\(A=1.3.5.7...99=\frac{\left(1.3.5.7...99\right)\left(2.4.6...100\right)}{2.4.6...100}=\frac{1.2.3...100}{\left(2.1\right)\left(2.2\right)...\left(2.50\right)}=\frac{\left(1.2.3...50\right)\left(51.52.53....100\right)}{\left(1.2.3...50\right)\left(2.2.2...2\right)}=\frac{51.52.53...100}{2.2...2}=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}=B\)