Test: The Powerful - GMAT MCQ

Method 1 to solve this GMAT Number Properties Question: Find the number of such integers existing for a lower power of 10 and extrapolate the results. 
Between 10 and 100, that is 10¹ and 10², we have 2 numbers, 11 and 20. 
Between 100 and 1000, that is 10² and 10³, we have 3 numbers, 101, 110 and 200. 
Therefore, between 10⁶ and 10⁷, one will have 7 integers whose sum will be equal to 2.

Alternative approach 
All numbers between 10⁶ and 10⁷ will be 7 digit numbers. There are two possibilities if the sum of the digits has to be '2'. 
Possibility 1: Two of the 7 digits are 1s and the remaining 5 are 0s. The left most digit has to be one of the 1s. That leaves us with 6 places where the second 1 can appear. 
So, a total of six 7-digit numbers comprising two 1s exist, sum of whose digits is '2'. 
Possibility 2: One digit is 2 and the remaining are 0s. The only possibility is 2000000. Total count is the sum of the counts from these two possibilities = 6 + 1 = 7

Take common (13!^8) from numerator
take common (13!^4) from denominator

now 13! has last digit 0
so 0-1 =-1=10+(-1)=9

Hence Option D is correct

While typing numbers from 1 to 1000, there are 9 single digit numbers: from 1 to 9. 
Each of these numbers requires one keystroke. 
That is 9 key strokes. 
There are 90 two-digit numbers: from 10 to 99. 
Each of these numbers requires 2 keystrokes. 
Therefore, 180 keystrokes to type the 2-digit numbers. 
There are 900 three-digit numbers: from 100 to 999. 
Each of these numbers requires 3 keystrokes. 
Therefore, 2700 keystrokes to type the 3-digit numbers. 
1000 is a four-digit number which requires 4 keystrokes. 
Totally, therefore, one requires 9 + 180 + 2700 + 4 = 2893 keystrokes.