Nếu là dấu cộng thì làm như sau
Ta có C = \(2+2^2+2^3+2^4+2^5+...+2^{14}+2^{15}\)
= \(\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+\left(2^{11}+2^{12}+2^{13}+2^{14}+2^{15}\right)\)
= \(62+2^5\left(2+2^2+2^3+2^4+2^5\right)+2^{10}\left(2+2^2+2^3+2^4+2^5\right)\)
= \(62+2^5.62+2^{10}.62=62.\left(1+2^5+2^{10}\right)⋮62\)