Question

[Trần Ngọc Linh]

Tìm x,y,z nguyên thỏa mãn 0<x<y<z: 2^x+2^y+2^z=4736

Answer 1

[Nguyễn Hoàng Minh]

\(x=7;y=9;z=12\)

\(2^x+2^y+2^z=4736\\ \Rightarrow2^x\left(1+2^{y-x}+2^{z-x}\right)=4736\)

Ta có \(0< x< y< z\Rightarrow y-z>0;x-z>0\)

\(\Rightarrow1+2^{y-x}+2^{z-x}\) lẻ 

\(\Rightarrow4736=2^7\cdot37=2^x\left(1+2^{y-x}+2^{z-x}\right)\\ \Rightarrow\left\{{}\begin{matrix}x=7\\2^{y-x}+2^{z-x}+1=37\left(1\right)\end{matrix}\right.\\ \Rightarrow2^{y-7}+2^{z-7}=36\\ \Rightarrow2^{y-7}\left(1+2^{z-y}\right)=36=2^2\cdot3^2\)

Mà \(0< y< z\Rightarrow z-y>0\Rightarrow1+2^{z-y}\) lẻ

\(\Rightarrow\left\{{}\begin{matrix}y-7=2\\1+2^{z-y}=3^2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=9\\2^{z-9}=8=2^3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=9\\z=12\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(7;9;12\right)\)