Question

Care este numarul maxim de numere
naturale consecutive a caror suma
este egala cu 2009?

Answer 1

$Sa~observam~pentru~inceput~ca~ 2009\ \textless \ 0+1+...+63,~deci~suma~ \\ \\ ceruta~are~mai~putin~de~64~de~termeni. \\ \\ Consideram~suma~a+(a+1)+...+(a+n)=2009....(n+1~termeni)\\ \\Avem:~2009=(1+2+...+(a+n))-(1+2+...+(a-1)) \\ \\ Echivalent~cu~2009= \frac{(a+n)(a+n+1)}{2}- \frac{(a-1)a}{2} \\ \\sau~inca:~4018=(a+n)(a+n+1)-(a-1)a \Leftrightarrow \\ \\ \Leftrightarrow 4018=^{calcule}=n^2+2an+2a+n \Leftrightarrow \boxed{4018=(n+1)(n+2a)}.$

$Numarul~de~termeni~ai~sumea~este~(n+1),~iar~acesta~trebuie \\ \\ sa~fie~maxim.~Din~relatia~din~chenar~rezulta~ca~(n+1)~este~divizor \\ \\ natural~al~lui~4018.~Dar~n+1 \leq 64,~iar~cel~mai~mare~divizor~al~lui \\ \\ 4018~mai~mic~sau~egal~cu~64~este~49. \\ \\ n+1=49 \Rightarrow n=48 \Rightarrow n+2a=82 \Rightarrow a=17. \\ \\Deci~numarul~maxim~cerut~de~problema~este~ \boxed{48},~suma~fiind: \\ \\ 17+18+...+64 =2009.$