A=\(2^0+2^1+....+2^{100}\)
=>2A=\(2^1+2^2+2^3+.....+2^{100}+2^{101}\)
=>2A-A=\(\left(2^1+2^2+....+2^{100}+2^{101}\right)-\left(2^0+2^1+....+2^{100}\right)\)
\(\Leftrightarrow A=2^{101}-1\)
Ta có \(2^{20}=4^{10}\equiv76\) (mod 100) =>\(\left(2^{20}\right)^5=2^{100}\equiv76\) (mod 100)
=>\(2^{100}\cdot2\)\(\equiv76\cdot2=152\) (mod 100)
Hay \(2^{101}\equiv152\) (mod 100)
=>\(2^{101}-1\equiv152-1=151\) (mod 100)
Lại có 151 chia cho 100 dư 51
Do đó A chia cho 100 dữ51
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