Explanation:
Okay, so Suzanne walks four miles every third day. The question is asking for the fewest number of miles she can walk in February. Hmm, let me think about how to approach this.

First, I need to figure out how many days are in February. Wait, February can have either 28 or 29 days depending on whether it's a leap year. But the problem doesn't specify if it's a leap year or not. Hmm, maybe I need to consider both possibilities? Or maybe there's a standard assumption here? Let me check. Usually, unless specified, February is considered to have 28 days. But since the question is about the fewest number of miles, maybe I should consider the shortest possible February, which is 28 days? Wait, but if it's a leap year, it's 29 days. But if we want the fewest number of miles, maybe we need to consider the minimum number of days? Wait, no. Wait, if February has 28 days, then she walks fewer times compared to 29 days? Wait, no. Wait, if she walks every third day, then the number of times she walks depends on how many days there are. Let me think.

So, if February has 28 days, how many times does she walk? Let me figure that out. Every third day means she walks on day 1, day 4, day 7, day 10, etc. So, starting from day 1, every third day. So, the number of times she walks is the number of days in February divided by 3, rounded up or down? Wait, let me think. For example, in 28 days, how many multiples of 3 are there?

Wait, if she walks every third day, starting on day 1, then the days she walks are 1, 4, 7, 10, 13, 16, 19, 22, 25, 28. Let me count that. Starting from 1, adding 3 each time. So, 1 + 3*(n-1) ≤ 28. Let me solve for n. 3*(n-1) ≤ 27. So n-1 ≤ 9, so n ≤ 10. So there are 10 days in February where she walks. Therefore, 10 times 4 miles is 40 miles. But wait, if February has 28 days, then the last day she walks is day 28. But if February had 29 days, then day 29 would be the next day after day 28. So, if February has 29 days, then the days she walks would be 1, 4, 7,..., 28, and then 31? Wait, 28 + 3 = 31, but February only has 29 days. So, the last day she walks would be day 25, then day 28, and then day 31 is beyond February. Wait, so in 29 days, how many times does she walk?

Let me calculate. Starting from day 1, adding 3 each time. Let's list them:

1, 4, 7, 10, 13, 16, 19, 22, 25, 28. Then the next one would be 31, which is beyond 29. So even in 29 days, she still walks 10 times. Wait, so whether it's 28 or 29 days, she walks 10 times? Wait, but if February has 29 days, then day 29 is the last day. So, does she walk on day 29? Let me check. If she walks every third day, starting on day 1, then the days are 1, 4, 7,..., 28, and then 31. But 31 is beyond 29, so she doesn't walk on day 29. Therefore, even in a leap year, she still walks 10 times. Therefore, regardless of whether February has 28 or 29 days, she walks 10 times. Therefore, the total miles would be 10 * 4 = 40 miles. But wait, the question says "the fewest number of miles she can walk in February." So, if February has 28 days, she walks 10 times. If it has 29 days, she still walks 10 times. Therefore, the fewest number of miles is 40? But wait, maybe there's a way to have fewer walks? Wait, maybe if she starts walking on a different day? Wait, the problem says "every third day." Does that mean starting from day 1, or can she choose the starting day? Wait, the problem says "Suzanne walks four miles every third day." So, does that mean that she walks on days that are multiples of three? Or does it mean that she walks every third day, starting from some day?

Wait, the problem is a bit ambiguous. Let me think. If it's "every third day," that could be interpreted as every third day starting from the first day. So, for example, if she starts on day 1, then day 1, day 4, day 7, etc. But if she starts on day 2, then day 2, day 5, day 8, etc. But the problem doesn't specify when she starts. So, maybe depending on when she starts, the number of walks can be different? Therefore, to find the fewest number of miles, we need to find the starting day that minimizes the number of walks in February.

Ah! That's probably the key here. The problem says "every third day," but doesn't specify the starting day. Therefore, depending on when she starts, she might walk fewer times. Therefore, to find the fewest number of miles, we need to find the minimal number of walks possible in February, given that she walks every third day, starting on some day.

Therefore, the problem is not about how many times she walks if she starts on day 1, but rather, what's the minimal number of walks possible in February, depending on her starting day.

Therefore, we need to figure out the minimal number of days she can walk in February, given that she walks every third day, starting on some day. Then multiply that by 4 to get the fewest number of miles.

So, how do we find the minimal number of walks?

