Trong A có 53 số hạng. Mà 53 chia 3 dư 2
Viết: A = (1 + 2) + (22 + 23 + 24) + (25 + 26 + 27) + ... + (250 + 251 + 252)
\(\Leftrightarrow\) A = 3 + 22 . (1 + 2 + 22) + 25 . (1 + 2 + 22) + ... + 250 . (1 + 2 + 22)
\(\Leftrightarrow\) A = 3 + 22 . 7 + 25 . 7 + ... + 250 . 7
\(\Leftrightarrow\) A = 3 + 7 . (22 + 25 + ... + 250)
Rõ ràng A chia 7 dư 3.
\(M=2^0+2^1+2^2+2^3+2^4+2^5+....+2^{50}+2^{51}+2^{52}\)
\(M=\left(2^0+2^1+2^2\right)+\left(2^3+2^4+2^5\right)+....+\left(2^{50}+2^{51}+2^{52}\right)\)
\(M=\left(1+2^1+2^2\right)+2^3\left(1+2^1+2^2\right)+....+2^{50}\left(1+2^1+2^2\right)\)
\(M=\left(1+2^1+2^2\right)\left(1+2^3+...+2^{50}\right)\)
\(M=7\left(1+2^3+...+2^{50}\right)⋮7\)
\(\Rightarrow\) Chia 7 dư 0