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aratati ca daca n=abc+bca+cab (a,b,c nu e 0) atunci n divizibil cu 37.Va rog repede.Multumesc.

Răspuns :

[tex]\overline{abc}+\overline{bca}+\overline{cab}=100a+10b+c+100b+10c+a+100c+10a+b=111a+111b+111c=111(a+b+c)=37\cdot 3(a+b+c)[/tex]