Đề bạn chỉnh là A chia hết cho 2 thì giải như sau:
17\(\equiv\)1(mod 2)
=>172\(\equiv\)1(mod 2)
13\(\equiv\)1(mod 2)
1321\(\equiv\)1(mod 2)
=>172-1321\(\equiv\)1-1\(\equiv\)0(mod 2)
Hay 172-1321 chia hết cho 2
24 chia hết cho 2
=>244 chia hết cho 2
=>A=172+244-1321 chia hết cho 2
Ta có: \(17^2=\left(...9\right)\) ; \(24^4=\left(24^2\right)^2=\left(...6\right)^2=\left(...6\right)\)
\(13^{21}=\left(13^4\right)^5.13=\left(...1\right)^5.13=\left(...1\right).13=\left(...3\right)\)
\(\Rightarrow A=17^2+24^4-13^{21}=\left(...9\right)+\left(...6\right)-\left(...3\right)\)
\(\Rightarrow A=\left(...5\right)-\left(...3\right)\)
\(\Rightarrow A=\left(...2\right)⋮2\)
\(\Rightarrow A⋮2\left(đpcm\right)\)
Sửa đề:
CMR \(A=17^2+24^4-13^{21}⋮2\)
Lời giải:
Ta có:
\(17^2=\overline{...9}\)
\(24^4=\left(24^2\right)^2=\overline{...6}^2=\overline{...6}\)
\(13^{21}=\left(13^4\right)^5.13=\overline{...1}^5.13=\overline{...1}.13=\overline{...3}\)
\(\Rightarrow A=17^2+24^2-13^{21}=\overline{...9}+\overline{...6}-\overline{...3}=\overline{...2}⋮2\)
\(\Rightarrow A⋮2\)
