Cho \(A = 3 + {3^2} + {3^3} + ... + {3^8} + {3^9}\) và \(B = 29 \cdot 47 - 29 \cdot 34.\)

(a)\(A\,\, \vdots \,\,3.\)

(b)\[B\,\, \vdots \,\,29.\]

(c)\(B\,\,\cancel{ \vdots }\,\,13.\)

(d)\(\left( {A + B} \right)\,\, \vdots \,\,13.\)

a) Đúng.

Ta có: \(A = 3\left( {1 + 3 + {3^2} + {3^3} + ... + {3^7} + {3^8}} \right)\) nên \(A\,\, \vdots \,\,3.\) Vậy \(A\,\, \vdots \,\,3.\)

b) Đúng.

Vì \(\left( {29 \cdot 47} \right)\,\, \vdots \,\,29;\;{\rm{ }}\left( {29 \cdot 34} \right)\,\, \vdots \,\,29\) nên \(\left( {29 \cdot 47 - 29 \cdot 34} \right)\,\, \vdots \,\,29\) hay \(B\,\, \vdots \,\,29.\) Vậy \(B\) chia hết cho 29.

c) Sai.

Ta có: \(B = 29 \cdot 47 - 29 \cdot 34 = 29\left( {47 - 34} \right) = \left( {29 \cdot 13} \right)\,\, \vdots \,\,13\) nên \(B\,\, \vdots \,\,13.\) Vậy \(B\,\, \vdots \,\,13.\)

d) Đúng.

\(A = \left( {3 + {3^2} + {3^3}} \right) + \left( {{3^4} + {3^5} + {3^6}} \right) + \left( {{3^7} + {3^8} + {3^9}} \right)\)

\(A = 3\left( {1 + 3 + {3^2}} \right) + {3^4}\left( {1 + 3 + {3^2}} \right) + {3^7}\left( {1 + 3 + {3^2}} \right)\)

\(A = 13\left( {3 + {3^4} + {3^7}} \right)\,\, \vdots \,\,13.\)

Do đó, \(A\,\, \vdots \,\,13.\) Mà \(B \vdots 13\) nên \(\left( {A + B} \right)\,\, \vdots \,\,13.\)