Chapter 1 Real Numbers R.D. Sharma Solutions for Class 10th Math Exercise 1.1
Exercise 1.1
Level 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) = n2 − n
We know that any positive integer is of the form 2q or 2q + 1, for some integer q
When n =2q,
n2 − n = (2q)2 − 2q = 4q2 − 2q
=2q(2q − 1)
Then n2 − n is divisible by 2.
When n = 2q + 1,
n2 − n = (2q + 1)2 − (2q + 1) = 4q2 + 4q + 1 − 2q − 1
= 4q2 + 2q
= 2q(2q + 1)
Then n2 − 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 n3 − n = n(n2 − 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, n3– 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
∴ n2 = (5n + 1)2
= 25q2 + 10q + 1
= 5(5q2 + 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 = 9q2
= 3 × 3q2
= 3m (m is any integer)
(3q + 1)2 = (3q)2 + 2(3q)(1) + (1)2
= 9q2 + 6q + 1
= 3(3q2 + 2q) + 1
= 3m + 1 (m is any integer)
(3q + 2)2 = (3q)2 + 2(3q)(2) + (2)2
= 9q2 + 12q + 4
= 9q2 + 12q + 4
= 3(3q2 + 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 = 16m2
= 4 × 4m2
= 4q (q is any integer)
(4m + 1)2 = (4m)2 + 2(4m)(1) + (1)2
= 16m2 + 8m + 1
= 4(4m2 + 2m) + 1
= 4q + 1 (q is any integer)
(4m + 2)2 = (4m)2 + 2(4m)(2) + (2)2
= 16m2 + 24m + 9
= 16m2 + 24m + 8 + 1
= 4(4m2 + 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)2 = 25m2
= 5(5m2)
= 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 = 25m2
= 5(5m2)
= 5q (q is any integer)
(5m + 1)2 = (5m)2 + 2(5m)(1) + (1)2
= 25m2 + 10m + 1
= 5(5m2 + 2m) + 1
= 5q + 1 (q is any integer)
(5m + 2)2 = (5m)2 + 2(5m)(2) + (2)2
= 25m2 + 20m + 4
= 5(5m2 + 4m) + 4
= 5q + 4 (q is any integer)
(5m + 3)2 = (5m)2 + 2(5m)(3) + (3)2
= 25m2 + 30m + 9
= 25m2 + 30m + 5 + 4
= 5(5m2 + 6m + 1) + 4
= 5(5m2 + 2m) + 1
= 5q + 1 (q is any integer)
(5m + 2)2 = (5m)2 + 2(5m)(2) + (2)2
= 25m2 + 20m + 4
= 5(5m2 + 4m) + 4
= 5q + 4 (q is any integer)
(5m + 3)2 = (5m)2 + 2(5m)(3) + (3)2
= 25m2 + 30m + 9
= 25m2 + 30m + 5 + 4
= 5(5m2 + 6m + 1) + 4
= 5q + 1 (q is any integer)
(5m + 4)2 = (5m)2 + 2(5m)(4) + (4)2
= 25m2 + 40m + 16
= 25m2 + 40m + 15 + 1
= 5(5m2) + 2(5m)(4) + (4)2
= 5q + 1 (q is any integer)
= 5(5m2) + 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
= 16q2 + 8q + 1
= 8(2q2 + 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)2 = 36q2
= 6(6q2)
= 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 = 36q2
= 6(6q2)
= 6m (m is any integer)
(6q + 1)2 = (6q)2 + 2(6q)(1) + (1)2
= 36q2 + 12q + 1
= 6(6q2 + 2q) + 1
= 6m + 1 (m is any integer)
= 6(6q2 + 2q) + 1
= 6m + 1 (m is any integer)
(6q + 2)2 = (6q)2 + 2(6q)(2) + (2)2
= 36q2 + 24q + 4
= 6(6q2 + 4q) + 4
= 6m + 4 (m is any integer)
= 6(6q2 + 4q) + 4
= 6m + 4 (m is any integer)
(6q + 3)2 = (6q)2 + 2(6q)(3) + (3)2
= 36q2 + 36q + 9
= 6(6q2 + 6q + 1) + 3
= 6m + 3 (m is any integer)
= 6(6q2 + 6q + 1) + 3
= 6m + 3 (m is any integer)
(6q + 4)2 = (6q)2 + 2(6q)(4) + (4)2
