Exercise 1.1

NCERT Solutions for Class 10 Maths
Chapter 1 Real Numbers Ex 1.1

Page 7

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

Answer :

(i) 135 and 225

In this given question, 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have,

225 = 135 × 1 + 90

Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get,

135 = 90 × 1 + 45

Again, 45 ≠ 0, repeating the above step for 45, we get,

90 = 45 × 2 + 0

Since, the remainder is now zero and the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.

Hence, the HCF of 225 and 135 is 45.

(ii) 196 and 38220

In this given question, 38220>196, therefore the by applying Euclid’s division algorithm and taking 38220 as divisor, we get,

38220 = 196 × 195 + 0

We have already got the remainder as 0 here. Therefore, HCF (196, 38220) = 196.

Hence, the HCF of 196 and 38220 is 196.

(iii) 867 and 225

In this given question, 867 is greater than 225. Let us apply now Euclid’s division algorithm on 867, to get,

867 = 225 × 3 + 102

Remainder 102 ≠ 0, therefore taking 225 as divisor and applying the division lemma method, we get,

225 = 102 × 2 + 51

Again, 51 ≠ 0. Now 102 is the new divisor, so repeating the same step we get,

102 = 51 × 2 + 0

Since the remainder is now zero, and the divisor is 51, therefore, HCF (867,225) = HCF(225,102) = HCF(102,51) = 51.

Hence, the HCF of 867 and 225 is 51.

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

Answer :

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

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.

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?

Answer :

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, therefore,

616 = 32 × 19 + 8

Since, 8 ≠ 0, therefore, taking 32 as 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.

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 the form 3m or 3m + 1.]

Answer :

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer q ≥ 0

and r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Or,

a² = (3q)² or (3q + 1)² or (3q + 2)²

a² = (9q)² or 9q² + 6q + 1 or 9q² + 12q + 4

= 3 × (3q²) or 3(3q² + 2q) + 1 or 3(3q² + 4q + 1) + 1

= 3k₁ or 3k₂ + 1 or 3k₃ + 1

Where k₁, k₂, and k₃ are some positive integers

Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.

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.

Answer :

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

∴ a = 3q or 3q + 1 or 3q + 2

Therefore, every number can be represented as these three forms. There are three cases.

a³ = (3q)³ = 27q³ = 9(3q)³ = 9m,

Where m is an integer such that m = 3q³

Case 2: When a = 3q + 1,

a³ = (3q +1)³

a³= 27q³ + 27q² + 9q + 1

a³ = 9(3q³ + 3q² + q) + 1

a³ = 9m + 1

Where m is an integer such that m = (3q³ + 3q² + q)

Case 3: When a = 3q + 2,

a³ = (3q +2)³

a³= 27q³ + 54q² + 36q + 8

a³ = 9(3q³ + 6q² + 4q) + 8

a³ = 9m + 8

Where m is an integer such that m = (3q³ + 6q² + 4q)