Originally Posted by
Quacky I'm unsure of the correct notation here so please bear with me.
Provided that x is a positive integer:
$\displaystyle 10^x$ will produce a number which is 1 followed by n-1 0's. What I mean is that:
$\displaystyle 10^2 = 10$, which is 1 followed by $\displaystyle (2-1)$ 0's
$\displaystyle 10^5 = 10000$ which is 1 followed by $\displaystyle (5-1)$ 0's
$\displaystyle 10^30$ will be a 1 followed by 29 zero digits.
If x is greater than one, as it is in your case, and an integer, it will end in 10 and be preceeded by (x-2) nines.
Eg:
$\displaystyle 10^6 - 90 $
$\displaystyle = 1,000,000 - 90 = 999910$
$\displaystyle x = 6$
x-2 = 4
The result is four 9's followed by 10.
$\displaystyle 10^9-90 = 999999910$
x=9
x-2=7
There are 7 nines followed by 10.
So in your example:
10^30-90
x=30
You will have 28 nine's followed by a 10.
The sum of the digits will be:
$\displaystyle 28\times9+1+0$
which is, as you say, 253