Question

Să se determine nr ab [cu bară de asupra] stiind că: aaa [cu bară de asupra]+37×(a+b) este pătrat perfect​

Answer 1

$\displaystyle\it\\\overline{aaa}+37(a+b)=k^2,~a,b~cifre,~k\in\mathbb{N}.\\\overline{aaa}+37(a+b)=100a+10b+a+37a+37b=111a+37a+37b=\\148a+37b=37(4a+b)=k^2.\\4\leq 4a+b\leq 45~(1).\\37(4a+b)~este~patrat~perfect~daca~si~numai~daca~exista~m\in\mathbb{N},~a.i.\\4a+b=37m^2,~dar~din~(1) \stackrel{(1)}{\Longrightarrow} m^2=1 \implies m=1.\\atunci,~4a+b=37,~de~unde~evident~daca~a\leq6~atunci~b\geq 10.\\daca~a=7 \implies b=9.\\daca~a=8 \implies b=5.\\daca~a=9 \implies b=1.$

$\displaystyle\it\\asadar,~\boxed{\it \overline{ab} \in \left\{79,85,91\right\}}}~.$