Interview Question for Quantitative Developer at Bank of America:

300 numbers contain a 3, but you counted numbers of the form x33, 3x3, and 33x *twice* so you must subtract them 300-30=270, but you subtracted 333 once too many, so add it back 300-30+1=271. Answer is 271.

271 does not seem right . I think it is 111. Aren't there just 11 numbers containing 3 between 0 an 100, like 3,13,23 30,33 etc. So between 0 and 1000 there are 111; after adding in for 300.

There are only 19 numbers between 1 and 100 containing '3': 3, 13, 23, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 43, 53, 63, 73, 83, 93.

The same logic holds for 101-200, 301-400, 401-500, 501-600, 601-700, 701-800, 801-900, and 901-1000. That is 19 * 8 = 152.

There are 20 numbers between 201-300 that contain a 3: 203, 213, 223, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 243, 253, 263, 273, 283, 293, 300. That brings the sum up to 152 + 20 = 172.

Finally, 99 numbers from 301-400 contain a 3. The final sum is 172+99=271.

Of course, there are much more intelligent ways to count. For example, instead of counting how many numbers contain a 3, count how many do NOT contain a 3. That means that there are 9 possibilities for the 1st number (0-9 except 3), 9 possibilities for the 2nd number, and 9 possibilities for the 3rd number. Thus, there are 9 * 9 * 9 = 729 that do not contain a 3 which means that there are 1000 - 729 = 271 numbers that do contain a 3.

Thanks Candidate, or more simply

For 300 to 399 = 100 numbers
For other x00 to x99 = 19 numbers each
= 19 x 9 = 171 numbers
Total of 271 numbers

I would go with elimination of one character from a base 10 numbering system gives you a base 9 numbering system.
9^3 = 729 permutations of a base 9 numbering system (a system with no number 3) with 3 digits, since 10^3 = 1000; 10^3 - 9^3 = 271 thus 271 numbers have a 3 in them.