Ta có n=Abcd=1000A+bcd=999A+A+bcd=3.37.9A+(A+bcd). Vì A+bcd chia hết 37, 3.37.9A chia hết cho 37=>1000A+bcd chia hết cho 37 hay Abcd chia hết cho 37. Vậy n chia hết cho 37

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\(\frac{7}{1x3}+\frac{7}{3x5}+...+\frac{7}{99x101}=\frac{7}{2}\left(\frac{2}{1x3}+\frac{2}{3x5}+...+\frac{2}{99x101}\right)\)

\(=\frac{7}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{99}-\frac{1}{101}\right)=\frac{7}{2}\left(1-\frac{1}{101}\right)=\frac{7}{2}.\frac{100}{101}=\frac{350}{101}\)