x:n=12 rest 5
n∈{1,2,3,4,5,6,7,8,9} Nu am luat si 0 pt. ca impartirea la 0 nu are sens.
REZOLVARE:
pt.n=1⇒x=12·1+5⇒x=17
pt.n=2⇒x=12·2+5⇒x=29
pt.n=3⇒x=12·3+5⇒x=41
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.
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pt.n=9⇒x=12·9+5⇒x=113 (observam ca sirul merge din 12 in 12)
⇒x∈{17,29,41,53,65,77,89,101,113}
Sper ca ai inteles,si ti-am fost de ajutor :))
