Among the first 1000 natural numbers , how many contain the digit 3 ?

The answer is 271.

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Clearly the number 1000 has no 3 in it . So now we are left with 999 numbers from 1 to 999 . Among the single-digit numbers , only one has a three in it that is 3, so 8 are without 3. Among the double-digit numbers , 72 (8 ×9 There are 9 digits so among the 9 digits, either one's or ten's place should have a 3 ) are without 3. And among the triple-digit numbers 648 (8 ×9×9) are without 3. So total numbers which contain the digit 3 are: 999-(8+72+648)=

271