Tìm số tự nhiên $\overline{abc} $ thỏa mãn:
$\overline{abc} =n^{2}-1$ và $\overline{cba} =(n-2)^{2} với n\in Z; n>2$
Ta có $100\leq \overline{abc}\leq 999\Rightarrow 10\leq \left|{n} \right|\leq 31$
Có $\overline{abc} -\overline{cba}=n^2-1-(n-2)^2\Leftrightarrow 99(a-c)=4n-5 \Rightarrow 4n-5$ là 1 bội của 99
Lại có $10\leq \left| {n} \right|\leq 31\Rightarrow 0<\left| {4n-5} \right|\leq 129$
Từ 2 điều trên suy ra  $\left| {4n-5} \right|=99\Rightarrow n=26$ hoặc $n=-23,5$ (loại)
$n=26\Rightarrow \overline{abc}=675 $
Bảo cái j thì bảo luôn đi –  Dark 04-10-15 10:01 PM
ê ddarrk vào chat riêng tao bảo –  ๖ۣۜTQT☾♋☽ 04-10-15 09:46 PM
hay,hay,lắm –  ๖ۣۜTQT☾♋☽ 04-10-15 09:45 PM

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