Lời giải:
Ta có:
\(A=3^{29}+2^{29}+3^{27}+2^{27}\)
\(=(3^{29}+3^{27})+(2^{29}+2^{27})\)
\(=(3^{27+2}+3^{27})+(2^{27+2}+2^{27})\)
\(=3^{27}(3^2+1)+2^{27}(2^2+1)\)
\(=10.3^{27}+5.2^{27}=10.3^{27}+10.2^{26}\)
\(=10(3^{27}+2^{26})\vdots 10\)
Vậy \(A\vdots 10\)