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Teorema impartirii cu rest (O.J.)
Posted: Sun Nov 02, 2008 1:59 pm
by Marcelina Popa
Gasiti numarul \( \overline{xyzt} \), stiind ca daca-l impartim la numarul \( \overline{yzt} \) obtinem catul \( x+1 \) si restul \( x+2 \).
(OJ - Mehedinti)
Posted: Thu Nov 06, 2008 7:27 am
by alex2008
Aplicam teorema impartirii cu rest si avem :
\( \overline{xyzt}=\overline{yzt}\cdot(x+1)+x+2 \)
Scriem numerele in baza 10 :
\( 1000x+100y+10z+t=100xy+100y+10xz+10z+tx+t+x+2 \)
\( 1000x=100xy+10xz+tx+x+2 \)
\( x(999-\overline{yzt})=2 \)
Dar x natural =>x=1 sau x=2
x=1 , rezulta\( \overline{yzt}=997 \) , rezulta \( \overline{xyzt}=1997 \)
x=2 , rezulta \( \overline{yzt}=998 \) , rezulta \( \overline{xyzt}=2998 \)