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OJM Neamt 2009 subiectul IV

 
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Tudor_C
Arhimede


Joined: 08 Mar 2009
Posts: 9

PostPosted: Mon Mar 09, 2009 8:04 pm    Post subject: OJM Neamt 2009 subiectul IV Reply with quote

Unui numar de 3 cifre distincte i se inverseaza pozitile ultimelor 2 cifre si nr. obtinut se aduna cu cel initial. Se obtine astfel un nr de patru cifre de forma \overline{177x} .Sa se afle x si nr de trei cifre.
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Laurian Filip
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Joined: 25 Nov 2007
Posts: 473
Location: Bucuresti

PostPosted: Mon Mar 09, 2009 9:44 pm    Post subject: Reply with quote

fie \overline{abc} acel numar.
\overline{abc}+\overline{acb}=\overline{177x}

a trebuie sa fie strict mai mic decat 9.
Daca a este mai mic sau egal ca 7, atunci \overline{bc}+\overline{cb}>300 , nu convine.

Asadar a=8

Deci ramane \overline{bc}+\overline{cb}=\overline{17x}
10b+c+10c+b=170+x
11(b+c)=170+x

deci 170+x e divizibil cu 11, de unde x=6
b+c=16

deci (a,b,c,x) \in {(8,7,9,6);(8,8,8,6);(8,9,7,6)}
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