Okay, let's try to figure out how many of the same digits are found in the base 7 and base 8 representations of 629 in decimal. Hmm, so first, I need to convert 629 from base 10 to both base 7 and base 8, right? Then, once I have those two numbers in their respective bases, I can compare their digits and count how many are the same. Let me break this down step by step.

Starting with base 7. To convert 629 into base 7, I need to divide 629 by 7 repeatedly and keep track of the remainders. Each remainder will be a digit in the base 7 number, starting from the least significant digit. Let me do that.

First division: 629 ÷ 7. Let's see, 7 times 80 is 560. 629 minus 560 is 69. So the first remainder is 69. Wait, but that's not possible because remainders should be less than 7. Oh, right, I need to correct that. So 7 times 80 is 560, 629 minus 560 is 69. Then 69 ÷ 7 is 9 with a remainder of 6. So the first remainder is 6, then 9 ÷ 7 is 1 with a remainder of 2, and then 1 ÷ 7 is 0 with a remainder of 1. Wait, maybe I did that wrong. Let me try again.

Wait, maybe a better approach is to find the highest power of 7 less than or equal to 629. Let's calculate the powers of 7:

7^0 = 1

7^1 = 7

7^2 = 49

7^3 = 343

7^4 = 2401. That's too big. So the highest power is 7^3 = 343.

Now, how many times does 343 go into 629? 629 ÷ 343 is approximately 1.83, so 1 time. 1 times 343 is 343. Subtract that from 629: 629 - 343 = 286.

Next, the next lower power is 7^2 = 49. 286 ÷ 49 is approximately 5.83, so 5 times. 5 times 49 is 245. Subtract that: 286 - 245 = 41.

Next power is 7^1 = 7. 41 ÷ 7 is about 5.85, so 5 times. 5 times 7 is 35. Subtract: 41 - 35 = 6.

Finally, 7^0 = 1. 6 ÷ 1 is 6, so 6 times. So the coefficients are 1 (for 343), 5 (for 49), 5 (for 7), and 6 (for 1). Therefore, 629 in base 7 is 1 5 5 6. Wait, so 1556 base 7. Let me check that. 1*343 + 5*49 + 5*7 + 6*1 = 343 + 245 + 35 + 6 = 343+245=588, 588+35=623, 623+6=629. Yes, that's correct. So base 7 representation is 1556.

Now, moving on to base 8. Let's do the same process. Find the highest power of 8 less than or equal to 629.

8^0 = 1

8^1 = 8

8^2 = 64

8^3 = 512

8^4 = 4096. That's too big. So the highest power is 8^3 = 512.

629 ÷ 512 = 1.228, so 1 time. 1*512 = 512. Subtract: 629 - 512 = 117.

Next lower power is 8^2 = 64. 117 ÷ 64 = 1.828, so 1 time. 1*64 = 64. Subtract: 117 - 64 = 53.

Next, 8^1 = 8. 53 ÷ 8 = 6.625, so 6 times. 6*8 = 48. Subtract: 53 - 48 = 5.

Finally, 8^0 = 1. 5 ÷ 1 = 5. So the coefficients are 1 (for 512), 1 (for 64), 6 (for 8), and 5 (for 1). Therefore, 629 in base 8 is 1 1 6 5. Let me verify that. 1*512 + 1*64 + 6*8 +5*1 = 512 + 64 = 576, 576 + 48 = 624, 624 +5=629. Correct. So base 8 representation is 1165.

Now, we have the two representations:

Base 7: 1 5 5 6

Base 8: 1 1 6 5

Now, we need to compare each digit in these two numbers. Let's list the digits:

Base 7 digits: 1, 5, 5, 6

Base 8 digits: 1, 1, 6, 5

Wait, but the order of digits in base 7 is 1 (thousands place), 5 (hundreds), 5 (tens), 6 (units). Similarly, base 8 is 1 (thousands), 1 (hundreds), 6 (tens), 5 (units). So when comparing, we need to compare each corresponding digit. But wait, the number of digits is the same here (4 digits each), so we can compare digit by digit.

But wait, actually, the digits are in different orders. Let me clarify. The base 7 number is 1556, which is 1*7^3 + 5*7^2 +5*7 +6. The base 8 number is 1165, which is 1*8^3 +1*8^2 +6*8 +5. So when written as numbers, they have the same number of digits (4), so each digit position corresponds to the same power of the base. However, the actual digits in each position are different. So to compare, we need to look at each digit in each position.

