Show that for every positive integer n, the total number of digits in the sequence 1,2,3,...,10^n is equal to the number of zero digits in the sequence 1,2,3,...,10^(n+1).
Let's find the first number first.
There are 9 1-digit numbers.
There are 90 2-digit numbers, 10 - 99, for 2*9*10^1 digits.
At any given number of digits, k, there are k*9*10^(k-1) digits. This holds true through k = n.
There is one (n+1)-digit number.
The total is 9 + 2*90 + ... + n*9*10^(n-1) + n + 1.
Now the second number to see if it's equal to the first:
There are no zeros among the 1-digit numbers.
Among the 2-digit numbers, there are 9 zeros.
Among k-digit numbers, in any given position except the first, a zero will occur 9*10^(k-2), the factor of 9 referring to the 9 possibilities for the first digit and the 10^(k-2) referring to the 10 possibilities in each of the remaining positions (other than the first and the one whose zeros you're considering). So for each of the k-1 positions a zero could occur, it will occur 9*10^(k-2) times. That adds up to (k-1)*9*10^(k-2).
This time we are going up to all the (n+1)-digit numbers plus one (n+2)-digit number. But that (n+2)-digit number has n+1 zeros, so we get
0 + 9 + 2*9*10^1 + ... + (n+1-1)*9*10^(n+1-2) + n + 1.
This simplifies to the same number as the first.
Posted by Charlie
on 2005-02-25 14:25:24