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Tuesday
, 29 november 2016
Determinati numerele prime n care au proprietatea ca...
Proposed by
Gabi
(3 comments)
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Comments (3)
Tuesday
, 29 nov 2016 11:40
[#]
Arckangel
Scriem relatia ca 5/p=1/m + 4/4n (4/4n <=> 1/n). De aici, p=m+4n. Cum m,n numere naturale nenule, cel mai mic numar prim care indeplineste conditia este 5 (5=1+4*1| m,n nu trebuie sa fie neaparat diferite). Deci, ca raspuns, p poate fi orice numar prin in afara de {1,2,3}.
Parerea mea, proprie, personala, dezvoltata de logica mea (de vineri)
1
Tuesday
, 29 nov 2016 18:53
[#]
Gabi
RE:
Azi e marti.
Wednesday
, 30 nov 2016 04:50
[#]
RE:
Si azi e miercuri. Si Sfantul Andrei, pe deasupra.
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