Question

Gasiti nr aa si bb astfel incat aa × aa+bb+c =2015

Answer 1

$aa*aa+bb=11a*11a+11b=11(11 a^{2}+b)= M_{11} . \\ 2015=183*11+2= M_{11} +2. \\ \\ aa*aa+bb+c=2015<=> M_{11} +c= M_{11}+2=>c= M_{11}+2,~iar \\ cum~c~este~cifra~=>~c=2.$

$aa*aa+bb+c=2015<=>(aa) ^{2}+bb+2=2015=> \\ =>(aa)^{2} +bb=2013<=>121 a^{2} +11b=2013<=> \\ <=>11(11 a^{2}+b)=11*183 => 11a^{2}+b=183 \\ \\ 11 a^{2}= M_{11}~si~183= 11*16+7=M_{11}+7=>b= M_{11}+7,~dar \\ b~este~cifra~=>b=7. \\ \\ 11 a^{2} +b=183<=>11 a^{2}+7=183=>11 a^{2}=176=> a^{2}=16=> \\ =>a=4.$

$Avem~a=4,~b=7~si~c=2. \\ aa=44~si~bb=77.$