\(A=7^1+7^2+...+7^{2013}\)
\(A=\left(7^1+7^2+7^3\right)+\left(7^4+7^5+7^6\right)+...+\left(7^{2011}+7^{2012}+7^{2013}\right)\)
\(A=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{2011}\left(1+7+7^2\right)\)
\(A=7.57+7^4.57+...+7^{2011}.57\)
\(A=57\left(7+7^4+...+7^{2011}\right)\)
\(A=19.3.\left(7+7^4+...+7^{2011}\right)\) chia hết cho 19
Vậy A chia 19 dư 0
Ta có: A=7+7^2+7^3+...+7^2013
=(7+7^2+7^3)+(7^4+7^5+7^6)+...+(7^2011+7^2012+7^2013)
=7.(1+7+7^2)+7^4.(1+7+7^2)+...+7^2011.(1+7+7^2)
=7.57+7^4.57+..+7^2011.57
=57.(7+7^4+..+7^2011) (chia hết cho 57)
Vì 57 chia hết cho 19
Nên A chia hết cho 19
A= 71 + 72 + ... + 72013
= (71 + 72 + 73) + ... + (72011 + 72012 + 72013)
= 71 . (1 + 7 + 49) + ... + 72011 . (1 + 7 + 49)
= 71 . 57 + ... + 72011 . 57
= 71 . 19 . 3 + ... + 72011 . 19 . 3
=> A chia hết cho 19 ( dư 0 )
Nhớ k nha !!!
