203
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This can be solved by looking at each set of digit lengths. From 1 through 9, it's obvious there are 9 digits. From 10 through 99, there are 2 digits for each of the 90 numbers, so that makes 90*2=180 digits. Next, from 100 to 400, there are 401 numbers with 3 digits each, making another 401*3=1,203 digits. So the final answer is 9+180+1,203 = 1,392 digits.
How many digits in a fed referemce number?
How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?
How many strings of three digits are there? 000 to 999, or a total of 1000. How many strings of three digits contain the same three digits? That's 000, 111, 222 ... 999! ten in total. The difference is your answer: 1000-10 = 990.
TWO digits are in the number 30, 3 and 0