# Real Numbers – Exercise 3.1 – Class X

#### Real Numbers – Exercise 3.1

1. Use Euclid’s division algorithm to find the HCF of the following numbers.

(i) 65 and 117

(ii) 237 and 81

(iii) 55 and 210

(iv) 305 and 793

1. Show that any positive even integer is of the form 4q or 4q + 1 , where q is a whole number.
2. 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, but not of the form 3m + 2
3. Prve that the prodct of three consecutive positive integers is divisible by 6
4. There are 75 roses and 45 lily flowers. These are to be made into bouquets containing both the flowers. All the bouquest should contain the same number of flowers . Find the number of flowers in them.
5. The length and breadth of a rectangular field is 110m and 30m respectively. Calculate the length of the longest rod which can measure the length and breadth of the field exactly.

#### Real numbers – Exercise 3.1  – Solutions

1. Use Euclids division algorithm to find the HCF of the following numbers.

(i) 65 and 117

(ii) 237 and 81

(iii) 55 and 210

(iv) 305 and 793

Solution:

(i) 65 and 117

Euclid’s division lemma stats that, for given positive integers a and b there exists unique integers q and r satisfying a = bq + r , 0 ≤ r < b.

By Euclid’s division algorithm,

Let us start from larger number , then we have

117 = 65 x 1 + 52

65 = 52 x 1 + 13

52 = 13 x 4 + 0

Therefore, HCF  of 117 and 65 is 13.

(ii) 237 and 81

Euclid’s division lemma stats that, for given positive integers a and b there exists unique integers q and r satisfying a = bq + r , 0 ≤ r < b.

By Euclid’s division algorithm,

Let us start from larger number , then we have

237 = 81 x 2 + 75

81 = 75 x 1 + 6

75 = 6 x 12 + 3

6 = 3×2 + 0

Therefore, HCF of 237 and 81 is 6

(iii) 55 and 210

Euclid’s division lemma stats that, for given positive integers a and b there exists unique integers q and r satisfying a = bq + r , 0 ≤ r < b.

By Euclid’s division algorithm,

Let us start from larger number , then we have

210 = 55 x 3 + 45

55 = 45 x 1+ 10

45 = 10 x 4 + 5

5 = 5 x 1 + 0

Therefore, HCF of 210 and 55 is 5

(iv) 305 and 793

Euclid’s division lemma stats that, for given positive integers a and b there exists unique integers q and r satisfying a = bq + r , 0 ≤ r < b.

By Euclid’s division algorithm,

Let us start from larger number , then we have

793 = 305 x 2 + 183

305 = 183 x 1 + 122

183 = 122 x 1 + 61

122 = 61 x 2 + 0

Therefore, HCF of 793 and 305 is 61

1. Show that any positive even integer is of the form 4q or 4q + 1 , where q is a whole number

Solution:

Let a be any positive integer.

Euclid’s division lemma stats that, for given positive integers a and b there exists unique integers q and r satisfying a = bq + r , 0 ≤ r < b.

By Euclid’s division algorithm, we have,

Let b = 4, then a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3

So the possible values of ‘a’  can be, a = 4q or 4q + 1 or 4q + 2 or 4q + 3 , since r = 0, 1, 2, 3

Since a is an even number. Then, 4q + 1 or  4q + 3 as they are both not divisible be 2.

Therefore,  a is an even positive integer.

1. 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, but not of the form 3m + 2

Solution:

Let a be any positive integer.

Euclid’s division lemma stats that, for given positive integers a and b there exists unique integers q and r satisfying a = bq + r , 0 ≤ r < b.

Let a and b be two positive integers and a > b

a  = (b x q) + r where q and r are positive integers and 0 ≤ r < b

Let b = 3

Therefore, a = 3q + r, where 0 ≤ r < 2

(i) if r  = 0 then a = 3q

(ii) If r = 1 then a = 3q + 1

(iii) if r = 2 then a = 3q + 2

Consider,

(i) if r = 0 then a = 3q

a2 = (3q)2 = 9q2 = 3(3q2), where 3q2 = m and m is an integer.[If we divide by 3 we get no remainder]

(ii) if  r = 1 then a =  3q  + 1

a2 =  (3q  + 1)2 = (3q)2 + 2(3q) + 1 = 9q2 + 6q + 1 , now divide this by 3.

We get remainder 1. So we can write it as 3(3q2 + 2q) + 1 = 3m + 1 , where m = 3q2 + 1

(iii) if r = 2 then a = 3q  + 2

a2 = (3q  + 2)2 =  9q2 +  2(2)(3q) + 4 = 9q2 + 12q +  4

Now if we divide  it  by 3, then we get 3(3q2 +  4q) +  4 , where remainder is 4. Therefore, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m, but not of the form 3m + 2.

1. Prove that the product of three consecutive positive integers is divisible by 6

Solution:

Let three consecutive integers be n , n + 1, n + 2.

Whenever a number is divided by 3 the remainder obtained is either 0 or 1 or 2.

Let n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So that n, n + 1 and n + 2 is always divisible by 3.
⇒ n (n + 1) (n + 2) is divisible by 3.

Similarly, whenever a number is divided 2 we will get the remainder is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.
If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So that n, n + 1 and n + 2 is always divisible by 2.
⇒ n (n + 1) (n + 2) is divisible by 2.

But n (n + 1) (n + 2) is divisible by 2 and 3.

∴ n (n + 1) (n + 2) is divisible by 6

1. There are 75 roses and 45 lily flowers. These are to be made into bouquets containing both the flowers. All the bouquets should contain the same number of flowers . Find the number of flowers in them.

Solution:

Given, there are 75 roses and 45 lily flowers.

75 = 45 x 1 + 30

45 = 30 x 1 + 15

15 = 3 x 5 + 0

Therefore, HCF of 75 and 45 is 45.

Thus, the number of bouquets that can be formed = 15

1. The length and breadth of a rectangular field is 110m and 30m respectively. Calculate the length of the longest rod which can measure the length and breadth of the field exactly.

Solution:

Given the length and breadth of a rectangular field is 110m and 30m respectively.

We have to calculate the length of the longest rod which can measure the length and breadth of the field exactly.

Resolving 110 and 30 into the prime factors. We get,

110 = 30 x 3 + 20

30 = 20 x 1 + 10

10 = 5 x 2 + 0

Therefore, the length of the longest rod which can measure the length and breadth  is10 m.