# Class 10 RD Sharma Solutions – Chapter 1 Real Numbers – Exercise 1.1 | Set 2

• Last Updated : 04 Dec, 2020

### Question 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:

a = bq+r ; where 0 < r < b

Putting b=6 we get,

⇒ a = 6q + r, 0 < r < 6

r = 0, a = 6q = 2(3q) = 2m, which is an even number. [m = 3q]

r = 1, a = 6q + 1 = 2(3q) + 1 = 2m + 1, which is an odd number. [m = 3q]

r = 2, a = 6q + 2 = 2(3q + 1) = 2m, which is an even number. [m = 3q + 1]

r = 3, a = 6q + 3 = 2(3q + 1) + 1 = 2m + 1, which is an odd number. [m = 3q + 1]

r = 4, a = 6q + 4 = 2(3q + 2) + 1 = 2m + 1, which is an even number. [m = 3q + 2]

r = 5, a = 6q + 5 = 2(3q + 2) + 1 = 2m + 1, which is an odd number. [m = 3q + 2]

Therefore, any odd positive integer can be of the form 6q +1,6q + 3,6q + 5, where q is some integer.

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

Solution:

a = 6q + r, where 0 ≤ r < 6 (Taking b=6 in Euclid’s division lemma)

a2 = (6q + r)2 = 36q2 + r+ 12qr
a2 = 6(6q2 + 2qr) + r2  0 ≤ r < 6

r = 0

a= 6 (6q2) = 6m, where, m = 6q2 is an integer.

r = 1

a2 = 6 (6q2 + 2q) + 1 = 6m + 1, where, m = (6q2 + 2q) is an integer.

r = 2,

a2 = 6(6q2 + 4q) + 4 = 6m + 4, where, m = (6q2 + 4q) is an integer.

r = 3,

a2 = 6(6q2 + 6q) + 9 = 6(6q2 + 6q) + 6 + 3

a2 = 6(6q2 + 6q + 1) + 3 = 6m + 3, where, m = (6q + 6q + 1) is integer.

r = 4,

a2 = 6(6q2 + 8q) + 16

= 6(6q2 + 8q) + 12 + 4

⇒ a= 6(6q+ 8q + 2) + 4 = 6m + 4, where, m = (6q2 + 8q + 2) is integer.

r = 5,

a= 6 (6q2 + 10q) + 25 = 6(6q+ 10q) + 24 + 1

a2 = 6(6q2 + 10q + 4) + 1 = 6m + 1, where, m = (6q2 + 10q + 4) is integer.

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

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

Solution:

For 6q,

(6q)3 = 216 q3 = 6(36q)3 + 0

= 6m + 0, (where m is an integer = (36q)3)

For 6q+1,

(6q+1)3 = 216q3 + 108q2 + 18q + 1

= 6(36q3 + 18q2 + 3q) + 1

= 6m + 1, (where m is an integer = 36q3 + 18q2 + 3q)

For 6q+2,

(6q+2)3 = 216q3 + 216q2 + 72q + 8

= 6(36q3 + 36q2 + 12q + 1) +2

= 6m + 2, (where m is an integer = 36q3 + 36q2 + 12q + 1)

For 6q+3,

(6q+3)3 = 216q3 + 324q2 + 162q + 27

= 6(36q3 + 54q2 + 27q + 4) + 3

= 6m + 3, (where m is an integer = 36q3 + 54q2 + 27q + 4)

For 6q+4,

(6q+4)3 = 216q3 + 432q2 + 288q + 64

= 6(36q3 + 72q2 + 48q + 10) + 4

= 6m + 4, (where m is an integer = 36q3 + 72q2 + 48q + 10)

For 6q+5,

(6q+5)3 = 216q3 + 540q2 + 450q + 125

= 6(36q3 + 90q2 + 75q + 20) + 5

= 6m + 5, (where m is an integer = 36q3 + 90q2 + 75q + 20)

Therefore, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

### Question 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.

Solution:

b=5

n = 5q+r

0 < r < 5

Therefore, n may be in the form of 5q, 5q+1, 5q+2, 5q+3, 5q+4

CASE 1:

When, n = 5q

n+4 = 5q+4

n+8 = 5q+8

n+12 = 5q+12

n+16 = 5q+16

n is only divisible by 5

CASE 2:

n = 5q+1

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

n+8 = 5q+9

n+12 = 5q+13

n+16 = 5q+17

n + 4 is only divisible by 5

CASE 3:

n = 5q+2

n+4 = 5q+6

n+8 = 5q+10 = 5(q+2)

n+12 = 5q+14

n+16 = 5q+18

n + 8 is only divisible by 5

CASE 4:

n = 5q+3

n+4 = 5q+7

n+8 = 5q+11

n+12 = 5q+15 = 5(q+3)

n+16 = 5q+19

n + 12 is only divisible by 5

CASE 5:

n = 5q+4

n+4 = 5q+8

n+8 = 5q+12

n+12 = 5q+16

n+16 = 5q+20 = 5(q+4)

Here, n + 16 is only divisible by 5

Therefore, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.

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

Solution:

b=6

a = 6m + r

0 ≤ r < 6.

a = 6m, 6m + 1, 6m + 2 , 6m + 3, 6m + 4, 6m + 5

Thus, we are choosing for a = 6m + 1 or, 6m + 3 or 6m + 5 for it to be an odd integer.

For a = 6m + 1,

(6m + 1)2 = 36m2 + 12m + 1

= 6(6m2 + 2m) + 1

= 6q + 1, where q is some integer and q = 6m2 + 2m.

For a = 6m + 3

(6m + 3)2 = 36m2 + 36m + 9

= 6(6m2 + 6m + 1) + 3

= 6q + 3, where q is some integer and q = 6m2 + 6m + 1

For a = 6m + 5,

(6m + 5)2 = 36m2 + 60m + 25

= 6(6m2 + 10m + 4) + 1

= 6q + 1, where q is some integer and q = 6m2 + 10m + 4.

Therefore, the square of an odd integer is of the form 6q + 1 or 6q + 3, for some integer q.

### Question 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:

No.

a = bq + r, 0 ≤ r < b

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

⇒ a = 3q + r

So, a can be of the form 3q, 3q + 1 or 3q + 2.

Now, for a = 3q

(3q)2 = 3(3q2) = 3m [where m = 3q2]

a = 3q + 1

(3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1 [where m = 3q2 + 2q]

a = 3q + 2

(3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1 = 3(3q2 + 4q + 1) + 1

= 3m + 1 [where m = 3q2 + 4q + 1]

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

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

Solution:

a = bm + r

b = 3

a = 3m + r

r = 0, 1, 2.

r = 0, a = 3m.

r = 1, a = 3m + 1.

r = 2, a = 3m + 2.

When a = 3m

a2 = (3m)2 = 9m2

a2 = 3(3m2) = 3q, where q = 3m2

When a = 3m + 1

a2 = (3m + 1)2 = 9m2 + 6m + 1

a2 = 3(3m2 + 2m) + 1 = 3q + 1, where q = 3m2 + 2m

When a = 3m + 2

a2 = (3m + 2)2

a2 = 9m2 + 12m + 4

a2 = 3(3m2 + 4m + 1) + 1

a2 = 3q + 1 where q = 3m2 + 4m + 1

Therefore, square of any positive integer cannot be of the form 3q + 2, where q is a natural number.

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