# R.D. Sharma Solutions Class 10th: Ch 1 Real Numbers Exercise 1.1

#### Chapter 1 Real Numbers R.D. Sharma Solutions for Class 10th Math Exercise 1.1

**Exercise 1.1**

**Level 1**

1. If a and b are two odd positive integers such that a > b, then prove that .... and the other is even.

**Solution**

2. Prove that the product of two .... is divisible by 2.

**Solution**

∴ Their product = n(n – 1) = n

^{2}− n

We know that any positive integer is of the form 2q or 2q + 1, for some integer q

When n =2q,

n

^{2}− n = (2q)

^{2}− 2q = 4q

^{2}− 2q

=2q(2q − 1)

Then n

^{2}− n is divisible by 2.

When n = 2q + 1,

n

^{2}− n = (2q + 1)

^{2}− (2q + 1) = 4q

^{2}+ 4q + 1 − 2q − 1

= 4q

^{2}+ 2q

= 2q(2q + 1)

Then n

^{2}− n is divisible by 2.

Hence the product of two consecutive positive integers is divisible by 2.

3. Prove that the product of three .... divisible by 6.

**Solution**

Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or, 6q + 3 or 6q + 4 or 6q + 5.

If n = 6q, then

n(n + 1)(n + 2) = (6q + 1)(6q + 2)(6q + 3)

= 6[(6q + 1)(3q + 1)(2q + 1)]

= 6m (divisible by 6)

If n = 6q + 1, then

n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4)

= 6[(6q + 1)(3q + 1)(2q + 1)]

= 6m (divisible by 6)

If n = 6q + 2, then

n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4)

= 6[(3q + 1)(2q + 1)(6q + 4)]

= 6m (divisible by 6)

If n = 6q + 3, then

n(n + 1)(n + 2) = (6q + 3)(6q + 4)(6q + 5)

= 6[(6q + 1)(3q + 2)(2q + 5)]

= 6m (divisible by 6)

If n = 6q + 4, then

n(n + 1)(n + 2) = (6q + 4)(6q + 5)(6q + 6)

= 6[(6q + 4)(3q + 5)(2q + 1)]

= 6m (divisible by 6)

If n = 6q + 5, then

n(n + 1)(n + 2) = (6q + 5)(6q + 6)(6q + 7)

= 6[(6q + 5)(q + 1)(6q + 7)]

= 6m (divisible by 6)

Hence, the product of three consecutive positive integer is divisible by 6.

4. For any positive integer n .... divisible by 6.

**Solution**

We have n

^{3}− n = n(n

^{2}− 1) = (n − 1) (n) (n + 1)

Let, n be any positive integer. Since any positive integer is of the form 6q or 6q + 1 or, 6q +2 or, 6q + 3 or, 6q + 4 or, 6q + 5.

If n = 6q, then

(n − 1)(n)(n + 1) = (6q − 1)(6q)(6q + 1)

= 6[(6q − 1)(q)(6q + 1)]

= 6m (divisible by 6)

If n = 6q + 1, then

(n − 1)(n + 1) = (6n)(6q + 1)(6q + 2)

= 6[(q)(6q + 1)(6q + 2)]

= 6m (divisible by 6)

If n = 6q + 2, then

(n − 1)(n)(n + 1) = (6𝑞 + 1)(6𝑞 + 2)(6𝑞 + 3)

= 6[(6q + 1)(3q + 1)(2q + 1)]

= 6m (divisible by 6)

If n = 6q + 3, then

(n − 1)(n)(n + 1) = (6q + 3)(6q + 4)(6q + 5)

= 6[(3q + 1)(2q + 1)(6q + 4)]

= 6m (divisible by 6)

If n = 6q + 4, then

(n − 1)(n)(n + 1) = (6q + 3)(6q + 4)(6q + 5)

= 6[(2q + 1)(3q + 2)(6q + 5)]

= 6m (divisible by 6)

If n = 6q + 5, then

(n − 1)(n)(n + 1) = (6q + 4)(6q + 5)(6q + 6)

= 6[(6q + 4)(6q + 5)(q + 1)]

= 6m (divisible by 6)

Hence, for any positive integer n, n

^{3}– n is divisible by 6.

