For the numbers in the 100's, 10 numbers (150, 151, ..., 159) contain the tens digit 5, and 10 numbers (105, 115, ..., 195) contain the units digit 5. However, we double counted 155. Therefore, 10 + 10 - 1 = 19 numbers in the 100's contain the digit 5 at least once. We can use the same reasoning to argue that there are 19 numbers that contain the digit 5 at least once in the 200's, 300's, etc. (excluding the 500's). That is because all the numbers in the 500's contain the digit 5 at least once; so there are 100 numbers in the 500's that contain the digit 5 at least once.swerve wrote:How many three digit numbers contain the digit at least once?
A. 52
B. 128
C. 252
D. 648
E. 900
The OA is C
Source: Economist GMAT
Therefore, in total, we have 19 x 8 + 100 = 152 + 100 = 252 three-digit numbers that contain the digit 5 at least once.
Alternate Solution:
We will use the formula:
#(three digit numbers) = #(three digit numbers that contain the digit 5 at least once) + #(three digit numbers that do not contain the digit 5).
Notice the number of three digit numbers, or equivalently, the number of terms between 100 and 999 inclusive, is 999 - 100 + 1 = 900.
To find the number of three digit numbers that do not contain the digit 5: There are 8 choices for the hundreds digit (all digits except 0 and 5), 9 choices for the tens digit (all digits except 5) and 9 choices for the units digit (all digits except 5); for a total of 8 x 9 x 9 = 648 numbers.
Therefore, there are 900 - 648 = 252 three digit numbers that contain the digit 5 at least once.
Answer: C
