Here is the solution of NCERT Class 10 Maths Chapter 1 Real Number Exercise 1.1. There are 5 Questions in Class 10 NCERT Maths Chapter 1 Real Numbers Exercise 1.1. All 5 Questions under Exercise 1.1 are based on Euclid's Division Lemma. You can go through each one of them one by one.
Question 1: Use Euclid’s division algorithm to find the HCF of:
(i)135 and 225
Solution 1 (i): Since 225 > 135,
225 = 135 × 1 + 90
Also,
135 = 90 × 1 + 45
90 = 2 × 45 + 0
Therefore, the HCF of 135 and 225 is 45.
(ii)196 and 38220
Solution 1 (ii): Since 38220 > 196,
38220 = 196 × 195 + 0
Therefore, HCF of 196 and 38220 is 196.
(iii)867 and 255
Solution 1 (iii): Since 867 > 255,
867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
Therefore, HCF of 867 and 255 is 51.
Question 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 2: By Euclid’s algorithm,
a = 6q + r, and r = 0, 1, 2, 3, 4, 5
Hence, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
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 not exactly divisible by 2.
Hence, these numbers are odd numbers.
Question 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 3: Euclid’s algorithm
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF of (616, 32) is 8.
Therefore, the maximum number of columns in which they can march is 8.
Question 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 4: a = 3q + r, r = 0, 1, 2
Therefore, a= 3q or 3q + 1 or 3q + 2
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m+1.
Question 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 5: a = 3q + r, 0 ≤ r < 3
a = 3q, 3q + 1, 3q + 2
There are three cases.
Case 1: When a = 3q,
a3 = (3q)3 = 27q3 = 9(3q3) = 9m
Case 2: When a = 3q + 1,
a3 = (3q + 1)3
a3 =27q3 + 27q2 + 9q + 1
a3 = 9(3q3 + 3q2 + q) + 1
a3 = 9m+ 1
Case 3: When a = 3q + 2,
a3 = (3q + 2)3
a3 = 27q3 + 54q2 + 36q + 8
a3= 9(3q3 + 6q2 + 4q) + 8
a3=9m+ 8
Therefore, the cube of any positive integer is of the form 9m, 9m + 1,or 9m + 8.
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