Alternatively, maybe the problem is just asking for the count of digits that are the same in the two representations, regardless of their position. Wait, the example given is $121_3$ and $413_5$ have one digit in common. Let's check: 121 in base 3 is 1,2,1 and 413 in base 5 is 4,1,3. The digits common are 1 and 3? Wait, no. Wait, 121 in base 3 is 1,2,1 and 413 in base 5 is 4,1,3. Comparing digit by digit: first digit 1 vs 4 (no), second digit 2 vs 1 (no), third digit 1 vs 3 (no). Wait, but the example says they have one digit in common. Wait, maybe the example is wrong? Or perhaps I'm misunderstanding the problem.

Wait, the example says $121_3$ and $413_5$ have one digit in common. Let's compute their base 10 values. 121 in base 3 is 1*9 + 2*3 +1 = 9+6+1=16. 413 in base 5 is 4*25 +1*5 +3=100+5+3=108. But 16 and 108 don't have any digits in common. Wait, maybe the example is talking about digits in their respective bases. So in base 3, 121 has digits 1,2,1; in base 5, 413 has digits 4,1,3. So the common digits are 1 and 3? Wait, 1 is in both, and 3 is only in the second number. Wait, 1 is in both. So maybe the example counts 1 as the common digit. So total of one digit. That makes sense. So the problem is asking for digits that are present in both base 7 and base 8 representations, regardless of their position. So in the example, digits 1 and 3 are present in both numbers, but the problem's example says one digit. Wait, maybe there's a mistake in the example. Wait, the problem says "For example, $121_{3}$ and $413_{5}$ would have one digit in common." Let's check. $121_3$ has digits 1,2,1. $413_5$ has digits 4,1,3. So the common digits are 1. So that's one digit. So the example is correct. So the problem is asking for the number of digits that appear in both base 7 and base 8 representations of 629. So we need to find the intersection of the digits in base 7 and base 8 representations.

Wait, so base 7 representation is 1556, which has digits 1,5,5,6. Base 8 representation is 1165, which has digits 1,1,6,5. So the digits in base 7 are {1,5,6} and digits in base 8 are {1,5,6}. So the intersection is {1,5,6}. Therefore, three digits. Wait, so the answer would be 3? But let me check again.

Wait, base 7 digits: first digit is 1, second 5, third 5, fourth 6. So the digits are 1,5,5,6. So the unique digits are 1,5,6.

Base 8 digits: first digit 1, second 1, third 6, fourth 5. So digits are 1,1,6,5. Unique digits 1,6,5.

So the intersection is 1,5,6. So three digits. Therefore, the answer is 3. But let me confirm again.

Base 7: 1556. So digits 1,5,5,6.

Base 8: 1165. Digits 1,1,6,5.

So both have 1,5,6. So three digits. Therefore, the answer is 3.

But let me make sure I didn't make a mistake in converting 629 to base 7 and base 8.

Converting 629 to base 7:

Divide 629 by 7:

629 ÷ 7 = 89 with remainder 6.

89 ÷ 7 = 12 with remainder 5.

12 ÷ 7 = 1 with remainder 5.

1 ÷ 7 = 0 with remainder 1.

So writing the remainders from last to first, we get 1 5 5 6. Correct.

Base 8:

629 ÷ 8 = 78 with remainder 5.

78 ÷ 8 = 9 with remainder 6.

9 ÷ 8 = 1 with remainder 1.

1 ÷ 8 = 0 with remainder 1.

So remainders from last to first: 1 1 6 5. Correct.

So digits in base 7: 1,5,5,6.

Digits in base 8: 1,1,6,5.

Intersection of digits: 1,5,6. So three digits. Therefore, the answer is 3.

But wait, the example given in the problem had two numbers with one digit in common. So in that case, digits are considered regardless of their position. So in our problem, 1556 (base 7) and 1165 (base 8) share three digits: 1,5,6. Therefore, the answer is 3.

I think that's correct. Let me just verify once more.

Base 7 digits: 1,5,5,6. So digits used are 1,5,6.

Base 8 digits: 1,1,6,5. So digits used are 1,6,5.

Therefore, the common digits are 1,5,6. So three digits. So the answer is 3.
Therefore, the number of same digits in the base 7 and base 8 representations of $629_{10}$ is $\boxed{3}$.
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