Class X
NCERT Text Book Solution
Exercise 1.1
1. Use Euclid's division algorithm to find the HCF of
(i)135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Solution:
(i)
From question it is clear that
225 > 135.
Apply Euclid's division lemma, to 135 and 225.
225 = 135x1 + 90
135 = 90x1 + 45
90 = 45x2 + 0
The remainder has now become zero.
The HCF of 225 and 135 is 45 Ans.
(ii)
From question it is clear that
38220>196
Apply Euclid's division lemma, to 38220 and 196.
38220 = 196x195 + 0
The remainder has now become zero.
The HCF of 38220 and 196 is 195 Ans.
(iii)
From question it is clear that
867>255
Apply Euclid's division lemma, to 867 and 255.
867 = 255x3 + 102
255 = 102x2 + 51
102 = 51x2 + 0
The remainder has now become zero.
The HCF of 867 and 255 is 51 Ans
2. 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 positive integer and b = 6.
Now, by Euclid's division algorithm,
a = 6q + r for some integer q ≥ 0,
r = 0, 1, 2, 3, 4, 5
Because 0 ≤ r < b
So, a can be
6q,or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5.
Since a is odd, a cannot be 6q, or 6q + 2, or 6q + 4 ( since they are all divisible by 2)
Therefore, any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5. Proved.
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:
For this, it will be find out the HCF of 616 and 32.
Apply Euclid's algorithm, to 616 and 32.
616 = 32x19 + 8
32 = 8x4 + 0
So, the HCF of 616 and 32 is 8.
Therefore, the maximum number of columns in which they can march = 8 Ans.
4. Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m, or 3m + 1 for some integer m.
Solution:
Let a be any positive integer and b = 3.
Now, Euclid's division lemma,
a = 3q + r for some integer q ≥ 0,
Since, 0 ≤ r < 3
a can be 3q, or 3q + 1, or 3q + 2
Therefore,
a² = 9q², or 9q² + 6q + 1, or 9q² + 12q + 1.
or, a² = 3x3q², or 3( 3q² + 2q) + 1, or 3(3q² + 4q) + 1
So, the square of any positive integer is either of the form 3m, or 3m + 1. Proved
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 a be any positive integer and b = 9.
Now, Euclid's division lemma,
a = 9q + r for some integer q ≥ 0,
Since, 0 ≤ r < 9
a can be 9q, or 9q + 1, or 9q + 2, 9q + 3, or ...... 9q + 8.
Therefore,
a³ = 9x81q³, or 9(81q³ + 27q² + 3q)+ 1, or 9(81q³ + 54q² + 6q) + 8 ..................
So, the cube of any positive integer is of the form 9m,or 9m + 1 or 9m + 8. Proved.
