Exercise 1.1
1. Use Euclid's division
algorithm to find the HCF of
(i) 135 and 225
Solution
: Since 225 > 135, we
apply Euclid's algorithm as follows
225 = 135 x
1 + 90
135 = 90
x 1 + 45
90 =
45 x 2
+ 0
\
The HCF of 225 and 135 is 45
(ii) 196 and 38220
Solution
: Since 38220 > 196, we apply Euclid's algorithm as follows
38220 = 196
x 195 + 0
\ The
HCF of 38220 and 196 is 196
(iii) 867 and 255
Solution
: Since 867
> 255, we apply Euclid's algorithm as follows
867 = 255 x 3 + 102
255
= 102 x 2 + 51
102 = 51 x
2 + 0
\
The HCF of 867 and 255 is 51
2. Show that
any positive odd integer is of the form 6 q + 1 or 6 q + 3 or 6 q + 5, where q
is some integer.
Solution
: Let a be a positive odd integer and b
= 6.
Since 0 ≤ r < 6, the
possible remainders are 0, 1, 2, 3, 4 and 5
i.e., a can be
6 q or 6 q + 1 or 6 q + 2 or 6 q
+ 3 or 6 q + 4 or 6 q + 5,where q is the quotient.
Since a is odd, it cannot be 6 q or 6 q + 2 or 6 q + 4 (all are divisible by 2)
\
any positive odd integer is of the form 6 q + 1 or 6 q + 3 or 6 q + 5.
3. An army contingent of 616 members is to march
behind an army band of 32 members in a
parade. The two groups are to march in the same number of columns. What is the
maximum number of columns in which they can march?
Solution : Now, we have to find the HCF of 616 and
32 using Euclid's algorithm so that we get the maximum number of columns ,
616 =
32 x 19
+ 8
32 =
8 x 4
+ 0
The HCF ( 616, 32)
is 8
\
The maximum number of columns in which army members can march is 8 columns.
4. Use
Euclid's division lemma to show that square of any positive integer is either
of the form 3m or 3m + 1 for some integer m
[Hint: Let x be any positive integer then it is of
the form 3q, 3q + 1 or 3q + 2. Now square each of these sides and show that
they can be rewritten in the form 3m or 3m + 1.]
Solution
: Let x be any positive integer which is of the form 3q, 3q + 1
or 3q + 2.
(i) If x = 3q,
on
squaring both sides we get,
x2
= (3q)2
= 9q2 = 3(3q2) = 3m , where m
= 3q2 and is an integer
(ii) If x = 3q +
1
on
squaring both sides we get,
x2 = (3q + 1)2 [( a + b)2 = a2 + 2ab + b2]
x2 =
9q2 + 2 (3q ) (1)
+ 1
x2 = 3 ( 3q2 + 2q)
+ 1
x2 =
3m + 1, where m = 3q2 +
2q and is an integer
(iii) If x = 3q +
2
on
squaring both sides we get,
x2 = (3q + 2)2 [(
a + b)2 = a2 + 2ab + b2]
x2 =
9q2 + 2 (3q ) (2)
+ 4
x2 = 3 ( 3q2 + 4q)
+ 3 + 1 [ 4 can be written as 3 + 1]
x2 = 3 [( 3q2 + 4q)
+ 1] + 1
x2 =
3m + 1, where m = 3q2 +
4q + 1 and is an integer
\
the square of any positive integer is either of the form 3m or 3m + 1
for some integer m
for some integer m
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.
Solution :
Let
x be any positive integer which is of the form 3q, 3q + 1 or 3q + 2.
(i) If x = 3q,
cubing
both sides we get,
x3 = (3q)3 = 27q3
=
9(3q3) = 9m , where
m =
3q3 and is an integer
(ii) If x = 3q +
1
on
cubing both sides we get,
x3 = (3q + 1)3 [( a + b)3 = a3
+ 3ab ( a + b) + b3]
x3 =
27q3 + 3 (3q ) (1)[(3q + 1)] + 1
x3 = 27q3 + 9q [(3q + 1)] + 1
x3 = 27q3 + 27q2 + 9q + 1
x3 = 9 (3q3 + 3q2 + q) + 1
x3 =
9m + 1, where m = 3q3 + 3q2 + q and is an integer
(iii) If x = 3q +
2
on
cubing both sides we get,
x3 = (3q + 2)3 [( a + b)3 = a3
+ 3ab ( a + b) + b3]
x3 =
27q3 + 3 (3q ) (2)[(3q + 2)] + 8
x3 = 27q3 + 18q [(3q + 2)] + 8
x3 = 27q3 + 54q2 + 36q + 8
x3 = 9 (3q3 + 6q2 + 4q) +
8
x3 =
9 m + 8, where m = 3q3 + 6q2 + 4q and
is an integer
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