Question

Aflati numerele abc dacă abc = a+b+c^3. Justificati!​

Answer 1

abc = a+b+c³

100a+10b+c=a+b+c³

99a+9b=c³-c

9(11a+b)=c(c²-1)

9(11a+b)=c(c-1)(c+1)

=> c-1=9, sau c=9; sau c+1=9

I. c-1=9 =>c=10; c+1=11

9(11a+b)=9•10•11   /:9

11a+b =110, nu convine, 11a+b ≤108; pentru a=b=9

II. c=9=>c-1=8; c+1=10

9(11a+b)=8•9•10    /:9

11a+b=80 => a=7 si b=3 => abc=739

739=7+3+9³  (A)

III. c+1=9=>c=8; c-1=7

9(11a+b)=7•8•9     /:9

11a+b=56 => a=5 si b=1 => abc=518

518=5+1+8³  (A)

abc∈{518; 739}