ta co nhan xet S=1+2+3+...+x se co \(\left(x-1\right):1+1\) =x so hang
va S= \(\frac{x\left(x+1\right)}{2}\)
mat khac dat so co 3 chu so can tim la aaa =a.111=a.3.37
=> \(\frac{x\left(x+1\right)}{2}=a.3.37\Rightarrow\frac{x\left(x+1\right)}{37}=6a\)
vi \(a\inℕ^∗\Rightarrow6a\inℕ^∗\Rightarrow\frac{x\left(x+1\right)}{37}\inℕ^∗\) va \(\frac{x\left(x+1\right)}{37}⋮6\) (*)
hay \(x\left(x+1\right)⋮37\)
mat khac \(x\left(x+1\right)⋮2\forall x\inℕ\)
va 37 la so nguyen to
=> \(\orbr{\begin{cases}x=37\\x+1=37\end{cases}\Leftrightarrow\orbr{\begin{cases}x=37\\x=36\end{cases}}}\)
thu lai vao dk (*) ta thay chi co x=36 thoa man
Vay can 36 so hang de S la mot so co 3 chu so giong nhau
