Okay, so I need to find the remainder when 2004 is divided by 12. Hmm, let me think. Remainders when dividing by 12... I remember that when a number is divided by 12, the remainder is the part that's left after subtracting the largest multiple of 12 that's less than the number. So, maybe I can start by dividing 2004 by 12 and see what the remainder is. 

Alternatively, I recall there's a rule for divisibility by 12. Since 12 is 3 times 4, a number is divisible by 12 if it's divisible by both 3 and 4. But here, I need the remainder, not just divisibility. So maybe that rule isn't directly helpful, but perhaps breaking down 2004 into parts that are easier to handle modulo 12. 

Let me try another approach. I know that 2004 divided by 12 is the same as (2000 + 4) divided by 12. So, maybe I can find the remainder of 2000 divided by 12 and the remainder of 4 divided by 12, then add them together. Wait, but when you add remainders, you have to be careful. For example, if the sum exceeds 12, you need to take the remainder again. 

First, let's find 2000 divided by 12. How many times does 12 go into 2000? Let's calculate 12 multiplied by 166 is 1992 because 12*160=1920 and 12*6=72, so 1920+72=1992. Then 2000 - 1992 = 8. So the remainder of 2000 divided by 12 is 8. Then 4 divided by 12 is 4 with a remainder of 4. So adding those remainders: 8 + 4 = 12. But 12 divided by 12 is 1 with a remainder of 0. So the total remainder would be 0. Wait, but that seems too straightforward. Let me check that again. 

Alternatively, maybe I should compute 2004 divided by 12 directly. Let's do that. 12 times 167 is 2004 because 12*160=1920, 12*7=84, so 1920+84=2004. So 12*167=2004 exactly, which means the remainder is 0. So that matches the previous result. 

But wait, maybe I should verify this another way. Let's break down 2004 into smaller parts. For example, 2004 is 2000 + 4. As before, 2000/12 gives a remainder of 8, and 4/12 gives a remainder of 4. Adding 8 + 4 gives 12, which is divisible by 12, so the total remainder is 0. That seems correct. 

Alternatively, maybe using modular arithmetic properties. Since 2004 = 2000 + 4, then 2004 mod 12 = (2000 mod 12 + 4 mod 12) mod 12. We already know 2000 mod 12 is 8, and 4 mod 12 is 4, so 8 + 4 = 12, and 12 mod 12 is 0. So that's another way to see it. 

Alternatively, maybe using the fact that 2004 is a multiple of 12. Let's check if 2004 is divisible by 12. Since 12 is 3*4, and 2004 is even (divisible by 2), so check divisibility by 3. The sum of the digits of 2004 is 2 + 0 + 0 + 4 = 6, which is divisible by 3. Also, 2004 divided by 4 is 501, which is an integer. Therefore, since it's divisible by both 3 and 4, it's divisible by 12. Hence, the remainder is 0. 

Wait, but maybe I should check this with another method just to be sure. Let's divide 2004 by 12 step by step. 

12 times 100 is 1200. 2004 - 1200 = 804. 

12 times 60 is 720. 804 - 720 = 84. 

12 times 7 is 84. So 100 + 60 + 7 = 167. 12*167 = 2004. So there's no remainder. 

Therefore, the remainder is 0. 

Alternatively, maybe using division algorithm. 2004 = 12*q + r, where 0 <= r <12. Since 12*167 = 2004, then q=167 and r=0. 

Yes, all methods lead to the same conclusion. So the remainder is 0. 

But just to be thorough, let me check with another approach. Let's consider that 2004 divided by 12 is equal to (2004 ÷ 12). Let's compute this division. 

12 into 2004. 

12*100=1200. 2004-1200=804. 

12*60=720. 804-720=84. 

12*7=84. 

So total is 100+60+7=167. So 12*167=2004. Therefore, remainder is 0. 

Another way: 2004 divided by 12. Let's compute 2004 ÷ 12. 

12 goes into 20 once (12), remainder 8. Bring down 0: 80. 12 goes into 80 six times (72), remainder 8. Bring down 4: 84. 12 goes into 84 seven times (84), remainder 0. So the quotient is 1 6 7, which is 167, and remainder 0. 

Yes, that's the standard long division method. 

Therefore, all methods confirm that the remainder is 0. 

I think that's sufficient. All approaches consistently give a remainder of 0. So I can be confident the answer is 0.
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