Question

Tìm số tự nhiên n để \(5^n\)- \(2^n\) chia hết cho 9

Answer 1

\(A=5^{3k+1}-2^{3k+1}=\left(5-2\right)\cdot\left(5^{3k}+5^{3k-1}\cdot2+...+2^{3k}\right)⋮̸9\)

(Vì \(5^{3k}+5^{3k-1}\cdot2+...+2^{3k}⋮̸3\))

TH2: n=3k+2

\(A=5^{3k+2}-2^{3k+2}\)

\(=\left(5-2\right)\cdot\left(5^{3k+1}+5^{3k}\cdot2+5^{3k-1}\cdot2^2+...+2^{3k+1}\right)\) không chia hết cho 9 vì \(5^{3k+1}+5^{3k}\cdot2+5^{3k-1}\cdot2^2+...+2^{3k+1}⋮̸3\)

TH3: n=3k

\(A=5^{3k}-2^{3k}=125^k-8^k=\left(125-8\right)\cdot B=117\cdot B⋮9\)