5. Prove that if a positive integer is of the form 6q + 5 .... but not conversely.

**Solution**

Let, n = 6q + 5, when q is a positive integer

We know that any positive integer is of the form 3k, or 3k + 1, or 3k + 2

∴ q = 3k or 3k + 1, or 3k + 2

If q = 3k then,

n = 6q + 5

= 6(3k) + 5

= 18k + 5

= 18k + 3 + 2

= 3(6k + 1) + 2

= 3m + 2 (m is any integer)

If q = 3k + 1 then,

n = 6q + 5

= 6(3k + 1) + 5

= 18k + 6 + 5

= 18k + 11

= 3(6k + 3) + 2

= 3m + 2 (m is any integer)

If q = 3k + 2 then,

n = 6q + 5

= 6(3k + 2) + 5

= 18k + 12 + 5

= 18k + 17

= 3(6k + 5) + 2

= 3m + 2 (m is any integer)

Hence, if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for some integer q.

Conversely,

Let n = 3q + 2

We know that a positive integer can be of the form 6k + 1, 6k + 2, 6k + 3, 6k + 4 or 6k + 5

If q = 6k + 1 then,

n = 3(6k + 1) + 2

= 18k + 5

= 6(3k) + 5

= 6m + 5 (m is any integer)

If q = 6k + 2 then,

n = 3(6k + 2) + 2

= 18k + 8

= 6 (3k + 1) + 2

= 6m + 2 (m is any integer)

Now, this is not of the form 6m + 5

Hence, if n is of the form 3q + 2, then it necessarily won’t be of the form 6q + 5 always.

6. Prove that the square of any .... 5q + 1 is of the same form.

**Solution**

∴ n

^{2}= (5n + 1)2

= 25q

^{2}+ 10q + 1

= 5(5q

^{2}+ 2q) + 1

= 5m + 1 (m is any integer)

Hence, the square of any positive integer of the form 5q + 1 is of the same form

7. Prove that the square of any .... 3m or, 3m + 1 but not of the form 3m +2.

**Solution**

By Euclid’s division algorithm,

a = bq + r, where 0 ≤ r ≤ b

Put b = 3

a = 3q + r, where 0 ≤ r ≤ 3

If r = 0, then a = 3q

If r = 1, then a = 3q + 1

If r = 2, then a = 3q + 2

Now,

(3q)

^{2}= 9q

^{2}

= 3 × 3q

^{2}

= 3m (m is any integer)

(3q + 1)

^{2}= (3q)

^{2}+ 2(3q)(1) + (1)

^{2}

= 9q

^{2}+ 6q + 1

= 3(3q

^{2}+ 2q) + 1

= 3m + 1 (m is any integer)

(3q + 2)

^{2}= (3q)

^{2}+ 2(3q)(2) + (2)

^{2}

= 9q

^{2}+ 12q + 4

= 9q

^{2}+ 12q + 4

= 3(3q

^{2}+ 4q + 1) + 1

= 3m + 1 m is any integer)

Hence the square of any positive integer is of the form 3m, or 3m+1 But not of the form 3m+2

8. Prove that the square of any positive .... 4q or 4q + 1 for some integer q.

**Solution**

a = bm + r, where 0 ≤ r ≤ b

Put b = 4

a = 4m + r, where 0 ≤ r ≤ 4

If r = 0, then a = 4m

If r = 1, then a = 4m + 1

If r = 2, then a = 4m + 2

If r = 3, then a = 4m + 3

Now,

(4m)

^{2}= 16m

^{2}

= 4 × 4m

^{2}

= 4q (q is any integer)

(4m + 1)

^{2}= (4m)

^{2}+ 2(4m)(1) + (1)

