a)
Ta có: \(222^{333}=\left(222^3\right)^{111}\equiv1^{111}=1\left(mod13\right)\)
\(\Rightarrow222^{333}+333^{222}\equiv1+333^{222}=1+\left(333^2\right)^{111}\)
\(\equiv1+12^{111}\equiv1+12^{110}\cdot12\equiv1+\left(12^2\right)^{55}\cdot12\)
\(\equiv1+1\cdot12\equiv13\equiv0\left(mod13\right)\)
Vậy $222^{333}+333^{222}$ chia hết cho $13.$
b) Ta có:
\(3^{105}\equiv\left(3^3\right)^{35}\equiv1^{35}\equiv1\) (mod13)
\(\Rightarrow3^{105}+4^{105}\equiv1+4^{105}\equiv1+\left(4^3\right)^{35}\)
\(\equiv1+12^{35}\equiv1+\left(12^2\right)^{17}\cdot12\equiv1+1\cdot12\equiv13\equiv0\left(mod13\right)\)
Vậy $3^{105}+4^{105}$ chia hết cho $13.$
Lại có:
\(3^{105}\equiv\left(3^3\right)^{35}\equiv5^{35}\equiv\left(5^5\right)^7\equiv1\left(mod11\right)\)
\(4^{105}\equiv\left(4^3\right)^{35}\equiv9^{35}\equiv\left(9^5\right)^7\equiv1\left(mod11\right)\)
Từ đây:\(3^{105}+4^{105}\equiv1+1\equiv2\left(mod11\right)\)
Vậy $3^{105}+4^{105}$ không chia hết cho $11.$
P/s: Rất lâu rồi không giải, không chắc.
