Let's calculate how many integers from 1-99 are there in which '7' appears at least once.
The integers in which '7' appears in units' place: 7, 17, 27, ..., 97: So there 10 numbers;
The integers in which '7' appears in tens' place: 70, 71, ... 79: So there 10 numbers
But, note that '77' is counted double.
So, from 1-99 are there are 10 + 10 – 1 = 19 integers in which '7' appears.
Same goes for 100-199; 200-299; 300-399, ..., and 900-999.
Total such integers = 19*10 = 190;
But this is an incomplete count. From 701-799, there are 100 – 19 = 81 integers that were not counted.
So, the required integers = 190 + 81 = 271.
Correct answer: E
Hope this helps!
-Jay
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