(Class X)

Exercise 1.1 

Question 1: 

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

Answer 1: 

(i) 135 and 225 

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain 

225 = 135 × 1 + 90 

Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to 

obtain 

135 = 90 × 1 + 45 

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) 196 and 38220 

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) 867 and 255 

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain 

867 = 255 × 3 + 102 

Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 

to obtain 

255 = 102 × 2 + 51 

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. 

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.

Answer 2: 

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. 

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 

Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer 

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer 

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer 

Clearly, 

6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer. 

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. 

Hence, these expressions of numbers are odd numbers. 

And therefore, any odd integer can be expressed in the form 6q + 1, or 

6q + 3, or 6q + 5 



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?

Answer 3: 

HCF (616, 32) will give the maximum number of columns in which they 

can march. 

We can use Euclid’s algorithm to find the HCF. 

616 = 32 × 19 + 8 

32 = 8 × 4 + 0 

The HCF (616, 32) is 8. 

Therefore, they can march in 8 columns each. 


Question 4: 

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 the form 3m or 3m + 1.] Answer 4: 

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 

= (3q)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. 



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. 

Answer 5: 

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. 

Case 1: When a = 3q, 

a3 = (3q)3 = 27q3 = 9(3q3)= 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) 

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 

9m + 8.