$10 \cdot 75$ product $\left(1+\frac{1}{1 \cdot 3}\right)\left(1+\frac{1}{2 \cdot 4}\right)\left(1+\frac{1}{3 \cdot 5}\right)\left(1+\frac{1}{4 \cdot 6}\right) \cdots$ $\left(1+\frac{1}{98 \cdot 100}\right)\left(1+\frac{1}{99 \cdot 101}\right)$ has an integer part of
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(China Guangzhou, Wuhan, Fuzhou and other five cities junior high school mathematics league, 1991)
[Solution] Notice that $1+\frac{1}{k(k+2)}=\frac{k^{2}+2 k+1}{k(k+2)}=\frac{(k+1)^{2}}{k(k+2)}$, then $\begin{aligned} \text { original expression } & =\frac{2^{2}}{1 \cdot 3} \cdot \frac{3^{2}}{2 \cdot 4} \cdot \frac{4^{2}}{3 \cdot 5} \cdot \frac{5^{2}}{4 \cdot 6} \cdots \cdot \frac{99^{2}}{98 \cdot 100} \cdot \frac{100^{2}}{99 \cdot 101} \\ & =\frac{2 \cdot 100}{101}=1 \frac{99}{101} . \text { (Note the cancellation of terms in the numerator and denominator) }\end{aligned}$ Therefore, the answer is $(A)$.