Okay, let's see. The problem is about redistributing coins from seven bags plus a new bag of 53 coins into eight bags with the same number of coins each. And we need to find the smallest number of coins originally had before finding the 53-coin bag, with the total being more than 200. Hmm. Let me break this down step by step.

First, let's denote the number of coins in each of the original seven bags as 'n'. So, the total number of coins before finding the 53-coin bag is 7n. Then, when we add the 53 coins, the total becomes 7n + 53. Now, we need to redistribute all these coins into eight bags, each with the same number of coins. Let's call the number of coins per bag after redistribution 'm'. Therefore, 8m = 7n + 53. Our goal is to find the smallest possible 7n such that 7n + 53 is divisible by 8, and 7n + 53 > 200. But since 7n + 53 > 200, that means 7n > 147, so n > 21. But n has to be an integer, so n ≥ 22. However, we also need 7n + 53 to be a multiple of 8. Let's formalize this.

We have the equation 7n + 53 ≡ 0 mod 8. Let's compute 53 mod 8 first. 8*6=48, so 53-48=5. So 53 ≡ 5 mod 8. Therefore, the equation becomes 7n + 5 ≡ 0 mod 8. Let's solve for n. 7n ≡ -5 mod 8, which is the same as 7n ≡ 3 mod 8. Now, 7 and 8 are coprime, so the inverse of 7 mod 8 exists. Let's find the inverse. 7*7=49 ≡ 1 mod 8. So inverse of 7 is 7. Therefore, multiplying both sides by 7: n ≡ 3*7 ≡ 21 ≡ 5 mod 8. So n ≡ 5 mod 8. Therefore, the smallest n satisfying n ≡5 mod8 and n ≥22. Let's find the smallest n ≥22 where n ≡5 mod8. 

Calculating numbers starting from 22: 22, 23, 24, 25. 25 divided by 8 is 3*8=24, remainder 1. So 25 ≡1 mod8. Next is 26: 26-24=2 → 2 mod8. 27: 3 mod8. 28:4. 29:5. So 29 is the smallest n≥22 with n≡5 mod8. Therefore, n=29. Then total coins before finding the 53-coin bag is 7*29=203. Then total after adding 53 is 203+53=256. 256 divided by 8 is 32, which is an integer. Also, 256>200, so that satisfies the condition. But wait, the problem says "the smallest number of coins you could have had before finding the bag of 53 coins". So 203? But let me check if there's a smaller n that satisfies n≡5 mod8 and 7n +53 >200. Let's see. n=29 gives 7*29=203. If we take n=29-8=21, but 21 is less than 22, which would make 7*21=147, 147+53=200, which is not more than 200. So 200 is not allowed. Therefore, n=29 is indeed the smallest possible. But wait, the problem says "the smallest number of coins you could have had before finding the bag of 53 coins". So 203 is the answer? Wait, but let me check again. The problem states "more than 200 coins" after redistribution. So 256>200, which is okay. But before redistribution, the total was 203. So 203 is the original total. So the answer should be 203. But let me confirm once more.

Alternatively, maybe there's a smaller n where 7n +53 is divisible by 8 and 7n +53 >200, but 7n +53 is just over 200. Let's compute 7n +53 >200 →7n>147 →n>21. So n≥22. We need 7n +53 ≡0 mod8. As we found, n≡5 mod8. The next n after 21 is 25, but 25 is 5 mod8. Wait, 25: 25-24=1 →1 mod8. Wait, no. Wait, 25 divided by8 is 3*8=24, remainder1. So 25≡1 mod8. Then 29 is 29-3*8=29-24=5 →5 mod8. So n=29 is the first n≥22 where n≡5 mod8. Therefore, 7*29=203. So yes, 203 is the smallest possible. Therefore, the answer is 203. But let me check if 203 is indeed the minimum. Suppose someone says n=5, which is 5 mod8, but 7*5=35. 35+53=88. 88 divided by8 is11. But 88>200? No. So 88 is not more than200. So 203 is the first one that's over 200. Hence, 203 is the answer. Therefore, the smallest number of coins before finding the 53-coin bag is 203. Wait, but let me check another way. Suppose we have 7n +53 =8m. So m=(7n+53)/8. We need m to be an integer, and 8m >200. So m>25. So m≥26. Therefore, 8m≥208. Therefore, 7n +53 ≥208 →7n≥155 →n≥155/7≈22.14. So n≥23. But n must be ≡5 mod8. Let's see, n=23: 23 mod8=7. Not 5. Next n=24: 24 mod8=0. Next 25:1. 26:2. 27:3. 28:4. 29:5. So n=29 is the first n≥23 with n≡5 mod8. Therefore, 29 is correct. Therefore, the answer is 7*29=203. So I think 203 is correct.

But let me verify once again. Suppose someone argues that maybe n=17, which is 17≡1 mod8, but 7*17=119. 119+53=172. 172/8=21.5, which is not integer. So no. n=13: 7*13=91. 91+53=144. 144/8=18. So 144 is the total, which is 18 per bag. But 144 is less than 200. So that's not allowed. So the next possible n is 29. Therefore, 203. So I think that's correct. Therefore, the answer is 203. So the final answer is \boxed{203}.
Therefore, the smallest number of coins you could have had before finding the bag of 53 coins is $\boxed{203}$.
</think>