\(222^{333}+333^{222}=\left(2^3\right)^{111}+\left(3^2\right)^{111}=8^{111}+9^{111}=\left(8+9\right)\cdot Q=17\cdot Q⋮17\)

Có thể mình làm sai hoặc bạn nhầm đề rồi nha!

cảm ơn bạn nhiều mình không chắc là mình viết đứng ko nữa dù sao cũng cảm ơn bạn vì đã giúp mình

Ừm k có j ^_^ 

#NguyễnMinhTrường

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

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.