Play the "bzzt" game Winner: Aditsu's CJam answer! A whopping 25 bytes! Nice!
You may continue to submit your answer, however you can no longer win. Original post kept for posterity:

The &quot;Bzzt&quot; game is a game where you must count to a number (in this case 500). However, if the number has a 3 in it, or is divisible by 3, you don't say the number. Instead, you say &quot;Bzzt&quot;.
Rules:

You can't hard-code the numbers.
The number only has to satisfy at least 1 of the following requirements

Divisible by 3
Number contains a 3


Some type of separator is mandatory (12bzzt14 doesn't count)
Score is measured in bytes.
You must count exactly to 500, starting at 1 or 0 (you chose).
The numbers must be output, but it doesn't matter how (e.g., stdout, writing to a text file, etc.).
0 can be divisible by 3 or not divisible. You can choose.
You can output the numbers one at a time (ex, output 1, then 2, then bzzt, then 4, etc) or all at once (e.g., output 1 2 bzzt 4 5).
You must replace the letter 3 with the word &quot;bzzt&quot;. This is not case sensitive (bZzt, Bzzt, bzzt are all okay).



This is a code-golf challenge, so the shortest code wins.
This contest ends June 30th 2014 (7 days from posting).
CJam - 25

501{3sI3%&lt;Is-I"bzzt"?N}fI


Thanks Howard :)

Try it at http://cjam.aditsu.net/

Explanation:

501{…}fI is basically for(int I=0; I&lt;501; ++I) {…}
3s converts 3 to string, i.e. "3"
I3% is I % 3
&lt; gets the left substring - "3".substring(0, I % 3) - which is "" for I % 3 == 0 and "3" otherwise
Is converts I to string
- with 2 strings does a set difference, resulting in an empty string iff I % 3 == 0 (the first string was empty) or I has a 3 digit in it
…I"bzzt"? is like … ? I : "bzzt"; the previous string is treated as a boolean value, where "" is false and any other string is true
N adds a newline