1. ## Calculating 10^30-90

Sum of the digits in the sum: (10^30)-90

Am I correct in saying it's 253?

2. Originally Posted by Mukilab
Sum of the digits in the sum: (10^30)-90

Am I correct in saying it's 253?
$\displaystyle 10^{30} >> 90$ so the 90 can be ignored.

$\displaystyle 10^{30} - 90 \approx 10^{30}$

I don't get the question though

3. 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 x 0's. What I mean is that:

$\displaystyle 10^3 = 1000$, which is 1 followed by $\displaystyle 3$ 0's
$\displaystyle 10^5 = 100000$ which is 1 followed by $\displaystyle 5$ 0's

$\displaystyle 10^{30}$ will be a 1 followed by 30 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 when 90 is subtracted.

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.

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

4. 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.

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
thank you, perfect answer. I'm quite shamed I didn't know the method!

EDIT: of course I didn't do it manually but it took me 7 minutes which is 3 minutes above my limited time per question :/

5. Actually, I'd made a mistake at the beginning. I've rectified it now.