How many 2002 digit numbers are there?

Answer 1

There are [9 x 10^2001] numbers which have 2002 digits. The [^] symbol means raise to the power. So 10^2 is 10² = 100 and 10^3 is 10³ = 1000, etc.

To see how this works, consider how many 2 digit numbers there are. The largest 2-digit number is 99 and there are 100 numbers [0 through 99] up to that. But we need to subtract off the 1-digit numbers [0 through 9]. There are ten of those, so we have 100 - 10 = 90.

Or write it this way: 10^2 - 10^1 = 9 x 10^1. So for a number of digits (n) we have 9 x 10^(n-1).

Let's check with n=3: 9 x 10^2 = 9 x 100 = 900. For 3-digit numbers, there is 100 through 999, so 0 through 999 is 1000 numbers [10^3], and subtract off 0 through 99 [100 or 10^2] which gives us 900 numbers, and satisfies the formula.