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Rezolvarea problemelor din num˘arul anterior                               15





                                            Clasa a VII-a



            MGO 131. Determinat ,i perechile (x, y) de numere naturale nenule cu proprietatea
            c˘a
                                       2
                                      x + y 2    · [7x, y] − 91xy = 0,
            unde [7x, y] reprezint˘a cel mai mic multiplu comun al numerelor 7x s , i y.

                                      Stelian Corneliu Andronescu s , i Costel B˘alc˘au, Pites , ti
                                                                  7x · y
                                                     2
                                                          2
                                a
                                               a
                                                                             a
            Solut ,ie. Ecuat , ia dat˘ poate fi rescris˘ ca x + y = 13 ·  , adic˘
                                                                  [7x, y]
                                                2
                                           2
                                          x + y = 13(7x, y)
            (unde (7x, y) reprezint˘ cel mai mare divizor comun al numerelor 7x s , i y).
                                  a
                                                            2
                                                       2
                Cazul 1. y 6= M7. Atunci ecuat , ia devine x +y = 13(x, y). Fie d = (x, y), deci

               x = da                               2  2   2                  2   2
                       , (a, b) = 1 s , i ecuat , ia devine d (a + b ) = 13d, adic˘a d(a + b ) = 13,
               y = db

                                                           x = a = 2       x = a = 3
                               2
                          2
            deci d = 1 s , i a + b = 13, deci avem solut , iile       s , i          .
                                                           y = b = 3       y = b = 2
                                                                     2
                                                                            2
                                          ∗
                Cazul 2. y = 7y 1 , y 1 ∈ N . Atunci ecuat , ia devine x + 49y = 91(x, y 1 ).
                                                                            1
                                       ∗
                                                                 2
                                                             2
            Rezult˘a c˘a x = 7x 1 , x 1 ∈ N s , i ecuat , ia devine 7(x + y ) = 13(7x 1 , y 1 ), de unde
                                                             1   1
                                                               2
                                                                      2
                                         ∗
            rezult˘a c˘a y 1 = 7y 2 , y 2 ∈ N , iar ecuat , ia devine x + 49y = 13(x 1 , y 2 ). Fie
                                                                      2
                                                               1

                                x 1 = da                               2  2      2
            d = (x 1 , y 2 ), deci       , (a, b) = 1 s , i ecuat , ia devine d (a + 49b ) = 13d,
                                 y 2 = db
                                                          ∗
                            2
                     2
                a
            adic˘ d(a + 49b ) = 13, care nu are solut , ii ˆın N .
            MGO 132. Demonstrat ,i c˘a fract ,ia
                              1 2019  + 2 2019  + 3 2019  + 4 2019  + 5 2019  + 6 2019
                              1 2020  + 2 2020  + 3 2020  + 4 2020  + 5 2020  + 6 2020
            este reductibil˘a.
                                                                   Ionel Tudor, C˘alug˘areni
            Solut ,ie. Avem 6 2019  = (7 − 1) 2019  = M7 − 1, 5 2019  = (7 − 2) 2019  = M7 − 2 2019  s , i
            4 2019  = (7−3) 2019  = M7−3 2019 , deci 1 2019 +2 2019 +3 2019 +4 2019 +5 2019 +6 2019  = M7.
            Pe de alt˘a parte, avem 2 2020  = 2 3·673+1  = 2 · 8 673  = 2(7 + 1) 673  = 2(M7 + 1) =
                                              2
            M7+2, 4  2020  = 2 2020   2  = (M7+2) = M7+4, 3 2020  = (7−4) 2020  = M7+4 2020  =
            M7+4, 5  2020  = (7−2) 2020  = M7+2 2020  = M7+2 s , i 6 2020  = (7−1) 2020  = M7+1,
            deci 1 2020  + 2 2020  + 3 2020  + 4 2020  + 5 2020  + 6 2020  = 2(1 + M7 + 2 + M7 + 4) = M7.
                   a
                     a
                                 a
                                             a
            Rezult˘ c˘ fract , ia dat˘ se simplific˘ prin 7, deci este reductibil˘a.
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