Ta có : \(\left(n-2\right)^2-\left(n^2-1\right)=\overline{cba}\) \(-\overline{abc}\)
\(\left(n-2\right)\left(n-2\right)-n^2+1=100c+10b+a-\left(100a+10b+c\right)\)
\(\left(n-2\right)n-2\left(n-2\right)-n^2+1=100c+10b+a-100a+10b+c\)
\(n^2-2n-2n+4-n^2+1=99c-99a\)
\(5-4n=99\left(c-a\right)\)
\(99\left(c-a\right)=4n-5\)
\(\Rightarrow4n-5⋮99\)
Ta có \(4n-5=99\Rightarrow n=26\)
\(\Rightarrow\overline{abc}\) \(=n^2-1=26^2-1\Rightarrow676-1=675\)
Vậy số cần tìm là 675
k cho mk nha !
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