a) Phân tích thành tích các thừa số nguyên tố: \(3072=2^{10}.3\)
Tổng các ước của \(3072\)là:
\(\sigma\left(3072\right)=1+3+\left(2^1+2^2+...+2^{10}\right)+3\left(2^1+2^2+...+2^{10}\right)\)
\(=4\left(1+2+2^2+...+2^{10}\right)\)
Ta có: \(A=1+2+...+2^{10}\)
\(2A=2+2^2+...+2^{11}\)
\(2A-A=\left(2+2^2+...+2^{11}\right)-\left(1+2+...+2^{10}\right)\)
\(A=2^{11}-1\)
Suy ra \(\sigma\left(3072\right)=4\left(2^{11}-1\right)=2^{13}-4\)
b) Tương tự.
\(4608=2^9.3^2\)
\(\sigma\left(4608\right)=1+3+3^2+\left(2^1+2^2+...+2^9\right)+3\left(2^1+2^2+...+2^9\right)+3^2\left(2^1+2^2+...+2^9\right)\)
\(=\left(1+3+3^2\right)\left(1+2^1+2^2+...+2^9\right)\)
\(=13299\)