Let me think. Let's consider February has 28 days. Let me think about how to schedule her walks so that she walks as few times as possible. Since she walks every third day, the number of walks depends on how the days align.

If she starts on day 1, then she walks on days 1, 4, 7,..., 28. As before, that's 10 days.

If she starts on day 2, then she walks on days 2, 5, 8,..., 29. But February only has 28 days, so the last day would be day 26 (since 2 + 3*(n-1) ≤ 28). Let me calculate:

Starting on day 2: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29. But 29 is beyond 28, so only 9 days. Wait, so starting on day 2, she walks 9 times? Wait, let me count:

Day 2, 5, 8, 11, 14, 17, 20, 23, 26. That's 9 days. Then the next would be 29, which is outside. So 9 walks. So that's fewer than 10. So that's better for minimizing.

Similarly, if she starts on day 3: days 3, 6, 9,..., 27, 30. So 3, 6, 9, 12, 15, 18, 21, 24, 27. Then next is 30, which is beyond. So that's 9 days as well. So 9 walks.

If she starts on day 4: days 4, 7, 10,..., 28, 31. So 4, 7, 10, 13, 16, 19, 22, 25, 28. Then next is 31. So that's 9 days. Wait, same as before.

Wait, so starting on day 1 gives 10 walks, starting on day 2, 3, 4 gives 9 walks. What about starting on day 5?

Starting on day 5: 5, 8, 11, 14, 17, 20, 23, 26, 29. But 29 is beyond 28, so 8 days? Wait, 5, 8, 11, 14, 17, 20, 23, 26. That's 8 days. Wait, so starting on day 5, she walks 8 times? Wait, that seems less. Wait, let me check again.

Starting on day 5:

First walk: day 5

Second: 5 + 3 = 8

Third: 11

Fourth: 14

Fifth: 17

Sixth: 20

Seventh: 23

Eighth: 26

Ninth: 29 (which is beyond 28). So only 8 walks? Wait, that's 8 walks? Wait, but 5 + 3*(n-1) ≤ 28. Let me solve for n:

3(n - 1) ≤ 23

n - 1 ≤ 23/3 ≈ 7.666...

So n - 1 = 7, so n = 8. So yes, 8 walks. So starting on day 5, she walks 8 times. That's even fewer. So that's better.

Similarly, starting on day 6: days 6, 9, 12, 15, 18, 21, 24, 27, 30. So 6, 9, 12, 15, 18, 21, 24, 27. Next is 30, which is beyond. So 8 walks. Same as starting on day 5.

Starting on day 7: days 7, 10, 13, 16, 19, 22, 25, 28, 31. So 7, 10, 13, 16, 19, 22, 25, 28. That's 8 walks. Next is 31, which is beyond. So 8 walks.

Starting on day 8: days 8, 11, 14, 17, 20, 23, 26, 29. So 8, 11, 14, 17, 20, 23, 26. Next is 29, which is beyond. So 7 walks? Wait, let me count:

Day 8, 11, 14, 17, 20, 23, 26. That's 7 days. Wait, so starting on day 8, she walks 7 times? Wait, that's even fewer. Wait, let me check the math again.

If starting on day 8, then the nth day is 8 + 3(n - 1). Let me solve 8 + 3(n - 1) ≤ 28.

So 3(n - 1) ≤ 20

n - 1 ≤ 20/3 ≈ 6.666...

So n - 1 = 6, so n = 7. Therefore, 7 walks. So that's 7 walks. So that's even fewer. So starting on day 8, she walks 7 times. So that's better.

Similarly, starting on day 9: days 9, 12, 15, 18, 21, 24, 27, 30. So 9, 12, 15, 18, 21, 24, 27. Next is 30. So 7 walks. Same as starting on day 8.

Starting on day 10: days 10, 13, 16, 19, 22, 25, 28, 31. So 10, 13, 16, 19, 22, 25, 28. That's 7 walks. Next is 31, which is beyond. So 7 walks.

Starting on day 11: days 11, 14, 17, 20, 23, 26, 29. So 11, 14, 17, 20, 23, 26. Next is 29, which is beyond. So 6 walks. Wait, that's even fewer. Let me check:

Starting on day 11: 11 + 3(n - 1) ≤ 28.

So 3(n - 1) ≤ 17

n - 1 ≤ 5.666...

So n - 1 = 5, so n = 6. Therefore, 6 walks. So starting on day 11, she walks 6 times. That's even fewer.

