Okay, let's see. I need to find the remainder when the sum of all odd numbers from 1 to 199 is divided by 7. Hmm, that sounds like a problem involving arithmetic series and modular arithmetic. Let me think step by step.

First, the sum of consecutive odd numbers. I remember that the sum of the first n odd numbers is n². Let me verify that. The first odd number is 1, which is 1². The sum of the first two odd numbers is 1 + 3 = 4, which is 2². The first three: 1 + 3 + 5 = 9, which is 3². Yeah, that seems right. So if I can figure out how many terms are in the sequence from 1 to 199, then the sum should be n², where n is the number of terms.

So, how many terms are there? The sequence is 1, 3, 5, ..., 199. Since they are consecutive odd numbers, each term increases by 2. The nth term of an arithmetic sequence is given by a_n = a_1 + (n - 1)d, where a_1 is the first term, d is the common difference. Here, a_1 = 1, d = 2. So, 199 = 1 + (n - 1)*2. Let's solve for n:

199 = 1 + 2(n - 1)
199 - 1 = 2(n - 1)
198 = 2(n - 1)
n - 1 = 99
n = 100

So there are 100 terms. Therefore, the sum is 100² = 10,000. Then, the problem is to find 10,000 mod 7. But maybe there's a smarter way than calculating 10,000 divided by 7 directly. Let me check.

Alternatively, since 7 is a prime number, maybe I can use properties of modular arithmetic to simplify. Let's see. 10,000 divided by 7. Let me compute 7 * 1428 = 9996. Then 10,000 - 9996 = 4. So 10,000 mod 7 is 4. Therefore, the remainder is 4. But wait, let me confirm that division. 7*1428: 7*1400=9800, 7*28=196; 9800+196=9996. Then 10,000-9996=4. Yes, so 4 is the remainder. So the answer is 4.

But wait, let me double-check. Maybe I made a mistake in the number of terms. Let me recount the terms. The first term is 1, the last term is 199. The formula for the number of terms is ((last - first)/difference) + 1. So ((199 - 1)/2) + 1 = (198/2) + 1 = 99 + 1 = 100. So that's correct. So the sum is indeed 100² = 10,000. Then, 10,000 divided by 7. Let me do the division again. 7*1428=9996, remainder 4. So yes, remainder 4. Therefore, the answer is 4. But maybe there's another way to compute this to confirm.

Alternatively, compute 100 mod 7 first, then square it, then take mod 7 again. Let's see. 100 divided by 7: 7*14=98, remainder 2. So 100 ≡ 2 mod 7. Then 100² ≡ 2² = 4 mod 7. So that's another way to get the same result. That's a good check. So regardless of the approach, the remainder is 4. Therefore, the answer is 4. I think that's solid. But let me think again if there's any chance I missed something.

Wait, another approach: The sum of the first n odd numbers is n². So if n=100, sum=100²=10,000. Then 10,000 divided by 7. 7*1000=7000, 7*200=1400, 7000+1400=8400. 10,000-8400=1600. 7*200=1400, 1600-1400=200. 7*28=196, 200-196=4. So total remainder is 4. Yep, same result. So all methods point to 4. I think that's correct.

Alternatively, maybe using the formula for the sum of an arithmetic series. The sum S = n/2*(first term + last term). Here, n=100, first term=1, last term=199. So S=100/2*(1 + 199)=50*200=10,000. Same as before. Then mod 7. So again, 10,000 mod 7=4. So consistent.

Therefore, I'm confident the answer is 4. But let me check once more with another method. Let's compute the sum modulo 7 directly. Since the sum is 10,000, but perhaps we can compute each term modulo 7 and sum those up, then take modulo 7. But that might be tedious, as there are 100 terms. However, maybe there's a pattern.

Wait, since all the terms are odd numbers, starting from 1 to 199. Each term can be written as 2k - 1, where k goes from 1 to 100. So the sum is Σ(2k - 1) from k=1 to 100. Which is 2Σk - Σ1. Σk from 1 to 100 is (100*101)/2=5050. So 2*5050=10100. Then Σ1 from 1 to 100 is 100. So 10100 - 100=10,000. Same result. But maybe compute 10,000 mod 7 again. 7*1428=9996, 10,000-9996=4. So yes. Alternatively, 10,000 divided by 7. 7*1000=7000, 10,000-7000=3000. 7*428=3000- (7*400=2800) => 3000-2800=200. Then 200 divided by 7 is 28*7=196, remainder 4. So total remainder is 4. Yep.

Alternatively, breaking down 10,000 into parts. 10,000 = 7*1428 + 4. So remainder 4. All methods confirm. Therefore, I think the answer is 4. I don't see any mistakes in the reasoning. It seems solid. So I can conclude the remainder is 4.
Therefore, the modulo $7$ remainder of the sum $1+3+5+7+9+\dots+195+197+199$ is $\boxed{4}$.
</think>