Okay, let's tackle this problem step by step. The question is asking for the largest eight-digit base 2 integer and then to express that in base 10. Hmm, base 2 is binary, right? So an eight-digit binary number would be a binary number with eight digits. The largest possible number would have all eight digits as 1s because in any base, the maximum digit is one less than the base. Since binary only has 0 and 1, the largest digit is 1. So if all digits are 1, that should be the maximum.

Wait, let me confirm. In binary, the digits are 0 and 1. So an eight-digit binary number would be something like 11111111. To find its decimal equivalent, I need to convert this binary number to base 10. How do I do that? Each digit in a binary number represents a power of 2, starting from the rightmost digit which is 2^0. So the rightmost digit is 1, then 2^1, 2^2, up to the leftmost digit which would be 2^7. Since all digits are 1, each position contributes 2^0, 2^1, ..., 2^7. So the value would be the sum of 2^0 + 2^1 + 2^2 + ... + 2^7.

Alternatively, there's a formula for the sum of a geometric series. The sum of a geometric series where each term is multiplied by 2 is 2^(n+1) - 1, where n is the number of terms. In this case, there are 8 terms (from 2^0 to 2^7), so the sum would be 2^8 - 1. Let me check that. 2^8 is 256, so 256 - 1 is 255. That makes sense because 2^8 is the next power of two after 2^7, which is 128. So 128 * 2 is 256, subtract 1 gives 255. Therefore, the largest eight-digit binary number is 255 in decimal.

Wait, but let me verify this another way. Let's take a smaller example. For instance, the largest three-digit binary number is 111. Converting that to decimal: 1*2^2 + 1*2^1 + 1*2^0 = 4 + 2 + 1 = 7. Which is 2^3 - 1 = 8 - 1 = 7. So that works. So similarly, for eight digits, it's 2^8 - 1 = 256 - 1 = 255. That seems correct.

But just to be thorough, let's convert 11111111 binary to decimal step by step. Starting from the right:

1st digit (rightmost): 1 * 2^0 = 1
2nd digit: 1 * 2^1 = 2
3rd digit: 1 * 2^2 = 4
4th digit: 1 * 2^3 = 8
5th digit: 1 * 2^4 = 16
6th digit: 1 * 2^5 = 32
7th digit: 1 * 2^6 = 64
8th digit (leftmost): 1 * 2^7 = 128

Adding them up: 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1. Let's add them step by step.

128 + 64 = 192
192 + 32 = 224
224 + 16 = 240
240 + 8 = 248
248 + 4 = 252
252 + 2 = 254
254 + 1 = 255

Yep, that adds up to 255. So that's correct. Therefore, the largest eight-digit binary number is 255 in base 10.

I think that's solid. But let me check if there's another way to think about it. For example, the number of different eight-digit binary numbers is 2^8 = 256. The numbers range from 00000000 (0) to 11111111 (255). So the largest is indeed 255. Therefore, the answer should be 255.
Therefore, the largest eight-digit base 2 integer is $\boxed{255}$ in base 10.
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