Question

aflati numerele de forma abc in baza zece ,stiind ca a,bc+c,ba=8,28

Answer 1

[tex]\overline{a,bc}+\overline{c,ba}=8,28|\cdot 100\\ \overline{abc}+\overline{cba}=828\\ 100a+10b+c+100c+10b+a=828\\ 101a+20b+101c=828\\ 101(a+c)=828-20b\\ Deorece\ membrul\ stang\ este\ multiplu\ de\ 101,atunci\ si\ membrul\\ drept\ trebuie\ sa\ fie\ multiplu\ de\ 101.Singura\ solutie\ convenabila\\ este\ b=1\\ 101(a+c)=808\\ a+c=8;\ a,c\neq 0\\ S:\overline{abc}\in \{117,216,315,414,513,612,711\}[/tex]