\(A=1+11+....+11^7+11^8+11^9\)
\(11A=11\left(1+11+.....+11^7+11^8+11^9\right)\)
\(\)\(11A=11+11^2+....+11^8+11^9+11^{10}\)
\(11A-A=\left(11+11^2+....+11^8+11^9+11^{10}\right)-\left(1+11+....+11^7+11^8+11^9\right)\)\(10A-11^{10}-1\)
\(A=\dfrac{11^{10}-1}{10}\)
Được biết:\(11^n=\overline{....1}\)
Nên: \(11^{10}-1=\overline{....1}-1=\overline{....0}\)
\(A=\dfrac{11^{10}-1}{10}=\dfrac{\overline{....0}}{10}=\overline{...0}⋮5\)
\(D=11^9+11^8+11^7+...+11+1\)
Đặt 11D= \(11^{10}+11^9+11^8+...+11^2+11\)
=> 11D-D= \(\left(11^{10}+11^9+11^8+...+11^2+11\right)-\left(11^9+11^8+11^7+...+11+1\right)\)=> 10D= \(11^{10}-1\)
=> D= \(11^{10}-1:10\)
Ta thấy: \(11^{10}\) có tận cùng là 1 mà \(11^{10}-1\) sẽ có tận cùng là 0
Mà 0 chia hết cho 5 =>\(11^{10}-1:10\) chia hết cho 5
Vậy....(đpcm)
Ta có :
\(D=11^9+11^8+.........+11+1\)
\(\Leftrightarrow11A=11^{10}+11^9+..........+11\)
\(\Leftrightarrow11A-A=\left(11^{10}+11^9+......+11\right)-\left(11^9+11^8+......+1\right)\)
\(\Leftrightarrow10A=11^{10}-1\)
\(\Leftrightarrow A=\dfrac{11^{10}-1}{10}\)
Ta thấy \(A=\dfrac{11^{10}-1}{10}=\dfrac{\left(....1\right)-1}{10}=\dfrac{\left(...0\right)}{10}=\left(...0\right)⋮5\)
\(\Leftrightarrowđpcm\)
