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Matematica E Simpla • View topic - Ecuatii diofantice

Matematica E Simpla

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 Post subject: Ecuatii diofantice
PostPosted: Wed Dec 03, 2008 9:41 am 
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Joined: Mon Dec 01, 2008 3:20 pm
Posts: 14
Location: Ialomita-Ograda
Enunt:
Intr-un cos sunt pere. Daca se grupeaza cate 6, ramane una, daca se grupeaza cate 5, ramane tot una.
Cate pere sunt in cos?


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 Post subject: Re: Ecuatii diofantice
PostPosted: Wed Dec 03, 2008 10:57 am 
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Joined: Wed Oct 22, 2008 6:27 pm
Posts: 29
Rezolvare:
Exista mai multe solutii la problema data.
Orice multiplu de 30, la care se adauga 1, satisface datele problemei.

De exemplu:
31 de pere, se pot grupa in :
- 5 grupe de cate 6, plus o para
- 6 grupe de cate 5, plus o para

61 de pere, se pot grupa in :
- 10 grupe de cate 6, plus o para
- 12 grupe de cate 5, plus o para

91 de pere, se pot grupa in :
- 15 grupe de cate 6, plus o para
- 18 grupe de cate 5, plus o para

181 de pere, se pot grupa in :
- 30 de grupe de cate 6, plus o para
- 36 de grupe de cate 5, plus o para

O sa spui ca intr-un cos nu pot sa incapa 61 de pere (sau 181). Depinde de cos. Asta nu e problema de matematica, ci este un fel de test de perspicacitate.

Daca notam cu x numarul de pere, cu a numarul de grupe de cate 6, si cu b numarul de grupe de cate 5, avem:
x = a * 6 + 1
x = b * 5 + 1
De unde, a * 6 = b * 5.
Cum 6 nu e divizibil cu 5, inseamna ca a este divizibil cu 5, adica a este de forma 5 * k
Deci x = 5 * k * 6 + 1 = 30 k + 1
Se vede imediat ca x este si de forma (6k) * 5 + 1, deci numerele de forma:
30 k + 1 sunt solutiile problemei.


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