**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.**

**Solution:**

**Show: a = 6q + 1, 6q+3 or 6q+5**

Let a be any positive odd integer; where b = 6,

When we divide a by 6 we get remainders 0, 1, 2, 3, 4 and 5 respectively;

where 0 ≤ r < b

Here a is an odd number, so the remainder is also an odd number.

remainder will be 1 or 3 or 5

Using Euclid division algorithm we get;

a = 6q + 1, 6q+3 or 6q+5

**Q3. In a parade, an army contingent of 616 members is to march behind an army band of 32 members. Both the groups are to march in equal number of columns. What is the maximum number of columns in which they can march?**

**Solution:**

**Maximum number of columns = HCF(616, 32)**

a = 616, b = 32 {let the largest number be a and the smallest number be b}

using Euclid’s division algorithm

a = bq + r (then)

616 = 32 ×19 + 8 {when we get r=0 we stop solving further}

32 = 8 × 4 + 0

b = 8 {value of b is HCF}

HCF = 8

Hence maximum number of columns = 8

**Question 4. Using Euclid’s division lemma, show that the square of any positive integer is of the form 3m or 3m + 1 for some integer m.**

**Solution:**

Show that : a^{2} = 3m or 3m + 1

*a = bq + r*

Let a be any positive integer where b = 3 and r = 0, 1, 2 because 0 ≤ r < 3

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

Therefore, a = 3q + 0 or 3q + 1 or 3q + 2

Now we get;

⇒ a^{2} = (3q + 0)^{2} or (3q + 1)^{2} or (3q +2)^{2}

⇒ a^{2} = 9q^{2} or 9q^{2} + 6q + 1 or 9q^{2} + 12q + 4

⇒ a^{2} = 9q^{2} or 9q^{2} + 6q + 1 or 9q^{2} + 12q + 3 + 1

⇒ a^{2} = 3(3q^{2}) or 3(3q^{2} + 2q) + 1 or 3(3q^{2} + 4q + 1) + 1

If m = (3q^{2}) or (3q^{2} + 2q) or (3q^{2} + 4q + 1) हो तो

We get That;

*a*^{2} = 3m or 3m + 1 or 3m + 1

**Q5. Using Euclid’s division lemma, show that the cube of a positive integer is of the form 9m, 9m + 1 or 9m + 8.**

**Solution:**

Let a be any positive integer;

Using Euclid’s division lemma;

a = bq + r where; 0 ≤ r < b

By keeping b = 9

a = 9q + r where; 0 ≤ r < 9

When r = 0;

a = 9q + 0 = 9q

*a*^{3} = (9q)^{3} = 9(81q^{3}) *or **9m **where **m = 81q*^{3}

When r = 1

*a = 9q + 1 *

*a*^{3} = (9q + 1)^{3} = 9(81q^{3} + 27q^{2} + 3q) + 1

* = 9m + 1 **where **m = 81q*^{3} + 27q^{2} + 3q

when r = 2

*a = 9q + 2 *

*a*^{3} = (9q + 2)^{3} = 9(81q^{3} + 54q^{2} + 12q) + 8

* = 9m + 2 **where **m = 81q*^{3} + 54q^{2} + 12q

Thus, the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.