Ta có
$A = 1 + 2^1 + 2^2 + 2^3 + \cdots + 2^{59} + 2^{60}$
$= (1+ 2^1) + 2^2(1+2^1) + \cdots + 2^{59}(1+2^1)$
$= 3+ 2^2.3 + \cdots + 2^{59}.3$
$= 3(1 + 2^2 + \cdots + 2^{59})$
Vậy A chia hết cho 3.
Tiếp theo, ta có
$A = (1 + 2^1 + 2^2) + 2^3(1 + 2^1 + 2^2) + \cdots + 2^{58}(1 + 2^1 + 2^2)$
$= 7 + 2^3 . 7 + \cdots + 2^{58} . 7$
$ = 7(1 + 2^3 + \cdots + 2^{58})$
Vậy A chia hết cho 7.
Mặt khác, ta có
$A = (1 + 2^1 + 2^2 + 2^3) + 2^4(1 + 2^1 + 2^2 + 2^3) + \cdots + 2^{57}(1 + 2^1 + 2^2 + 2^3)$
$= 15 + 2^4.15 + \cdots + 2^{57} . 15$
$= 15(1 + 2^4 + \cdots + 2^{57})$
Vậy A chia hết cho 15.
