a,\(A=2+2^2+2^3+.....+2^{20}=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+....+\left(2^{16}+2^{17}+2^{18}+2^{19}+2^{20}\right)=1.60+2^6.\left(2+2^2+2^3+2^4+2^5\right)+......+2^{16}.\left(2+2^2+2^3+2^4+2^5\right)=1.60+2^6.60+....+2^{16}.60=\left(1+2^6+....+2^{16}\right).60=>A⋮60⋮5\)
b,\(3⋮n+2=>n+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Xét TH
TH1:n+2=-1
=>n=-1-2
=>n=-3(Loại)
TH2:n+2=-3
=>n=-3-2
=>n=-5(Loại)
TH3:n+2=1
=>n=1-2
=>n=-1(Loại)
TH4:n+2=3
=>n=3-2
=>n=1(Thoả mãn)
Vậy n=1