Chứng minh rằng: A = $2+2^2+2^3+\mathrm{...}+2^{60}$  chia hết cho 3, 7, 15.

A =   $\left(2+2^2\right)+\left(2^3+2^4\right)+\mathrm{...}+\left(2^{59}+2^{60}\right)=3.2+{3.2}^2+\mathrm{....}+{3.2}^{59}$

Do đó, A chia hết cho 3

A = $\left(2+2^2+2^3\right)+\left(2^4+2^6+2^7\right)\mathrm{...}+\left(2^{58}+2^{59}+2^{60}\right)=7.2+{7.2}^4+\mathrm{...}+{7.2}^{58}$

Do đó, A chia hết cho 7

A = $\left(2+2^2+2^3+2^4\right)+\mathrm{...}+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)=2.15+\mathrm{...}+2^{57}.15$

Do đó, A chia hết cho 15.