^{2}

= 16m

^{2}+ 8m + 1

= 4(4m

^{2}+ 2m) + 1

= 4q + 1 (q is any integer)

(4m + 2)

^{2}= (4m)

^{2}+ 2(4m)(2) + (2)

^{2}

= 16m

^{2}+ 24m + 9

= 16m

^{2}+ 24m + 8 + 1

= 4(4m

^{2}+ 6m + 2) + 1

= 4q + 1 (q is any integer)

Hence,the square of any positive integer is of the form 4q or 4q+1 for some integer m

By Euclid’s division algorithm,

a = bm + r, where 0 ≤ r ≤ b

Put b = 5

a = 5m + r, where 0 ≤ r ≤ 4

If r = 0, then a = 5m

If r = 1, then a = 5m + 1

If r = 2, then a = 5m + 2

If r = 3, then a = 5m + 3

If r = 4, then a = 5m + 4

Now,

(5m)

= 5(5m

= 5q (q is any integer)

9. Prove that the square of any positive .... 5q, 5q + 1, 5q + 4 for some
integer q.

**Solution**

a = bm + r, where 0 ≤ r ≤ b

Put b = 5

a = 5m + r, where 0 ≤ r ≤ 4

If r = 0, then a = 5m

If r = 1, then a = 5m + 1

If r = 2, then a = 5m + 2

If r = 3, then a = 5m + 3

If r = 4, then a = 5m + 4

Now,

(5m)

^{2}= 25m^{2}= 5(5m

^{2})= 5q (q is any integer)

(5m + 1)

^{2}= (5m)

^{2}+ 2(5m)(1) + (1)

^{2}

= 25m

= 5(5m

= 5q + 1 (q is any integer)

(5m + 2)

= 25m

= 5(5m

= 5q + 4 (q is any integer)

(5m + 3)

= 25m

= 25m

= 5(5m

^{2}+ 10m + 1= 5(5m

^{2}+ 2m) + 1= 5q + 1 (q is any integer)

(5m + 2)

^{2}= (5m)^{2}+ 2(5m)(2) + (2)^{2}= 25m

^{2}+ 20m + 4= 5(5m

^{2}+ 4m) + 4= 5q + 4 (q is any integer)

(5m + 3)

^{2}= (5m)^{2}+ 2(5m)(3) + (3)^{2}= 25m

^{2}+ 30m + 9= 25m

^{2}+ 30m + 5 + 4= 5(5m

^{2}+ 6m + 1) + 4
= 5q + 1 (q is any integer)

(5m + 4)

^{2}= (5m)

^{2}+ 2(5m)(4) + (4)

^{2}

= 25m

^{2}+ 40m + 16

= 25m

= 5(5m

= 5q + 1 (q is any integer)

^{2}+ 40m + 15 + 1= 5(5m

^{2}) + 2(5m)(4) + (4)^{2}= 5q + 1 (q is any integer)

Hence, the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.

10. Show that the square of an odd ..... 8q + 1, for some integer q.

**Solution**

By Euclid’s division algorithm

a = bq + r, where 0 ≤ r ≤ b

Put b = 4

a = 4q + r, where 0 ≤ r ≤ 4

If r = 0, then a = 4q even

If r = 1, then a = 4q + 1 odd

If r = 2, then a = 4q + 2 even

If r = 3, then a = 4q + 3 odd

Now,

(4q + 1)

^{2}= (4q)^{2}+ 2(4q)(1) + (1)^{2}
= 16q

^{2}+ 8q + 1
= 8(2q

^{2}+ q) + 1
= 8m + 1 where m is some integer.

Hence the square of an odd integer is of the form 8q + 1, for some integer q

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

**Solution**

Let a be any odd positive integer we need to prove that a is of the form 6q + 1, or 6q +3, 6q + 5, where q is any integer.

Since a is an integer consider b = 6 another integer applying Euclid’s division lemma we get,

a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 since 0 ≤ r < 6.

