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
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Count the 7 in the ones place
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100 to 109 = one 7
110 to 119 = one 7
so by that logic, 100 to 299 = 30 7s
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Count the 7 in the tens place
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170 to 179 = 10 7s
270 to 279 = 10 7s
Total 7s = 20
Total 7s so far = 30 + 20 = 50
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Take away duplicates
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Two of the number 177 and 277 are already accounted for when we calculate the 7 in the ones place.
50 - 2 = 48
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Ans: 48 numbers have at least a digit 7 in them
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