Question

tìm số dư trong phép chia ( theo Ptp đồng dư) a)1^5+3^5+5^5+....+75^5+99^5:4 b) 35^2 -35^3+35^4-35^8+35^16+35^32:425

Answer 1

a) \(S=1^5+3^5+....+75^5+99^5\)

\(\left(2a+1\right)^5-\left(2a+1\right)=2a\left(2a+1\right)\left(2a+2\right)\left[\left(2a+1\right)^2+1\right]\)

\(\left(2a+1\right)^5-\left(2a+1\right)=4a\left(2a+1\right)\left(a+1\right)\left[\left(2a+1\right)^2+1\right]⋮4\)

\(S=\left(1^5-1\right)+\left(3^5-3\right)+....+\left(75^5-75\right)+\left(99^5-99\right)+\left(1+3+5+...+75+99\right)\)

\(\Leftrightarrow\begin{matrix}1^5-1⋮4\\3^5-3⋮4\\5^5-5⋮4\\...........\\75^5-5⋮4\\99^5-99⋮4\end{matrix}\)

\(S_1=1+3+5+7+...+75+99=\frac{\left(1+75\right)\left[\frac{75-1}{2}+1\right]}{2}+99=38.38+96+3\)

\(\Rightarrow S_1:4\) dư 3

\(\Leftrightarrow S\) chia 4 dư 3