Áp dụng hằng đẳng thức sau
an−1=(a−1).[an−1+an−2+...+1]=(a−1).pan−1=(a−1).[an−1+an−2+...+1]=(a−1).p (nn là 1 số nguyên dương)
an+1=(a+1).[an−1−an−2+..+1]=(a+1).qan+1=(a+1).[an−1−an−2+..+1]=(a+1).q (nn là 1 số nguyên dương lẻ)

Thay vào ta được như sau:

+) 222333−1=(222−1).p=13.17.p222333−1=(222−1).p=13.17.p

+) 333222+1=(3332)111+1=110889111+1=(110889+1).q=13.8530.q333222+1=(3332)111+1=110889111+1=(110889+1).q=13.8530.q

=>=> 222333+333222=222333−1+333222+1=13(17p+8530q)⋮13222333+333222=222333−1+333222+1=13(17p+8530q)⋮13

Vậy: 222333+333222⋮13222333+333222⋮13 (đpcm)(đpcm) 

\(\left(222^{333}+333^{222}\right)⋮13\)

Áp dụng hằng đẳng thức sau
$a^n-1=(a-1).[a^{n-1}+a^{n-2}+...+1]=(a-1).p$ ($n$ là 1 số nguyên dương)
$a^n+1=(a+1).[a^{n-1}-a^{n-2}+..+1]=(a+1).q$

a) \(222^{333}+333^{222}\)

\(=\left(111.2\right)^{333}+\left(111.3\right)^{222}\)

\(=111^{333}.2^{333}+111^{222}.3^{222}\)

\(=111^{222}.\left(111^{111}.2^{333}+3^{222}\right)\)

\(=111^{222}.\left(111^{111}.2^{3.111}+3^{2.111}\right)\)

\(=111^{222}.\left[111^{111}.\left(2^3\right)^{111}+\left(3^2\right)^{111}\right]\)

\(=111^{222}.\left(111^{111}.8^{111}+9^{111}\right)\)

\(=111^{222}.\left[\left(111.8\right)^{111}+9^{111}\right]\)

\(=111^{222}.\left(888^{111}+9^{111}\right)\)

\(=111^{222}.\left(888+9\right)\left[888^{110}-888^{109}.9+.....-888.9^{109}+9^{110}\right]\)

\(=111^{222}.7992\left[888^{110}-888^{109}.9+.....-888.9^{109}+9^{110}\right]\)

\(=111^{222}.897\left[888^{110}-888^{109}.9+.....-888.9^{109}+9^{110}\right]\)

\(=111^{222}.13.69\left[888^{110}-888^{109}.9+.....-888.9^{109}+9^{110}\right]⋮13\)

Vậy \(222^{333}+333^{222}⋮13\left(dpcm\right)\)

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.

a: \(222^{333}-1=\left(222-1\right)\left(222^{332}+222^{331}+\cdots+1\right)\)

\(=221\cdot\left(222^{332}+222^{331}+\cdots+1\right)=13\cdot17\cdot\left(222^{332}+222^{331}+\cdots+1\right)\) ⋮13

=>\(222^{333}\) =1(mod 13)

333 chia 13 dư 12

=>333=-1(mod 13)

=>\(333^{222}\) =-1(mod 13)

Do đó: \(222^{333}+333^{222}\) ⋮13

b: \(3^{105}+4^{105}\)

\(=\left(3^3\right)^{35}+\left(4^3\right)^{35}\)

\(=\left(3^3+4^3\right)\left(3^{102}-3^{99}\cdot4^3+\cdots+4^{102}\right)\)

\(=\left(27+64\right)\left(3^{102}-3^{99}\cdot4^3+\cdots+4^{102}\right)\)

\(=91\left(3^{102}-3^{99}\cdot4^3+\cdots+4^{102}\right)=13\cdot7\cdot\left(3^{102}-3^{99}\cdot4^3+\cdots+4^{102}\right)\) ⋮13

du 2 h cho minh nha