Soln.
Q.5.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m , 9m + 1 or 9m + 8 for some integer m.
Let x be any positive integer
and b = 3
Here, divisor is 9
(x)3
= m
= m + 1
= m + 8
9
9
9
∴ Possible remainders are
0, 1, 2, 3, 4, 5, 6, 7, 8
i.e. 9q
9q + 1
9q + 2
9q + 3
9q + 4
9q + 5
9q + 6
9q + 7
9q + 8
Lets consider divisor as 3
∴ Possible remainders are
0, 1, 2
i.e. 3q
3q + 1
3q + 2
Applying Euclid’s Division Algorithm, we get
∴
x = 3q,
x = 3q + 1
or x = 3q + 2
i)
If
x
3q
=
x3
⇒
=
(3q)3
=
27q3
=
9
9m
=
for some integer m, where
m
3q3
=
(3q3)
ii)
If
x
=
3q
+
1
⇒
x3
=
(3q
+
1)
3
(3q)3
=
+
3
(3q)2
(1)
+
3
(3q)
(1)2
+
(1)3
9
=
(3q3 +
3q2
+
q)
+
1
Apply,
(a + b)3 = a3 + 3a2b + 3ab2 + b3
27q3
=
+
27q2
+
9q
+
1
9m
=
+ 1
for some integer m, where
m
3q3 + 3q2 + q
=
iii)
If
x
=
3q
+
2
⇒
x3
=
(3q
+
2)
3
(3q)3
=
+
3
(3q)2
(2)
+
3
(3q)
(2)2
+
(2)3
9
=
(3q3 +
6q2
+
4q)
+
8
Apply,
(a + b)3 = a3 + 3a2b + 3ab2 + b3
27q3
=
+
54q2
+
36q
+
8
9m
=
+ 8
for some integer m, where
m
3q3 + 6q2 + 4q
=
9m , 9m + 1 or 9m + 8 for some integer m.
Exercise 1.1