B2. Determine the number of factors 2 in $2001 !=1 \times 2 \times 3 \times \ldots \times 2000 \times 2001$.
B2. Each even number provides at least one factor of 2. There are 1000 even numbers in the product. This already gives at least 1000 factors of 2. However, all multiples of 4 provide at least one additional factor of 2, and there are 500 multiples of 4 (2001/4=500.25, which rounds down to 500). Each multiple of 8 provides another additional factor of 2, and there are 250 multiples of 8 (2001/8=125, ... again rounding down). Then, each multiple of 16 provides yet another additional factor of 2, and there are 125 multiples of 16. Similarly for multiples of 32, 64, ... and 1024. In total, we find: 1000 + 500 + 250 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 1994 factors of 2.