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

However since a is odd so a cannot take the values 6q, 6q + 2 and 6q + 4

(since all these are divisible by 2)

Also, 6q + 1 = 2 × 3q + 1 = 2k 1 + 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 odd numbers.

Therefore, any odd integer can be expressed is of the form 6q + 1, or 6q + 3, 6q + 5 where q is any integer

Concept insight: In order to solve such problems Euclid’s division lemma is applied to two integers a and b the integer b must be taken in accordance with what is to be proved, for example here the integer b was taken 6 because a must be of the form 6q + 1, 6q +3, 6q + 5. Basic definition of even (divisible by 2) and odd numbers (not divisible by 2) and the fact that addiction and multiplication of integers is always an integer are applicable here.

**Level 2**

12. Show that square of any positive integer cannot be of the form 6m + 2 or 6m +5 for any integer m.

**Solution**

By Euclid’s division algorithm,

a = bq + r, where 0 ≤ r ≤ b

Put b = 5

a = 5m + r, where 0 ≤ r ≤ 4

If r = 0, then a = 5q

If r = 1, then a = 5q + 1

If r = 2, then a = 5q + 2

If r = 3, then a = 5q + 3

If r = 4, then a = 5q + 4

Now,

(6q)

= 6(6q

= 6m (m is any integer)

a = bq + r, where 0 ≤ r ≤ b

Put b = 5

a = 5m + r, where 0 ≤ r ≤ 4

If r = 0, then a = 5q

If r = 1, then a = 5q + 1

If r = 2, then a = 5q + 2

If r = 3, then a = 5q + 3

If r = 4, then a = 5q + 4

Now,

(6q)

^{2}= 36q^{2}= 6(6q

^{2})= 6m (m is any integer)

(6q + 1)

^{2}= (6q)

^{2}+ 2(6q)(1) + (1)

^{2}

= 36q

= 6(6q

= 6m + 1 (m is any integer)

^{2}+ 12q + 1= 6(6q

^{2}+ 2q) + 1= 6m + 1 (m is any integer)

(6q + 2)

^{2}= (6q)^{2}+ 2(6q)(2) + (2)^{2}
= 36q

= 6(6q

= 6m + 4 (m is any integer)

^{2}+ 24q + 4= 6(6q

^{2}+ 4q) + 4= 6m + 4 (m is any integer)

(6q + 3)

^{2}= (6q)^{2}+ 2(6q)(3) + (3)^{2}
= 36q

= 6(6q

= 6m + 3 (m is any integer)

^{2}+ 36q + 9= 6(6q

^{2}+ 6q + 1) + 3= 6m + 3 (m is any integer)

(6q + 4)

^{2}= (6q)^{2}+ 2(6q)(4) + (4)^{2}
= 36q

= 6(6q

= 6m + 4 (m is any integer)

^{2}+ 48q + 16= 6(6q

^{2}+ 8q + 2) + 4= 6m + 4 (m is any integer)

(6q + 5)

^{2}= (6q)^{2}+ 2(6q)(5) + (5)^{2}
= 36q

= 6(6q

= 6m + 1 (m is any integer)

^{2}+ 60q + 25= 6(6q

^{2}+ 10q + 4) + 1= 6m + 1 (m is any integer)

Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5.

13. Show that the cube of a positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.

**Solution**
By Euclid’s division algorithm,

a = bq + r, where 0 ≤ r ≤ b

Put b = 6

a = 6m + r, where 0 ≤ r ≤ 5

If r = 0, then a = 6m

If r = 1, then a = 6m + 1

If r = 2, then a = 6m + 2

If r = 3, then a = 6m + 3

If r = 4, then a = 6m + 4

If r = 5, then a = 6m + 5

Now,

(6m)