Similarly, starting on day 12: days 12, 15, 18, 21, 24, 27, 30. So 12, 15, 18, 21, 24, 27. Next is 30. So 6 walks. Same as starting on day 11.

Starting on day 13: days 13, 16, 19, 22, 25, 28, 31. So 13, 16, 19, 22, 25, 28. Next is 31. So 6 walks.

Starting on day 14: days 14, 17, 20, 23, 26, 29. So 14, 17, 20, 23, 26. Next is 29. So 5 walks. Let me check:

14 + 3(n - 1) ≤ 28

3(n - 1) ≤ 14

n - 1 ≤ 4.666...

So n - 1 = 4, n = 5. So 5 walks. So starting on day 14, she walks 5 times. That's even fewer.

Similarly, starting on day 15: days 15, 18, 21, 24, 27, 30. So 15, 18, 21, 24, 27. Next is 30. So 5 walks.

Starting on day 16: days 16, 19, 22, 25, 28, 31. So 16, 19, 22, 25, 28. Next is 31. So 5 walks.

Starting on day 17: days 17, 20, 23, 26, 29. So 17, 20, 23, 26. Next is 29. So 4 walks. Let me check:

17 + 3(n - 1) ≤ 28

3(n - 1) ≤ 11

n - 1 ≤ 3.666...

So n - 1 = 3, n = 4. So 4 walks. So starting on day 17, she walks 4 times.

Starting on day 18: days 18, 21, 24, 27, 30. So 18, 21, 24, 27. Next is 30. So 4 walks.

Starting on day 19: days 19, 22, 25, 28, 31. So 19, 22, 25, 28. Next is 31. So 4 walks.

Starting on day 20: days 20, 23, 26, 29. So 20, 23, 26. Next is 29. So 3 walks. Let me check:

20 + 3(n - 1) ≤ 28

3(n - 1) ≤ 8

n - 1 ≤ 2.666...

So n - 1 = 2, n = 3. So 3 walks. So starting on day 20, she walks 3 times.

Starting on day 21: days 21, 24, 27, 30. So 21, 24, 27. Next is 30. So 3 walks.

Starting on day 22: days 22, 25, 28, 31. So 22, 25, 28. Next is 31. So 3 walks.

Starting on day 23: days 23, 26, 29. So 23, 26. Next is 29. So 2 walks.

Starting on day 24: days 24, 27, 30. So 24, 27. Next is 30. So 2 walks.

Starting on day 25: days 25, 28, 31. So 25, 28. Next is 31. So 2 walks.

Starting on day 26: days 26, 29. So 26. Next is 29. So 1 walk.

Starting on day 27: days 27, 30. So 27. Next is 30. So 1 walk.

Starting on day 28: days 28, 31. So 28. Next is 31. So 1 walk.

Starting on day 29: but February only has 28 days, so starting on day 29 is not possible.

So, depending on the starting day, the number of walks varies from 1 to 10. Therefore, the minimal number of walks is 1? Wait, but if she starts on day 26, she walks on day 26 and 29. But day 29 is beyond February, so only 1 walk. Similarly, starting on day 27, she walks on day 27. Starting on day 28, she walks on day 28. So, if she starts on day 26, she walks once. But is that possible? Wait, the problem says "every third day." So if she starts on day 26, then the next walk would be day 26 + 3 = day 29, which is outside of February. Therefore, she only walks once. So, in that case, the fewest number of miles would be 4 miles. But that seems too low. Is there a constraint that she must walk every third day starting from the first day? Or can she choose any starting day?

Wait, the problem says "Suzanne walks four miles every third day." So, does that mean that she walks on days that are every third day, but not necessarily starting from day 1? So, if she can choose when to start, then she can minimize the number of walks. Therefore, the minimal number of walks is 1, so 4 miles. But that seems counterintuitive. Maybe there's a misunderstanding here.

Wait, maybe "every third day" is meant to be interpreted as every third day starting from the first day of February. So, if she starts on day 1, then walks every third day. But the problem doesn't specify. However, in common language, "every third day" usually means that she walks on days that are spaced three days apart, but the starting point could be any day. However, if the problem is from a math competition or something, maybe there's a standard interpretation.

Alternatively, maybe the problem is referring to walking every third day, meaning that she walks on days 3, 6, 9, etc. But that would be every third day starting from day 3. But that's not clear.

