Question

[Ninh Nguyễn Anh Ngọc]

CMR: A = 3¹⁰⁵ + 4¹⁰⁵ chia hết cho 13 nhưng không chia hết cho 11

Answer 1

[soyeon_Tiểubàng giải]

Ta có:

• \(3^3=27\equiv1\left(mod13\right)\Rightarrow\left(3^3\right)^{35}=3^{105}\equiv1\left(mod13\right)\)

\(4^3=64\equiv-1\left(mod13\right)\Rightarrow\left(4^3\right)^{35}=4^{105}\equiv-1\left(mod13\right)\)

Vậy \(A=3^{105}+4^{105}\equiv1+\left(-1\right)\left(mod13\right)\) hay \(A⋮13\left(1\right)\)

• \(4^3\equiv-2\left(mod11\right)\Rightarrow\left(4^3\right)^5=4^{15}\equiv\left(-2\right)^5\left(mod11\right)\) hay \(4^{15}\equiv1\left(mod11\right)\)

\(3^5=243\equiv1\left(mod11\right)\Rightarrow\left(3^5\right)^{21}=3^{105}\equiv1\left(mod11\right)\)

Vậy \(A=3^{105}+4^{105}\equiv1+1\left(mod11\right)\) hay \(A=3^{105}+4^{105}\equiv2\left(mod11\right)\)

=> A không chia hết cho 11 (2)

Từ (1) và (2) => đcpm

Answer 2

[Nguyễn Anh Duy]

Chứng minh chia hết cho 13:

\(A=3^{105}+4^{105}\\ A=\left(3^3\right)^{35}+\left(4^3\right)^{35}\\ A=27^{35}+64^{35}\\ A=\left(27+64\right)\left(27^{34}-27^{33}.35+.......+35^{34}\right)\)

\(A=91\left(27^{34}-27^{33}.35+........+35^{34}\right)\)

\(A=13.7\left(27^{34}-27^{33}.35+........+35^{34}\right)\) chia hết cho 13

Chứng minh không chia hết cho 11

\(3^{105}=243^{21}=\left(242+1\right)^{21}=242^{21}+2.242+1^{21}=242^{21}+2.242+1\)

Vì \(242\) chia hết cho 11 nên \(242^{21}+2.242+1\) chia 11 dư 1

\(4^{105}=1024^{21}=\left(1023+1\right)^{21}=1023^{21}+2.1023+1\)

Vì \(1023\) chia hết cho 11 nên \(1023^{21}+2.1023+1\) chia 11 dư 1

Vậy tổng \(A=3^{105}+4^{105}\) chia 11 dư 2 \(\left(1+1\right)\)

Vậy A không chia hết cho 11 (2)