Question

Fie x si y 2 numere nat care verifica egalitatea: 5x+7y=2011

Aratati ca 285<x+y<403​

Answer 1

Răspuns:

$2011=5x+7y\leq 7x+7y=7(x+y)$, de unde $x+y\geq \frac{2011}{7}=287,28...$. Rezulta ca x+y>287, deci x+y>285.

$2011=5x+7y\geq 5x+5y=5(x+y)$, de unde $x+y\leq \frac{2011}{5}=402.2$. Rezulta x+y<403.