Question

aflati a, b, c numere prime cu proprietatea 287a + 82b + 14c = 2009

Answer 1

287a + 82b + 14c = 2009

7·41a+82b+7·2c = 7·287 ⇒ b=7

Relatia  devine:

7·41a+82·7+7·2c = 7·287 |:7  ⇒  41a+82+2c = 287 ⇒

⇒ 41a+41·2 + 2c = 41·7 ⇒  c = 41

Relatia devine :

41a+41·2 + 2·41 = 41·7 |:41 ⇒  a + 2 +2 = 7 ⇒ a+4=7 ⇒a =3

Deci, numerele cerute sunt:

 a = 3,  b = 7,  c = 41.