Question

[Nguyễn Hoàng]

Cho \(A=1^n+2^n+3^n+4^n\).Tìm n để A chia hết cho 5

Answer 1

[Đoàn Đức Hà]

\(A\left(n\right)=1^n+2^n+3^n+4^n\)

\(n\)chẵn: \(n=2k\)

\(A=1^{2k}+2^{2k}+3^{2k}+4^{2k}=1+4^k+9^k+16^k\equiv\left[1+\left(-1\right)^k+\left(-1\right)^k+1^k\right]\left(mod5\right)\)

Với \(k\)lẻ thì \(1+\left(-1\right)^k+\left(-1\right)^k+1^k=0\)do đó \(A⋮5\).

Với \(k\)chẵn thì \(1+\left(-1\right)^k+\left(-1\right)^k+1^k=4\)do đó \(A⋮̸5\).

\(n\)lẻ: \(n=2k+1\)

\(A=1^{2k+1}+2^{2k+1}+3^{2k+1}+4^{2k+1}=1+2.4^k+3.9^k+4.16^k\)

\(\equiv1+2.\left(-1\right)^k+3.\left(-1\right)^k+4.1\left(mod5\right)\)

\(\equiv5\left(-1\right)^k\left(mod5\right)\)

\(\equiv0\left(mod5\right)\).

Vậy \(n⋮̸4\)thì \(A⋮5\).