Okay, so I need to find an integer n such that 0 ≤ n < 18 and n is congruent to -11213141 modulo 18. Hmm, let me think about how to approach this. 

First, I remember that when dealing with congruences modulo 18, it's helpful to find the remainder when the given number is divided by 18. But since the number here is negative, I need to be careful. Maybe I can convert the negative number to a positive equivalent modulo 18. 

Wait, so n ≡ -11213141 mod 18. That means n is the remainder when -11213141 is divided by 18. But since negative remainders are a bit tricky, perhaps I can add a multiple of 18 to make it positive. Alternatively, I can find the equivalent positive value first. Let me recall that for any integer a, a mod 18 is the same as (a + k*18) mod 18 for any integer k. So, maybe I can add 18 multiple times until I get a positive number within the range 0 to 17.

But 11213141 is a huge number. Adding 18 each time would take forever. There must be a smarter way. Oh, right! The modulo operation can be applied to the absolute value first, then handle the sign. Since -11213141 mod 18 is the same as (18 - (11213141 mod 18)) mod 18. Wait, is that correct? Let me check.

If I have a negative number a, then a mod m is equal to (m - (|a| mod m)) mod m. So yes, that seems right. So first, I need to compute 11213141 mod 18, then subtract that from 18, and the result will be the positive equivalent modulo 18. But if the subtraction results in a negative number, I need to add 18 again to get it into the 0-17 range. Let me verify that with a small example. For instance, -5 mod 3. 5 mod 3 is 2, so 3 - 2 = 1, which is correct because -5 ≡ 1 mod 3. Another example: -7 mod 5. 7 mod 5 is 2, so 5 - 2 = 3. -7 ≡ 3 mod 5. That's correct. So the formula works. 

Therefore, my plan is:

1. Compute 11213141 mod 18. Let's call this remainder R.
2. Then, compute (18 - R) mod 18. This will give me the equivalent positive remainder for -11213141 mod 18.
3. Since 18 - R could be 0, which is already in the range, but if R is 0, then 18 - 0 is 18, which mod 18 is 0. So that's fine.

Alternatively, if R is 0, then -11213141 ≡ 0 mod 18. But let's check. For example, if a number is divisible by 18, then its negative is also divisible by 18, so the remainder would be 0. So that's consistent.

So, first step: compute 11213141 mod 18.

But how do I compute such a large number modulo 18? There's a trick where you can break down the number digit by digit, using the fact that 10 ≡ 10 mod 18, 10^2 ≡ 100 mod 18, etc. Wait, but maybe there's a simpler method. 

Alternatively, since 18 = 2 * 9, and 2 and 9 are coprime, by the Chinese Remainder Theorem, computing modulo 18 is equivalent to computing modulo 2 and modulo 9, then combining the results. But I'm not sure if that helps here. Alternatively, perhaps using the fact that 10 ≡ 10 mod 18, 100 ≡ (10*10) ≡ 100 mod 18. Let's compute 100 mod 18: 18*5=90, so 100-90=10. So 100 ≡ 10 mod 18. Similarly, 1000 ≡ 10*10 ≡ 100 ≡ 10 mod 18. Wait, that's a pattern. So each power of 10 is congruent to 10 mod 18. Therefore, each digit in the number contributes its value times 10^k ≡ 10 mod 18. Hmm, that might not be helpful. Wait, maybe let's see.

But perhaps a better approach is to note that 10 ≡ 10 mod 18, 10^2 ≡ 10*10 = 100 ≡ 10 mod 18, 10^3 ≡ 10*10^2 ≡ 10*10 = 100 ≡ 10 mod 18, and so on. Therefore, 10^k ≡ 10 mod 18 for any k ≥ 1. Therefore, each digit in the number (except the units digit) is multiplied by 10, which is equivalent to multiplying by 10 mod 18. Wait, but 10 mod 18 is 10, so multiplying by 10 each time. So maybe we can compute the number modulo 18 by processing each digit and updating the result accordingly.

Let me recall that for a number d_n d_{n-1} ... d_1 d_0, the value mod m can be computed as ((...((d_n * 10 + d_{n-1}) * 10 + d_{n-2}) * 10 + ... ) * 10 + d_0) mod m. So, since 10 ≡ 10 mod 18, perhaps we can use this recursive formula.

Let me test this with a small number. Let's take 1234 mod 18. 

Start with 1. Then 1*10 + 2 = 12. 12 mod 18 is 12. Then 12*10 + 3 = 123. 123 mod 18: 18*6=108, 123-108=15. Then 15*10 +4=154. 154 mod 18: 18*8=144, 154-144=10. So 1234 mod 18 is 10. Let's check with actual division: 18*68=1224, 1234-1224=10. Correct. So the method works.

Therefore, the algorithm is to process each digit from left to right, updating the remainder each time by multiplying by 10 and adding the next digit, then taking mod 18 each time. 

So, for the number 11213141, let's process each digit step by step:

Start with remainder = 0.

