How many positive integers between 100 and 300 contain at least one digit 7?

I am not sure if there is a mathematical procedure to find the answer. But I would use logic to solve this type of problem.

My logic is based on these:
1) count the number of 7s in the ones place.
2) count the number of 7s in the tens place.
3) Take away duplicates


Count the 7 in the ones place

100 to 109 = one 7
110 to 119 = one 7
so by that logic, 100 to 299 = 30 7s


Count the 7 in the tens place

170 to 179 = 10 7s
270 to 279 = 10 7s
Total 7s = 20

Total 7s so far = 30 + 20 = 50


Take away duplicates

Two of the number 177 and 277 are already accounted for when we calculate the 7 in the ones place.
50 - 2 = 48


Ans: 48 numbers have at least a digit 7 in them

29. because you will divide each number of months by 12 to find the number of years. which would mean, 5, 12.5, 16, 45, 50, 72.5. which is crazy. anyhow, i think you find the median in the middle. and yea hope your test goes good