Question

Aflați numerele abc dacă abc= a+b+c³. Justificați!​

Answer 1

$\displaystyle\bf\\\overline{abc}=a+b+c^3 \Leftrightarrow 100a+10b+c=a+b+c^3 \Leftrightarrow 99a+9b=c^3 -c \Leftrightarrow \\9(11a+b)=c(c^2-1)\Leftrightarrow 9(11a+b)=c(c-1)(c+1) \implies c =9,~c-1=9,~sau~c+1=9 \implies c~poate~fi~9,~10,~8,~dar~c~este~cifra~\implies c~poate~fi~9~sau~8.\\\boxed{\bf c=9} \implies 9(11a+b)=9\cdot8\cdot10 \Leftrightarrow 11a+b=80 \Leftrightarrow \\11a=80-b,~cum~b~este~cifra~iar~80-b=\mathcal{M}_{11} \implies 80-b=77 \implies b =3,~a=7.$

$\displaystyle\bf\\\boxed{\bf c=8} \implies 9(11a+b)=8\cdot 7\cdot 9 \implies 11a+b=56 \implies \\11a=56-b \Leftrightarrow 56-b=\mathcal{M}_{11},~dar~b~este~cifra~\implies 56-b=55\implies b=1,~a=5.\\\overline{abc}~poate~fi~518~sau~739.$