Wait, let me check the exact wording: "Suzanne walks four miles every third day." So, the phrase "every third day" is a bit ambiguous. In some contexts, it could mean that she walks on days that are multiples of three, i.e., days 3, 6, 9, etc. In other contexts, it could mean that she walks on days that are spaced three days apart, starting from some day. For example, if she walks on day 1, then day 4, day 7, etc. So, depending on the interpretation, the answer could vary.

But given that the problem is asking for the fewest number of miles, and if we take the interpretation that she can choose the starting day to minimize the number of walks, then the answer could be as low as 4 miles (if she walks only once). However, that seems too low. Maybe there's a different interpretation.

Alternatively, maybe "every third day" is meant to be that she walks on the third day of each week, but that's not necessarily related to February.

Alternatively, maybe the problem is referring to walking every third day of the month, meaning that she walks on the 3rd, 6th, 9th, etc., days of February. In that case, the number of walks would be floor(28 / 3) = 9 (since 3*9=27, 3*10=30 which is beyond 28). Therefore, 9 walks, 9*4=36 miles. But if February has 29 days, then floor(29/3)=9 (since 3*9=27, 3*10=30). So still 9 walks. But then the fewest number of miles would be 36. However, this is assuming that she walks on the 3rd, 6th, 9th, etc., days, which is a different interpretation.

But the problem says "every third day," not "every third day of the month." So, if it's every third day, starting from day 1, then it's days 1, 4, 7,... as before. But if she can choose the starting day, then she can minimize the number of walks.

But maybe the problem assumes that she starts walking on the first day of February. In that case, the number of walks is 10 for 28 days, as calculated earlier. But then the answer would be 40 miles. However, the problem says "the fewest number of miles she can walk in February," implying that depending on the starting day, the number of walks can vary, so we need to find the minimum.

But if we take the interpretation that "every third day" means that she walks on days that are multiples of three, i.e., days 3, 6, 9,..., then the number of walks is floor(28/3)=9, so 9*4=36. But if she can choose the starting day, then she can have fewer walks. For example, if she starts on day 1, walks on days 1,4,7,... which is different from multiples of three.

This is confusing. Maybe I need to check standard interpretations of "every third day."

In common usage, "every third day" usually means that the interval between walks is three days. So, if she walks on day 1, then the next walk is day 4 (three days later), then day 7, etc. So, the starting day is not specified, but the interval is three days. Therefore, depending on when she starts, the number of walks can vary. Therefore, to find the fewest number of miles, we need to find the starting day that results in the fewest number of walks in February.

Earlier, I found that if she starts on day 26, she walks once. But that seems odd. Let me verify with an example. Suppose February has 28 days. If she starts on day 26, then she walks on day 26. The next walk would be day 26 + 3 = day 29, which is beyond February. Therefore, only one walk. So, 4 miles. But is that acceptable? If she walks every third day, starting on day 26, then yes, she walks once. But maybe the problem assumes that she walks on at least one day? Or maybe the problem assumes that she walks every third day starting from the first day of February? The problem doesn't specify, so it's ambiguous.

However, given that the problem asks for the fewest number of miles, and if we take the interpretation that she can choose the starting day, then the answer could be as low as 4 miles. But that seems too low. Maybe there's a different way to interpret "every third day."

Alternatively, maybe "every third day" means that she walks on days that are the third day of each week, i.e., every Wednesday. But that would be a fixed day of the week, not dependent on the month. However, February can have 28 or 29 days, which is a multiple of 7 or not. For example, 28 days is exactly 4 weeks, so if she walks every Wednesday, she would walk 4 times in February (if 28 days) or 5 times (if 29 days). But the problem doesn't mention weeks or days of the week, so this is probably not the intended interpretation.

Alternatively, maybe the problem is from a math competition where "every third day" is meant to be every third day starting from day 1, and the answer is 40 miles. However, given that the problem says "the fewest number of miles," it's possible that the answer is lower if we consider different starting days.

But maybe there's a constraint that she must walk on at least one day? Or maybe the problem assumes that she walks every third day starting from the first day of February. In that case, the answer would be 10 walks, 40 miles. However, since the problem says "fewest," maybe we need to consider that she could have fewer walks if she starts later.

But if we take the starting day as day 26, then she walks once. But maybe the problem assumes that she walks every third day without skipping days? Wait, no. If she walks every third day, starting on day 26, then she walks on day 26, and then the next would be day 29, which is outside. So, only one walk. But maybe the problem assumes that she walks every third day, meaning that she walks on days 3, 6, 9,..., which would be different.

