2a. Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.1)
- 1. PEDAGOGY OF MATHEMATICS – PART II BY Dr. I. UMA MAHESWARI Principal Peniel Rural College of Education,Vemparali, Dindigul District [email protected]
- 2. X STD Ex 2.1
- 25. Answer: The positive integers when divided by 3 leaves remainder 2. By Euclid’s division lemma a = bq + r, 0 ≤ r < b. Here a = 3q + 2, where 0 ≤ q < 3, a leaves remainder 2 when divided by 3. ∴ 2, 5, 8, 11 ……………..
- 26. Answer: Here a = 532, b = 21 Using Euclid’s division algorithm a = bq + r 532 = 21 × 25 + 7 Number of completed rows = 21 Number of flower pots left over = 7
- 27. Solution: Let n – 1 and n be two consecutive positive integers.Then their product is (n – 1)n. (n – 1)(n) = n2 – n. We know that any positive integer is of the form 2q or 2q + 1 for some integer q. So, following cases arise. Case I.When n = 2q. In this case, we have n2 – n = (2q)2 – 2q = 4q2 – 2q = 2q(2q – 1) ⇒ n2 – n = 2r, where r = q(2q – 1) ⇒ n2 – n is divisible by 2.
- 28. Answer: Let the positive integer be a, b, and c We know that by Euclid’s division lemma a = bq + r a = 13q + 9 ….(1) b = 13q + 7 ….(2) c = 13q + 10 ….(3) Add (1) (2) and (3) a + b + c = 13q + 9 + 13q + 7 + 13q + 10 = 39q + 26 a + b + c = 13 (3q + 2) This expansion will be divisible by 13 ∴ a + b + c is divisible by 13
- 29. Solution: Let x be any integer. The square of x is x2. Let x be an even integer. x = 2q + 0 then x2 = 4q2 + 0 When x be an odd integer When x = 2k + 1 for some integer k. x2 = (2k + 1 )2 = 4k2 + 4k + 1 = 4k (k + 1) + 1 = 4q + 1 where q = k(k + 1) is some integer. Hence it is proved.
- 30. (i) 340 and 412 Answer: To find the HCF of 340 and 412 using Euclid’s division algorithm.We get 412 = 340 × 1 + 72 The remainder 72 ≠ 0 Again applying Euclid’s division algorithm to the division of 340 340 = 72 × 4 + 52 The remainder 52 ≠ 0 Again applying Euclid’s division algorithm to the division 72 and remainder 52 we get 72 = 52 × 1 + 20 The remainder 20 ≠ 0 Again applying Euclid’s division algorithm 52 = 20 × 2 + 12
- 31. The remainder 12 ≠ 0 Again applying Euclid’s division algorithm 20 = 12 × 1 + 8 The remainder 8 ≠ 0 Again applying Euclid’s division algorithm 12 = 8 × 1 + 4 The remainder 4 ≠ 0 Again applying Euclid’s division algorithm 8 = 4 × 2 + 0 The remainder is zero ∴ HCF of 340 and 412 is 4
- 32. (ii) 867 and 255 Answer: To find the HCF of 867 and 255 using Euclid’s division algorithm.We get 867 = 255 × 3 + 102 The remainder 102 ≠ 0 Using Euclid’s division algorithm 255 = 102 × 2 + 51 The remainder 51 ≠ 0 Again using Euclid’s division algorithm 102 = 51 × 2 + 0 The remainder is zero ∴ HCF = 51 ∴ HCF of 867 and 255 is 51
- 33. (iii) 10224 and 9648 Answer: Find the HCF of 10224 and 9648 using Euclid’s division algorithm.We get 10224 = 9648 × 1 + 576 The remainder 576 ≠ 0 Again using Euclid’s division algorithm 9648 = 576 × 16 + 432 The remainder 432 ≠ 0 Using Euclid’s division algorithm 576 = 432 × 1 + 144 The remainder 144 ≠ 0 Again using Euclid’s division algorithm 432 = 144 × 3 + 0 The remainder is 0 ∴ HCF = 144 The HCF of 10224 and 9648 is 144
- 34. (iv) 84,90 and 120 Answer: Find the HCF of 84, 90 and 120 using Euclid’s division algorithm 90 = 84 × 1 + 6 The remainder 6 ≠ 0 Using Euclid’s division algorithm 4 = 14 × 6 + 0 The remainder is 0 ∴ HCF = 6 The HCF of 84 and 90 is 6 Find the HCF of 6 and 120 120 = 6 × 20 + 0 The remainder is 0 ∴ HCF of 120 and 6 is 6 ∴ HCF of 84, 90 and 120 is 6
- 35. Solution: The required number is the H.C.F. of the numbers. 1230 – 12 = 1218, 1926 – 12 = 1914 First we find the H.C.F. of 1218 & 1914 by Euclid’s division algorithm. 1914 = 1218 × 1 + 696 The remainder 696 ≠ 0.
- 36. Again using Euclid’s algorithm 1218 = 696 × 1 + 522 The remainder 522 ≠ 0. Again using Euclid’s algorithm. 696 = 522 × 1 + 174 The remainder 174 ≠ 0. Again by Euclid’s algorithm 522 = 174 × 3 + 0 The remainder is zero. ∴The H.C.F. of 1218 and 1914 is 174. ∴The required number is 174.
- 37. Answer: Find the HCF of 32 and 60 60 = 32 × 1 + 28 ….(1) The remainder 28 ≠ 0 By applying Euclid’s division lemma 32 = 28 × 1 + 4 ….(2) The remainder 4 ≠ 0 Again by applying Euclid’s division lemma 28 = 4 × 7 + 0….(3) The remainder is 0
- 38. HCF of 32 and 60 is 4 From (2) we get 32 = 28 × 1 + 4 4 = 32 – 28 = 32 – (60 – 32) 4 = 32 – 60 + 32 4 = 32 × 2 -60 4 = 32 x 2 + (-1) 60 When compare with d = 32x + 60 y x = 2 and y = -1 The value of x = 2 and y = -1
- 39. Solution: Let a (+ve) integer be x. x = 88 × y + 61 61 = 11 × 5 + 6 (∵ 88 is multiple of 11) ∴ 6 is the remainder. (When the number is divided by 88 giving the remainder 61 and when divided by 11 giving the remainder 6).
- 40. Answer: Let the consecutive positive integers be x and x + 1. The two number are co – prime both the numbers are divided by 1. If the two terms are x and x + 1 one is odd and the other one is even. HCF of two consecutive number is always 1. Two consecutive positive integer are always coprime. FundamentalTheorem of Arithmetic Every composite number can be written uniquely as the product of power of prime is called fundamental theorem of Arithmetic.