Răspuns :
[tex]\displaystyle\it\\\overline{abc}=b^b-b-c,~a,b,c~cifre.\\evident,~100\leq \overline{abc}\leq 999 \implies 100\leq b^b-b-c\leq 999 \implies b\leq 4.\\daca~b\leq 3 \implies b^b-b-c\leq 100,~deci~evident~\boxed{\it b=4}~.\\\overline{abc}=b^b-b-c \Leftrightarrow 100a+10b+c=b^b-b-c \Leftrightarrow 100a+2c=b^b-b-10b,~\\dar~b=4 \implies 100a+2c=4^4-4-40=212,~impartim~membrii~ecuatiei~la~2.\\50a+c=106,~evident~a~va~fi~2,~atlfel~50a+c\leq 100.\\100+c=106 \implies c=6.\\asadar,~\boxed{\it \overline{abc}=246}~.[/tex]
Răspuns:
multumesc colegului pseudoecho, solutia lui buna m-a ajutat sa vad unde gresisem eu pusesem un b in loc de 10b , la abc numar
!
abc numar=246
Explicație pas cu pas:
100a+10b+c=b^b-b-c
100a+11b+2c=b^b
b cifra, pt b=1, b^b o cifra,
pt b=2 , 2²=4 .... 1 cifra
b=3, 3³=27 ....2 cifre
pt b4.4^4=256 trei cifre
b=5, 5^5=3125, 4 cifre
convine doar b=4
fie b=4
100a+11b+2c=4^4=256
la bunul simt a=2
11*4+2c=56
44+2c=56
2c=12
c=6
abc numar=246