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OJM Neamt 2009 subiectul I

 
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Tudor_C
Arhimede


Joined: 08 Mar 2009
Posts: 9

PostPosted: Mon Mar 09, 2009 7:45 pm    Post subject: OJM Neamt 2009 subiectul I Reply with quote

Consideram un patrat de 3X3 pe care il completam cu nr. naturale dupa modelul

\overline{\underline{\|1 \|3 \| 6 \|}}
{{\underline{\|2 \|5 \| 8 \|}}}
{\underline{\|4 \|7 \| 9 \|}}

Stergem nr. si recompletam patratul cu aceeasi regula cu nr. 10,11,12,13......18 si asa mai departe.

A. Pe ce pozitie a patratului va fi numarul 2005? Justificati gasirea pozitiei.
B. Care este configuratia patratului si care sunt nr. daca suma lor este 1260?
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Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Fri Jun 12, 2009 8:55 pm    Post subject: Reply with quote

a) De la stanga la dreapta, patratul este ocupat cu cate 9 numere grupate pe diagonala in 5 siruri.
Deoarece ultimul numar este intotdeauna multiplu de 9, inseamna ca numarul 2005 apartine patratului cu numerele 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007. Configuratia patratului este urmatoarea:

\overline{\underline{\|1999\|2001\|2004\|}}
{{\underline{\|2000\|2003\|2006\|}}}
 {{\underline{\|2002\|2005\|2007\|

b)Fie x numarul din coltul stanga sus al patratului.

\overline{\underline{\| x \|x+2\|x+5\|}}
{{\underline{\|x+1\|x+4\|x+7\|}}}
 {{\underline{\|x+3\|x+6\|x+8\|

Suma numerelor din patrat este x+x+1+x+2+x+3+x+4+x+5+x+6+x+7+x+8=9x+\frac{8\cdot 9}{2}=9x+36=1260 \Rightarrow 9x=1224; x=136.

Deci numerele sunt 136, 137, 138, 139, 140, 141, 142, 143, 144.
Configuratia patratului este:

\overline{\underline{\|136\|138\|141\|}}
{{\underline{\|137\|140\|143\|}}}
 {{\underline{\|139\|142\|144\|
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