= 6(36m

= 6q (q is any integer)

a = bq + r, where 0 ≤ r ≤ b

Put b = 6

a = 6m + r, where 0 ≤ r ≤ 5

If r = 0, then a = 6m

If r = 1, then a = 6m + 1

If r = 2, then a = 6m + 2

If r = 3, then a = 6m + 3

If r = 4, then a = 6m + 4

If r = 5, then a = 6m + 5

Now,

(6m)

^{3}= 216m^{3}= 6(36m

^{3})= 6q (q is any integer)

(6m + 1)

^{3}= (6m)^{3}+108m^{2}+ 18m + (1)^{3}
= 6(36m

= 6q + 1 (q is any integer)

^{3}+ 18m^{2}+ 3m) + 1= 6q + 1 (q is any integer)

(6m + 2)

^{3}= (6m)^{3}+ 216m^{2}+ 72m + (2)^{3}
= 216m

= 6(36m

= 6q + 2 (q is any integer)

^{3}+ 216m^{2}+ 72m + 6 + 2= 6(36m

^{3}+ 36m^{2}+ 12m + 1) + 2= 6q + 2 (q is any integer)

(6m + 3)

^{3}= (6m)^{3}+ 324m^{2}+ 162m + (3)^{3}
= 216m

= 6(36m

= 6q + 3 (q is any integer)

^{3}+ 324m^{2}+ 162m + 24 + 3= 6(36m

^{3}+ 54m^{2}+ 27m + 4) + 3= 6q + 3 (q is any integer)

(6m + 4)

^{3}= (6m)^{3}+ 432m^{2}+ 288m + (4)^{3}
= 216m

= 6(36m

= 6q + 4 (q is any integer)

Hence, the cube of any positive integer can be of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.

^{3}+ 432m^{2}+ 288m + 60 + 4= 6(36m

^{3}+ 72m^{2}+ 48m + 10) + 4= 6q + 4 (q is any integer)

(6m + 5)

^{3}= (6m)^{3}+ 432m^{2}+ 288m + (4)^{3}
= 216m

= 6(36m

= 6q + 5 (q is any integer)

^{3}+ 540m^{2}+ 450m + 120 + 5= 6(36m

^{3}+ 90m^{2}+ 75m + 20) + 5= 6q + 5 (q is any integer)

Hence, the cube of any positive integer can be of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.

14. Show that one and only one out of n, n+4, n+8, n+12 and n+16 is divisible by 5, where n is any positive integer.

Consider the numbers n, (n + 4), (n + 8), (n + 12) and (n + 16) , where n is any positive integer. Suppose n = 5q + r, where 0 ≤ r < 5

n = 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4

(By Euclid's division algorithm)

Case 1:

When n = 5q.

n = 5q is divisible by 5.

n + 4 = 5q + 4 is not divisible by 5.

n + 8 = 5q + 5 + 5 + 3 = 5(q + 1) + 3 is not divisible by 5.

n + 12 = 5q + 10 + 2 = 5(q + 2) + 2 is not divisible by 5.

n + 16 = 5q + 15 + 1 = 5(q + 3) + 1 is not divisible by 5.

Case 2:

When n = 5q + 1.

n = 5q + 1 is not divisible by 5.

n + 4 = 5q + 1 + 4 = 5(q + 1) is divisible by 5.

n + 8 = 5q + 1 + 5 + 3 = 5(q + 1) + 4 is not divisible by 5.

n + 12 = 5q + 1 + 12 = 5(q + 2) + 3 is not divisible by 5.

n + 16 = 5q + 1 + 16 = 5(q + 3) + 2 is not divisible by 5.

Case 3:

When n = 5q + 2.

n = 5q + 2 is not divisible by 5.

n + 4 = 5q + 2 + 4 = 5(q + 1) + 1 is not divisible by 5.

n + 8 = 5q + 2 + 8 = 5(q + 2) is divisible by 5.

n + 12 = 5q + 2 + 12 = 5(q + 2) + 4 is not divisible by 5.

n + 16 = 5q + 2 + 16 = 5(q + 3) + 3 is not divisible by 5.