= 36q2 + 48q + 16
= 6(6q2 + 8q + 2) + 4
= 6m + 4 (m is any integer)
= 6(6q2 + 8q + 2) + 4
= 6m + 4 (m is any integer)
(6q + 5)2 = (6q)2 + 2(6q)(5) + (5)2
= 36q2 + 60q + 25
= 6(6q2 + 10q + 4) + 1
= 6m + 1 (m is any integer)
= 6(6q2 + 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
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)3 = 216m3
= 6(36m3)
= 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 = 216m3
= 6(36m3)
= 6q (q is any integer)
(6m + 1)3 = (6m)3 +108m2 + 18m + (1)3
= 6(36m3 + 18m2 + 3m) + 1
= 6q + 1 (q is any integer)
= 6q + 1 (q is any integer)
(6m + 2)3 = (6m)3 + 216m2 + 72m + (2)3
= 216m3 + 216m2 + 72m + 6 + 2
= 6(36m3 + 36m2 + 12m + 1) + 2
= 6q + 2 (q is any integer)
= 6(36m3 + 36m2 + 12m + 1) + 2
= 6q + 2 (q is any integer)
(6m + 3)3 = (6m)3 + 324m2 + 162m + (3)3
= 216m3 + 324m2 + 162m + 24 + 3
= 6(36m3 + 54m2 + 27m + 4) + 3
= 6q + 3 (q is any integer)
= 6(36m3 + 54m2 + 27m + 4) + 3
= 6q + 3 (q is any integer)
(6m + 4)3 = (6m)3 + 432m2 + 288m + (4)3
= 216m3 + 432m2 + 288m + 60 + 4
= 6(36m3 + 72m2 + 48m + 10) + 4
= 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.
= 6(36m3 + 72m2 + 48m + 10) + 4
= 6q + 4 (q is any integer)
(6m + 5)3 = (6m)3 + 432m2 + 288m + (4)3
= 216m3 + 540m2 + 450m + 120 + 5
= 6(36m3 + 90m2 + 75m + 20) + 5
= 6q + 5 (q is any integer)
= 6(36m3 + 90m2 + 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.
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 = 36m2 + 12m + 1 = 6 (6m2 + 2m) + 1= 6q + 1, where q = 6m2 + 2m is an integer.
(6m+3)2 = 36m2 ± 36m + 9 = 6 (6m2 + 6m + 1) + 3 = 6q + 3, where q = 6m2 + 6m +1 is an integer.
(6m+5)2 = 36m2 ± 60m + 25 = 6 (6m2 + 10m + 4) + 1 = 6q + 1, where q = 6m2 +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 = 9q2 = 3m, where m = 3q2
(3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1, where m = 3q2 + 2q
(3q + 2)2 = 9q2 + 12q + 4 = 3 (3q2 + 4q + 1) + 1 = 3m + 1, where m = 3q2 + 4q + 1
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 = 36m2 + 12m + 1 = 6 (6m2 + 2m) + 1= 6q + 1, where q = 6m2 + 2m is an integer.
(6m+3)2 = 36m2 ± 36m + 9 = 6 (6m2 + 6m + 1) + 3 = 6q + 3, where q = 6m2 + 6m +1 is an integer.
(6m+5)2 = 36m2 ± 60m + 25 = 6 (6m2 + 10m + 4) + 1 = 6q + 1, where q = 6m2 +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 = 9q2 = 3m, where m = 3q2
(3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1, where m = 3q2 + 2q
(3q + 2)2 = 9q2 + 12q + 4 = 3 (3q2 + 4q + 1) + 1 = 3m + 1, where m = 3q2 + 4q + 1
Therefore, 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 = 9q2 = 3m, where m = 3q2
(3q + 1)2 = 9q2 + 6q + 1 = 3 (3q2 + 2q) + 1 = 3m + 1, where m = 3q2 + 2q
(3q + 2)2 = 9q2 + 12q + 4 = 3(3q2 + 4q + 1) -I- 1 = 3m + 1, where m = 3q2 + 4q + 1
Therefore, the square of a positive integer cannot be of the form 3m + 2, where m is a natural number.