Page 57 - Start Up Mathematics_8 (Non CCE)
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VIII.  The natural numbers containing all digits as 6 with units digit as 7 follow another interesting pattern.
                                                            2
                                                           7   =  49
                                                            2
                                                         67   =  4489
                                                            2
                                                        667   =  444889
                                                            2
                                                       6667   =  44448889    and so on.
                    Example 3:      Which of the following natural numbers are not perfect squares? Give reasons.

                                    (a)  2,037       (b)  1,024     (c)  5,298      (d)  33,222    (e)  13,456
                    Solution:       (a)  2,037       (c)  5,298     (d)  33,222 are not perfect squares because numbers ending
                                    with digits 2, 3, 7, 8 are never perfect squares.
                    Example 4:      Write the units digit of the squares of: (a) 61      (b) 372      (c) 2,134
                    Solution:       (a)  Since 61 has 1 at the units place, its square will have 1 as the units digit.
                                    (b)  Since 372 has 2 at the units place, its square will have 4 as the units digit.
                                    (c)  Since 2,134 has 4 at the units place, its square will have 6 as the units digit.
                    Example 5:      Write the Pythagorean triplet, whose members are formed by (a) 10 and (b) 16.
                    Solution:       (a)  m = 10
                                        2m = 2 × 10 = 20
                                                    2
                                          2
                                        m  – 1 = (10)  – 1 = 100 – 1 = 99
                                                     2
                                          2
                                        m  + 1 = (10)  + 1 = 100 + 1 = 101
                                        So,  the Pythagorean triplet is 20, 99, 101.
                                    (b)   m = 16
                                        2m = 2 × 16 = 32
                                          2
                                                    2
                                        m  – 1 = (16)  – 1 = 256 – 1 = 255
                                          2
                                                     2
                                        m  + 1 = (16)  + 1 = 256 + 1 = 257
                                        So, the Pythagorean triplet is 32, 255, 257.
                    Example 6:      Without adding, find the following sum:
                                    (a)  1 + 3 + 5 + 7              (b)  1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17

                    Solution:       (a)  1 + 3 + 5 + 7 is the sum of the first 4-odd natural numbers.
                                                              2
                                        So,  1 + 3 + 5 + 7 = (4)  = 16.
                                    (b)   1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 is the sum of the first 9-odd natural numbers.
                                                                                     2
                                        So,  1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = (9)  = 81.
                    Example 7:      Write (a) 25 and (b) 144 as a sum of odd natural numbers.
                    Solution:       (a)  25 = 5 2

                                        So,  25 can be written as a sum of the first 5-odd natural numbers.
                                        \ 25 = 1 + 3 + 5 + 7 + 9
                                    (b)  144 = (12) 2
                                        So,  144 can be written as a sum of the first 12-odd natural numbers.
                                        \ 144 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
                                            2
                    Example 8:      Write 15  as the sum of two consecutive natural numbers.
                                                              2
                                                  2
                                           2
                    Solution:       (2n + 1)  = (2n  + 2n) + (2n  + 2n + 1) where 2n + 1 is an odd natural number.
                                          15 = 2 × 7 + 1                                 {Compare 2 × 7 + 1 with 2n + 1}
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