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Page 1 of 4, Mathematics, (Chapter-1) (Real Numbers), (Class X), Exercise 1.1, Question 1:, Use Euclid's division algorithm to find the HCF of:, (i) 135 and 225, (ii) 196 and 38220, (iii) 867 and 255, Answer 1:, (i) 135 and 225, Since 225 > 135, we apply the division lemma to 225 and 135 to obtain, 225 = 135 x 1 + 90, Since remainder 90 0, we apply the division lemma to 135 and 90 to, obtain, 135 = 90 x 1 + 45, We consider the new divisor 90 and new remainder 45, and apply the, division lemma to obtain, 90 = 2 x 45 + 0, Since the remainder is, zero,, the process stops., Since the divisor at this stage is 45,, Therefore, the HCF of 135 and 225 is 45., (ii) 196 and 38220, Since 38220 > 196, we apply the division lemma to 38220 and 196 to, obtain, 38220 = 196 x 195 + 0, Since the remainder is zero, the process stops., Since the divisor at this stage is 196,, Therefore, HCF of 196 and 38220 is 196., (iii) 867 and 255, Since 867 > 255, we apply the division lemma to 867 and 255 to obtain, 867 = 255 × 3 + 102, Since remainder 102 0, we apply the division lemma to 255 and 102, to obtain, 1 | Page
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Page 2 of 4, 255 = 102 x 2 + 51, We consider the new divisor 102 and new remainder 51, and apply the, division lemma to obtain, 102 = 51 x 2 + 0, Since the remainder is zero, the process stops., Since the divisor at this stage is 51, Therefore,, HCF of 867 and 255 is 51., Question 2:, Show that any positive odd integer is of the form 6q + 1, or 69 + 3, or, 6q + 5, where q is some integer., Answer 2:, Let a be any positive integer and b 6., Then, by Euclid's algorithm, a =, 6q + r for some integer q 2 0, and, r = 0, 1, 2, 3, 4, 5 because 0sr< 6., Therefore, a =, 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, Also, 69 + 1 = 2 x 3q + 1 = 2kı + 1, where ki 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, 69 + 3, 6q + 5 are of the form 2k + 1, where k is an integer., Therefore, 69 + 1, 6q + 3, 6q + 5 are not exactly divisible by 2., Hence, these expressions of numbers are odd numbers., And therefore, any odd integer can be expressed in the form 6q + 1, or, 6q + 3, or 69 + 5, 2 | Page
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Page 3 of 4, 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?, Answer 3:, HCF (616, 32) will give the maximum number of columns in which they, can march., We can use Euclid's algorithm to find the HCF., 616 = 32 x 19 + 8, 32 = 8 x 4 + 0, The HCF (616, 32) is 8., Therefore, they can march in 8 columns each., Question 4:, Use Euclid's division lemma to show that the square of any positive, integer is either of 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 and show that they can be rewritten, in the form 3m or 3m + 1.], Answer 4:, Let a be any positive integer and b = 3., Then a = 39 +r for some integer q z 0, And r= 0, 1, 2 because 0 sr<3, Therefore, a = 3q or 3q + 1 or 3q + 2, Or,, a? = (3q)? or (3q + 1)2 or (3q + 2)2, = (3q)? or 9q2 + 6q+1 or 9q2 + 12q +4, = 3 x (3q2) or 3 x (3q2 + 2q) + 1 or 3 x (3q2 +4q + 1) + 1, = 3k, or 3k2+1 or 3k,+ 1, Where kı, k2, and k3 are some positive integers, %3D, Hence, it can be said that the square of any positive integer is either of, the form 3m or 3m + 1., 3 | Page
