Question

  Chứng minh rằng    ₂3⁴ⁿ⁺¹+3 chia hết cho 11

Answer 1

3⁴ⁿ⁺¹ = (3⁴)ⁿ.3 = 81ⁿ.3

Ta có 81 = 1 (mod 10) 

=> 81ⁿ = 1ⁿ (mod 10)

=> 81^{n .}3 = 3 (mod 10)

=> 3⁴ⁿ⁺¹ = 3 (mod 10) Hay 3⁴ⁿ⁺¹ chia cho 10 dư 3 => 3⁴ⁿ⁺¹ = 10.k + 3

Vậy \(2^{3^{4n+1}}=2^{10k+3}=\left(2^{10}\right)^k.2^3=1024^k.8\)

Ta có 1024 = 1(mod 11) => 1024^k = 1(mod 11) => \(2^{3^{4n+1}}=1024^k.8\) = 8 (mod 11)

=> \(2^{3^{4n+1}}+3\) = (8 + 3) (mod 11) = 11 (mod 11) => \(2^{3^{4n+1}}+3\) chia hết cho 11