10^n has n+1 digits: 1 and n zeros. For example: 10^2=100 --> 3 digits: 1 and 2 zeros;
10^n-49 will have n digits: n-2 9's and 51 in the and. For example: 10^4-49=10,000-49=9,951 --> 4 digits: 4-2=two 9's and 51 in the end;
We are told that the sum of all the digits of 10^n-49 is equal to the three-digit number x13 --> 9(n-2)+5+1=x13 --> 9n-12=x13 --> 9n=x25 --> x25 is divisible by 9 --> the sum of its digits must be divisible by 9 --> x=2 --> 9n=225 --> n=25.
Hope it's clear.
I'm just wondering about this approach: I tried to find a pattern.
n= 2 --> sum of digits = 6
n= 3 --> sum of digits = 15
n= 4 --> sum of digits = 24
n= 5 --> sum of digits = 33
(an increase of 9 for every n)
So for every odd n we get an odd units digit in the sum. And since x13 ends with an odd (3) number, I eliminated ans. choices A, C and E.
As for B and D - I was guessing/hoping that the units digit would stay on 3 for multiples of 5. So I picked ans. choice B).
But after thinking about it. I think I got lucky. So I'm wondering if there is something/similar that relates to this approach that yields a correct answer every time?