Question

Determinaţi numărul natural abc cu proprietatea că
abc = b^b - b - c. (abc este in baza 10) .

Va rog frumos un pic de ajutor la rezolvarea acestei probleme, multumesc!​

Answer 1

$\overline{abc}=b^b-b-c.\\plecam~de~la~faptul~ca~100\leq \overline{abc} \leq 999 \Leftrightarrow 100\leq b^b-b-c\leq 999 \implies \\0\leq b\leq 4,~daca~b\geq 5,~atunci~b^b-b-c\geq 1000.\\\overline{abc}=b^b-b-c \Leftrightarrow 100a+10b+c=b^b-b-c.\\daca~b\leq 3,~atunci~b^b-b\leq 100,~prin~urmare~\boxed{b=4}~.\\100a+40+2c=4^4-4=252~\bigg|-40 \implies 100a+2c=212~\bigg|:2 \implies\\50a+c=106,~evident~\boxed{a=2} \implies \boxed{c=5}~.$