Case 4:

When n = 5q + 3.

n = 5q + 3 is not divisible by 5.

n + 4 = 5q + 3 + 4 = 5(q + 1) + 2 is not divisible by 5.

n + 8 = 5q + 3 + 8 = 5(q + 2) + 1 is not divisible by 5.

n + 12 = 5q + 3 + 12 = 5(q+ 3) is divisible by 5.

n + 16 = 5q + 3 + 16 = 5(q+ 3) + 4 is not divisible by 5.

Case 5:

When n = 5q + 4.

n = 5q + 4 is not divisible by 5.

n + 4 = 5q + 4 + 4 = 5(q + 1) + 3 is not divisible by 5.

n + 8 = 5q + 4 + 8 = 5(q + 2) + 2 is not divisible by 5.

n + 12 = 5q + 4 + 12 = 5(q + 3) + 1 is not divisible by 5.

n + 16 = 5q + 4 + 16 = 5(q + 4) is divisible by 5.

Hence, in each case, one and only one out of n, n + 4, n+ 8, n + 12 and n + 16 is divisible by 5.

15. Show that the square of an odd positive integer can be of the form 6q+1 or 6q+3 for some integer q.

Any positive integer can be written in the form of 6m, 6m+1, 6m+2, 6m+3, 6m+4, 6m+5 for some integer m.

Thus, an odd positive integer can be of the form 6m+1, 6m+3, 6m+5.

We have, (6m+1)

(6m+3)

(6m+5)

Thus, the square of an odd positive integer can be of the form 6q+ 1 or 6q+ 3 for some integer q.

16. A positive integer is of the form 3q+1, q being a natural number. Can you write its square in any form other than 3m+1, 3m or 3m+2 for some integer m? Justify your answer.

By Euclid's lemma, b = aq + r, 0 ≤ r < a.

Here, b is a positive integer and a = 3.

∴ b = 3q + r, for 0 ≤ r < 3

This must be in the form 3q, 3q + 1 or 3q + 2.

Now,

(3q)

(3q + 1)

(3q + 2)

Therefore, the square of a positive integer 3q + 1 is always in the form of 3m or 3m + 1 for some integer m.

**Solution**Consider the numbers n, (n + 4), (n + 8), (n + 12) and (n + 16) , where n is any positive integer. Suppose n = 5q + r, where 0 ≤ r < 5

n = 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4

(By Euclid's division algorithm)

Case 1:

When n = 5q.

n = 5q is divisible by 5.

n + 4 = 5q + 4 is not divisible by 5.

n + 8 = 5q + 5 + 5 + 3 = 5(q + 1) + 3 is not divisible by 5.

n + 12 = 5q + 10 + 2 = 5(q + 2) + 2 is not divisible by 5.

n + 16 = 5q + 15 + 1 = 5(q + 3) + 1 is not divisible by 5.

Case 2:

When n = 5q + 1.

n = 5q + 1 is not divisible by 5.

n + 4 = 5q + 1 + 4 = 5(q + 1) is divisible by 5.

n + 8 = 5q + 1 + 5 + 3 = 5(q + 1) + 4 is not divisible by 5.

n + 12 = 5q + 1 + 12 = 5(q + 2) + 3 is not divisible by 5.

n + 16 = 5q + 1 + 16 = 5(q + 3) + 2 is not divisible by 5.

Case 3:

When n = 5q + 2.

n = 5q + 2 is not divisible by 5.

n + 4 = 5q + 2 + 4 = 5(q + 1) + 1 is not divisible by 5.

n + 8 = 5q + 2 + 8 = 5(q + 2) is divisible by 5.

n + 12 = 5q + 2 + 12 = 5(q + 2) + 4 is not divisible by 5.

n + 16 = 5q + 2 + 16 = 5(q + 3) + 3 is not divisible by 5.

