**Exercise 1.1**

**Q1. Use Euclid’s division algorithm to find the HCF of:**

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

**Answer **(i) 225 > 135 we always divide greater number with smaller one.

Divide 225 by 135 we get 1 quotient and 90 as remainder so that

225= 135 × 1 + 90

Divide 135 by 90 we get 1 quotient and 45 as remainder so that

135= 90 × 1 + 45

Divide 90 by 45 we get 2 quotient and no remainder so we can write it as

90 = 2 × 45+ 0

As there are no remainder so divisor 45 is our HCF.

**(ii) 38220 > 196 we always divide **greater** number with smaller one.**

Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as

38220 = 196 × 195 + 0

As there is no remainder so divisor 196 is our HCF.

**(iii) 867 > 255 we always divide **greater** number with smaller one.**

Divide 867 by 255 then we get quotient 3 and remainder is 102 so we can write it as

867 = 255 × 3 + 102

Divide 255 by 102 then we get quotient 2 and remainder is 51 so we can write it as

255 = 102 × 2 + 51

Divide 102 by 51 we get quotient 2 and no remainder so we can write it as

102 = 51 × 2 + 0

As there is no remainder so divisor 51 is our HCF.

**Q2. 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 take a as any positive integer and b = 6.

Then using Euclid’s algorithm we get a = 6q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < b and the value of b is 6

So total possible forms will 6q + 0 , 6q + 1 , 6q + 2,6q + 3, 6q + 4, 6q + 5

6q + 0

6 is divisible by 2 so it is a even number

6q + 1

6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number

6q + 2

6 is divisible by 2 and 2 is also divisible by 2 so it is a even number

6q +3

6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number

6q + 4

6 is divisible by 2 and 4 is also divisible by 2 it is a even number

6q + 5

6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number

So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5.

**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?**

**Answer**

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.

**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 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.

**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.**

**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.

Case 1: When a = 3q,

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)

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

or 9m + 8.