Ta có:

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

\(=111^{222}\left[\left(111.2^3\right)^{111}+\left(3^2\right)^{111}\right]\)

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

\(\Rightarrow888^{111}+9^{111}\)

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

\(=13.69.\left(888^{110}-888^{109}.9+...-9^{109}+9^{110}\right)\)

\(=13.69.Q\)

\(\Rightarrow222^{333}+333^{222}⋮13\) (Đpcm)

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) 

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

an−1=(a−1).[an−1+an−2+...+1]=(a−1).p" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">an−1=(a−1).[an−1+an−2+...+1]=(a−1).pn" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">nan+1=(a+1).[an−1−an−2+..+1]=(a+1).q" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">an+1=(a+1).[an−1−an−2+..+1]=(a+1).qn" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">n222333−1=(222−1).p=13.17.p" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">222333−1=(222−1).p=13.17.p
333222+1=(3332)111+1=110889111+1=(110889+1).q=13.8530.q" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">333222+1=(3332)111+1=110889111+1=(110889+1).q=13.8530.q
=>" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">=>222333+333222=222333−1+333222+1=13(17p+8530q)⋮13" role="presentation" style="border:0px; color:rgb(40, 40, 40); direction:ltr; display:inline-block; float:none; font-family:helvetica,arial,sans-serif; font-size:18.06px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-wrap:normal" class="MathJax_CHTML mjx-chtml">\(222\)333+333222=222333−1+333222+1=13(17p+8530q)⋮13

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

Áp dụng công thức :\(a^n+b^n\) chia hết cho a+b

\(VT=\left(222^3\right)^{111}+\left(333^2\right)^{111}\) chia hết cho \(222^3+333^2\)

\(222^3\) chia 13 dư 1 (bấm máy tính )

\(333^2\) chia 13 dư 12 

\(\Rightarrow222^3+333^2\) chia hết cho 13 

\(\Rightarrow\) đpcm

Ta có 222 ≡ 1(mod 13) nên 222^333 ≡ 1 (mod 13) 
Và 333^2 ≡ -1 (mod 13) nên 333^222 ≡ -1 (mod 13) 
Cộng lại ta có: 
222^333 + 333^222 ≡ 0 (mod 13) đpcm 
Bài 2: 
Ta có 109^3 ≡ 1 (mod 7) nên 109^345 ≡ 1( mod 7) 
Vậy số dư của phép chia trên là 1

cho mình hỏi mod là j???

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.