Question

Chứng minh rằng: 43¹⁰¹ +23¹⁰¹ chia hết cho 66

Giúp mình giải bài này với!

Answer 1

\(43^{101}+23^{101}=43\cdot43^{100}+23\cdot23^{100}=\left(66-23\right)\cdot43^{100}+23\cdot23^{100}\)

\(=66\cdot43^{100}-23\cdot43^{100}+23\cdot23^{100}=66\cdot43^{100}-23\left(43^{100}-23^{100}\right)\)

\(=66\cdot43^{100}-23\left(43-23\right)\left(43^{99}+43^{98}\cdot23+43^{97}\cdot23^2+43^{96}\cdot23^3+...+43\cdot23^{98}+23^{99}\right)\)

\(=66\cdot43^{100}-23\cdot20\left(43^{98}\left(43+23\right)+43^{96}\cdot23^2\left(43+23\right)+...+23^{98}\left(43+23\right)\right)\)

\(=66\cdot43^{100}-460\left(4^{98}\cdot66+4^{96}\cdot23^2\cdot66+...+23^{98}\cdot66\right)\)

\(=66\cdot43^{100}-460\cdot66\left(4^{98}+4^{96}\cdot23^2+...+23^{98}\right)\)

\(=66\left(43^{100}-460\left(4^{98}+4^{96}\cdot23^2+...+23^{98}\right)\right)⋮66\Rightarrow43^{100}+23^{100}⋮66\)(đpcm)

Answer 2

cái chỗ \(43^{100}-23^{100}=\left(43-23\right)\left(43^{99}+43^{98}\cdot23+43^{97}\cdot23^2+43^{96}\cdot23^3+...+43\cdot23^{98}+23^{99}\right)\)

là áp dụng hđt \(a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+a^{n-4}b^3+...+b^{n-1}\right)\)