Euclid’s lemma, for positive integers c and d,

there exist unique integers q and s, such that c = dq + s, 0 ≤ s < d.

2) If we keep the value of d = 3, then 0 ≤ s < 3, i.e. s = 0, 1, or 2. It can’t be 3

because s<3. 

3) So, the possible values for c = 3q, (3q + 1), or (3q + 2). 

4) Now, find the square of all the possible values of c. 

5) If q is any positive integer, its square will also be a positive integer. 

Solution:

1) Let 'c' be any positive integer and d = 3.

2) So, c = 3q + s for some integer q ≥ 0 and s = 0,1,2 as 0 ≤ s < 3.

3) ∴ c = (3q + 0), (3q + 1), (3q + 2)

4) Now we will find c² , for c = (3q + 0), (3q + 1), (3q + 2)

c² = (3q + 0)²

= 9 q² 

= 3 (3 q²)  take (3 q²) = m

= (3 m)

c² = (3q + 1)²

= 9 q² + 6q + 1

= 3 (3 q² + 2q) + 1   take (3 q² + 2q) = m

= (3 m +1)

c² = (3q + 2)²

= 9 q² + 12q + 4

= 3 (3 q² + 4q +1) + 1   take (3 q² + 2q + 1) = m

= (3 m +1)

5) So, the square of any positive integer is of form 3m or (3m + 1) for some integer m.

Q-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.

Explanation:

1) Let 'c' be a positive integer. Using Euclid’s lemma, for positive integers c and d,

there exist unique integers q and s, such that c = dq + s, 0 ≤ s < d.

2) If we keep the value of d = 3, then 0 ≤ s < 3, i.e. s = 0, 1, or 2. It can’t be 3

because s<3. 

3) So, the possible values for c = 3q, (3q + 1), or (3q + 2). 

4) Now, find the cube of all the possible values of c. 

5) If q is any positive integer, its square will also be a positive integer. 

Solution:

1) Let 'c' be any positive integer and d = 3.

2) So, c = 3q + s for some integer q ≥ 0 and s = 0,1,2 as 0 ≤ s < 3.

3) ∴ c = (3q + 0), (3q + 1), (3q + 2)

4) Now we will find c³ , for c = (3q + 0), (3q + 1), (3q + 2)

c³ = (3q + 0)³

= 27 q³ 

= 9 (3 q³)  take (3 q³) = m

= (9 m)

c³ = (3q + 1)³

= 27 q³ +27 q² + 9q + 1

= 9 (3 q³ + 3 q² + q) + 1   take (3 q² + 3 q² + q) = m

= (9 m +1)

c³ = (3q + 2)³

= 27 q³ +54 q² + 36q + 8

= 9 (3 q³ + 6 q² + 4q) + 8   take (3 q³ + 6 q² + 4q) = m

= (9 m + 8)

5) So, the cube of any positive integer is of the form (9m), (9m + 1), or (9m + 8) for some integer m.

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