1. First digit: 1
   remainder = (0 * 10 + 1) mod 18 = 1 mod 18 = 1
2. Second digit: 1
   remainder = (1 * 10 + 1) mod 18 = 11 mod 18 = 11
3. Third digit: 2
   remainder = (11 * 10 + 2) mod 18 = 112 mod 18. Let's compute 18*6=108, 112-108=4. So remainder=4
4. Fourth digit: 1
   remainder = (4 * 10 + 1) mod 18 = 41 mod 18. 18*2=36, 41-36=5. remainder=5
5. Fifth digit: 3
   remainder = (5 * 10 + 3) mod 18 = 53 mod 18. 18*2=36, 53-36=17. remainder=17
6. Sixth digit: 1
   remainder = (17 * 10 + 1) mod 18 = 171 mod 18. Let's divide 171 by 18: 18*9=162, 171-162=9. remainder=9
7. Seventh digit: 4
   remainder = (9 * 10 + 4) mod 18 = 94 mod 18. 18*5=90, 94-90=4. remainder=4
8. Eighth digit: 1
   remainder = (4 * 10 + 1) mod 18 = 41 mod 18. 18*2=36, 41-36=5. remainder=5

So after processing all digits, the remainder is 5. Therefore, 11213141 mod 18 is 5.

Therefore, according to the earlier plan, n ≡ -11213141 mod 18 ≡ (18 - 5) mod 18 = 13 mod 18. So n should be 13. Let me verify this.

Alternatively, let's check if 11213141 + 13 is divisible by 18. 11213141 +13 = 11213154. Let's check if this is divisible by 18.

Divisibility by 18 requires the number to be divisible by both 2 and 9. Since 11213154 is even, it's divisible by 2. Now check divisibility by 9: sum the digits.

1+1+2+1+3+1+5+4 = 1+1=2, 2+2=4, 4+1=5, 5+3=8, 8+1=9, 9+5=14, 14+4=18. The sum is 18, which is divisible by 9. Therefore, 11213154 is divisible by 18, which means 11213141 ≡ -13 mod 18. Wait, but according to our earlier calculation, 11213141 ≡5 mod 18, so -11213141 ≡ -5 mod 18, which is 13 mod 18. But when we add 13 to 11213141, we get 11213154, which is divisible by 18. So that checks out. Therefore, n=13 is the correct answer.

But let me double-check the modulo 18 calculation again. Maybe I made a mistake in processing the digits. Let me go through the steps again.

Number: 1 1 2 1 3 1 4 1

Processing each digit:

1. Start with 0.
2. 1: 0*10 +1 =1 mod18=1
3. 1: 1*10 +1=11 mod18=11
4. 2: 11*10 +2=112 mod18. Let's compute 18*6=108, 112-108=4. So 4
5. 1: 4*10 +1=41 mod18. 18*2=36, 41-36=5. 5
6. 3: 5*10 +3=53 mod18. 18*2=36, 53-36=17. 17
7. 1: 17*10 +1=171 mod18. 18*9=162, 171-162=9. 9
8. 4: 9*10 +4=94 mod18. 18*5=90, 94-90=4. 4
9. 1: 4*10 +1=41 mod18. 18*2=36, 41-36=5. 5

Wait, so after processing all digits, the remainder is 5. Therefore, 11213141 ≡5 mod18. Therefore, -11213141 ≡-5 mod18. To make it positive, 18-5=13. So n=13. That seems correct. 

But let me try another approach to confirm. Let's compute 11213141 divided by 18 and find the remainder. But since the number is 11,213,141, dividing by 18 directly is tedious. Alternatively, note that 18=2*9, so if I can check divisibility by 9, but since we already did that, maybe it's not helpful. 

Alternatively, perhaps use the fact that 10 ≡10 mod18, and 10^k ≡10 mod18 for k≥1. Wait, but then each digit's contribution is digit *10^k ≡ digit*10 mod18. So the entire number mod18 is equivalent to the sum of (digit_i *10) for all digits except the last one, plus the last digit. Wait, let me see.

Wait, 11213141 can be written as:

1*10^7 + 1*10^6 + 2*10^5 +1*10^4 +3*10^3 +1*10^2 +4*10^1 +1*10^0

Each 10^k for k≥1 is ≡10 mod18. So:

1*10^7 ≡1*10 mod18=10
1*10^6 ≡1*10 mod18=10
2*10^5 ≡2*10 mod18=20≡2 mod18 (since 20-18=2)
1*10^4 ≡1*10 mod18=10
3*10^3 ≡3*10 mod18=30≡12 mod18 (30-18=12)
1*10^2 ≡1*10 mod18=10
4*10^1 ≡4*10 mod18=40≡4 mod18 (40-36=4)
1*10^0 ≡1*1 mod18=1

So adding all these up:

10 (from 10^7) +10 (10^6) +2 (10^5) +10 (10^4) +12 (10^3) +10 (10^2) +4 (10^1) +1 (10^0)

Sum: 10+10=20; 20+2=22; 22+10=32; 32+12=44; 44+10=54; 54+4=58; 58+1=59.

Now, 59 mod18: 18*3=54, 59-54=5. So remainder is 5. Which matches the previous result. Therefore, 11213141 ≡5 mod18. Therefore, -11213141 ≡-5 mod18. Since -5 mod18 is 13, n=13.

Therefore, the answer is 13.

**Final Answer**
The integer \( n \) that satisfies the given condition is \boxed{13}.
Therefore, the integer $n$ that satisfies $0\le n<18$ and $n\equiv -11213141\pmod{18}$ is $\boxed{13}$.
</think>