Đặt \(A=10^n+18n-1\)
\(=\left(10^n-1\right)+18n\)
\(=9999999...99+18n\)
| n chữ số 9 |
\(\Rightarrow\frac{A}{9}=11111...1+2n\)
| n chữ số 1 |
Có :
\(111111...111\text{≡}1.n\text{≡}n\left(mod3\right)\)
| n chữ số 1|
\(\Rightarrow111111....11+2n\text{≡}n+2n\text{≡}3n\text{≡}0\left(mod3\right)\)
\(\Rightarrow\frac{A}{9}\text{≡}0\left(mod3\right)\)
\(\Rightarrow A\text{≡}0\left(mod27\right)\)
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