Case 4:

When n = 5q + 3.

n = 5q + 3 is not divisible by 5.

n + 4 = 5q + 3 + 4 = 5(q + 1) + 2 is not divisible by 5.

n + 8 = 5q + 3 + 8 = 5(q + 2) + 1 is not divisible by 5.

n + 12 = 5q + 3 + 12 = 5(q+ 3) is divisible by 5.

n + 16 = 5q + 3 + 16 = 5(q+ 3) + 4 is not divisible by 5.

Case 5:

When n = 5q + 4.

n = 5q + 4 is not divisible by 5.

n + 4 = 5q + 4 + 4 = 5(q + 1) + 3 is not divisible by 5.

n + 8 = 5q + 4 + 8 = 5(q + 2) + 2 is not divisible by 5.

n + 12 = 5q + 4 + 12 = 5(q + 3) + 1 is not divisible by 5.

n + 16 = 5q + 4 + 16 = 5(q + 4) is divisible by 5.

Hence, in each case, one and only one out of n, n + 4, n+ 8, n + 12 and n + 16 is divisible by 5.

15. Show that the square of an odd positive integer can be of the form 6q+1 or 6q+3 for some integer q.

**Solution**Any positive integer can be written in the form of 6m, 6m+1, 6m+2, 6m+3, 6m+4, 6m+5 for some integer m.

Thus, an odd positive integer can be of the form 6m+1, 6m+3, 6m+5.

We have, (6m+1)

^{2}= 36m^{2}+ 12m + 1 = 6 (6m^{2}+ 2m) + 1= 6q + 1, where q = 6m^{2}+ 2m is an integer.(6m+3)

^{2}= 36m^{2}± 36m + 9 = 6 (6m^{2}+ 6m + 1) + 3 = 6q + 3, where q = 6m^{2}+ 6m +1 is an integer.(6m+5)

^{2}= 36m^{2}± 60m + 25 = 6 (6m^{2}+ 10m + 4) + 1 = 6q + 1, where q = 6m^{2}+10m+4 is an integer.Thus, the square of an odd positive integer can be of the form 6q+ 1 or 6q+ 3 for some integer q.

16. A positive integer is of the form 3q+1, q being a natural number. Can you write its square in any form other than 3m+1, 3m or 3m+2 for some integer m? Justify your answer.

**Solution**By Euclid's lemma, b = aq + r, 0 ≤ r < a.

Here, b is a positive integer and a = 3.

∴ b = 3q + r, for 0 ≤ r < 3

This must be in the form 3q, 3q + 1 or 3q + 2.

Now,

(3q)

^{2}= 9q^{2}= 3m, where m = 3q^{2}(3q + 1)

^{2}= 9q^{2}+ 6q + 1 = 3(3q^{2}+ 2q) + 1 = 3m + 1, where m = 3q^{2}+ 2q(3q + 2)

^{2}= 9q^{2}+ 12q + 4 = 3 (3q^{2}+ 4q + 1) + 1 = 3m + 1, where m = 3q^{2}+ 4q + 1Therefore, the square of a positive integer 3q + 1 is always in the form of 3m or 3m + 1 for some integer m.

17. Show that the square of any positive integer cannot be of the form 3m+2, where m is a natural number.

**Solution**

By Euclid's lemma, b = aq + r, 0 ≤ r < a.

Here, b is a positive integer and a = 3.

∴ b = 3q + r, for 0 < r < 3

This must be in the form 3q, 3q + 1 or 3q + 2 .

Now,

(3q)

^{2}= 9q^{2}= 3m, where m = 3q2
(3q + 1)

^{2}= 9q^{2}+ 6q + 1 = 3 (3q^{2}+ 2q) + 1 = 3m + 1, where m = 3q^{2}+ 2q
(3q + 2)

^{2}= 9q^{2}+ 12q + 4 = 3(3q^{2}+ 4q + 1) -I- 1 = 3m + 1, where m = 3q^{2}+ 4q + 1
Therefore, the square of a positive integer cannot be of the form 3m + 2, where m is a natural number.