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Bạn xem lại đề bài nhé. Với \(a=1,b=9\) thì \(111a+25b=336⋮12\) nhưng \(9a+13b=126⋮̸12\). Mình nghĩ đề bài là chứng minh \(9a+3b⋮12\). Vì \(111a+25b⋮12\) nên \(108a+24b+3a+b⋮12\) hay \(3a+b⋮12\) hay \(9a+3b⋮12\).
Chứng minh: a,222^333+333^222 chia hết cho 13
b, 3^105+4^105 chai hết cho 13 nhưng ko chia hết cho 11
![](https://rs.olm.vn/images/avt/0.png?1311)
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.
Chứng minh: a,222^333+333^222 chia hết cho 13
b, 3^105+4^105 chai hết cho 13 nhưng ko chia hết cho 11
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9^{34}-27^{22}+81^{16}.\)
\(=\left(3^2\right)^{34}-\left(3^3\right)^{22}+\left(3^4\right)^{16}\)
\(=3^{68}-3^{66}+3^{64}\)
\(=3^{64}.\left(3^4-3^2+1\right)\)
\(=3^{64}.\left(81-9+1\right)\)
\(=3^{64}.73\)
\(=3^{62}.3^2.73\)
\(=3^{62}.9.73\)
\(=3^{62}.657\)
Vì \(657⋮657\) nên \(3^{62}.657⋮657.\)
\(\Rightarrow9^{34}-27^{22}+81^{16}⋮657\left(đpcm\right).\)
Chúc bạn học tốt!
\( {9^{34}} - {27^{22}} + {81^{16}}\\ = {\left( {{3^2}} \right)^{34}} - {\left( {{3^3}} \right)^{22}} + {\left( {{3^4}} \right)^{16}}\\ = {3^{68}} - {3^{66}} + {3^{64}}\\ = {3^{62}}\left( {{3^6} - {3^4} + {3^2}} \right)\\ = {3^{62}}\left( {729 - 81 + 9} \right)\\ = {3^{63}}.657\)
chia hết cho $657$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(9^{34}-27^{22}+81^{16}=9^{34}-\left(3^3\right)^{22}+\left(9^2\right)^{16}\)
\(=9^{34}-3^{66}+9^{32}=9^{34}-9^{33}+9^{32}\)
\(=9^{32}\left(9^2-9+1\right)=9^{32}.73\)
\(=9^{31}.\left(8.73\right)=9^{31}.657⋮657\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a: \(3^x=243\)
nên \(3^x=3^5\)
hay x=5
b: \(x^5=32\)
nên \(x^5=2^5\)
hay x=2
c: \(x^6=729\)
\(\Leftrightarrow x^2=9\)
=>x=3 hoặc x=-3
a) Sai đề.
b) \(9^{34}-27^{22}+81^{16}\)
\(=3^{68}-3^{66}+3^{64}\)
\(=3^{64}\left(3^4-3^2+1\right)=3^{64}.73=3^{62}.9.73\)
= \(3^{62}.657⋮657\)