Alternatively, maybe the problem is in the style of the SAT or similar, where "every third day" is meant to be every third day of the month, i.e., days 3, 6, 9, etc. In that case, for February with 28 days, the number of walks is floor(28 / 3) = 9, so 9 * 4 = 36 miles. But if February has 29 days, it's still floor(29 / 3) = 9, so same answer. But the problem says "fewest number of miles," so maybe if February has 28 days, it's 9 walks, but if it has 29 days, it's still 9. So 36 miles. But then why mention February? Maybe the problem is from a specific context where February is considered to have 28 days, and the answer is 36. But I need to check.

Alternatively, maybe the problem is from a math competition where "every third day" is meant to be every third day starting from day 1, and the answer is 40. But then why ask for the fewest? Maybe there's a different way.

Wait, let me think again. If the problem is from a math competition, maybe the answer is 36. Let me check with both interpretations.

First interpretation: every third day starting from day 1. For February with 28 days, number of walks is ceil(28 / 3) = 10? Wait, no. Wait, if you start on day 1, then the number of walks is floor((28 - 1)/3) + 1 = floor(27/3) + 1 = 9 + 1 = 10. So 10 walks. So 10 * 4 = 40.

Second interpretation: every third day meaning days that are multiples of 3, i.e., days 3, 6, 9,... So number of walks is floor(28 / 3) = 9. So 9 * 4 = 36.

Third interpretation: choosing the starting day to minimize the number of walks, which could be as low as 1 walk, 4 miles.

But which interpretation is correct? Since the problem says "every third day," without specifying starting day, but asking for the fewest number of miles, it's likely that the intended interpretation is the second one, where she walks on days that are multiples of three, i.e., days 3, 6, 9,..., leading to 9 walks in February, hence 36 miles. However, if the problem is from a source that considers "every third day" as every third day starting from day 1, then it's 40 miles. But since the problem asks for the fewest number of miles, maybe the answer is 36. But I need to verify.

Alternatively, maybe the problem is from a math competition where "every third day" is meant to be every third day starting from day 1, and the answer is 40. However, if you can choose the starting day, then the answer could be lower. But maybe there's a standard way to interpret "every third day" in math problems.

Looking up similar problems, usually, "every third day" is interpreted as every third day starting from the first day. For example, if someone says "I walk every third day," it usually means they walk on day 1, day 4, day 7, etc. So, in that case, the number of walks would be 10 for 28 days, leading to 40 miles. However, since the problem asks for the fewest number of miles, maybe there's a different way.

Wait, but if you can choose the starting day, then you can have fewer walks. For example, if she starts on day 2, then walks on days 2, 5, 8,..., which in 28 days would be 9 walks. If she starts on day 3, same thing. If she starts on day 4, same. If she starts on day 5, 8 walks. If she starts on day 6, 8 walks. If she starts on day 7, 8 walks. If she starts on day 8, 7 walks. If she starts on day 9, 7 walks. If she starts on day 10, 7 walks. If she starts on day 11, 6 walks. If she starts on day 12, 6 walks. If she starts on day 13, 6 walks. If she starts on day 14, 5 walks. If she starts on day 15, 5 walks. If she starts on day 16, 5 walks. If she starts on day 17, 4 walks. If she starts on day 18, 4 walks. If she starts on day 19, 4 walks. If she starts on day 20, 3 walks. If she starts on day 21, 3 walks. If she starts on day 22, 3 walks. If she starts on day 23, 2 walks. If she starts on day 24, 2 walks. If she starts on day 25, 2 walks. If she starts on day 26, 1 walk. If she starts on day 27, 1 walk. If she starts on day 28, 1 walk.

So, the minimal number of walks is 1, leading to 4 miles. But is this acceptable? If the problem allows Suzanne to choose when to start walking, then yes. But maybe the problem assumes that she starts walking on the first day of February. In that case, the answer would be 10 walks, 40 miles. But since the problem says "fewest number of miles," it's possible that the answer is 4 miles. However, this seems too low, and maybe there's a misunderstanding.

Alternatively, maybe the problem is referring to walking every third day of the month, meaning that she walks on the 3rd, 6th, 9th, etc., days of February. In that case, for February with 28 days, the number of walks is floor(28 / 3) = 9, so 9 * 4 = 36 miles. This is a common interpretation in math problems where "every third day" refers to the day number being a multiple of three.

Given that ambiguity, but considering that the problem asks for the fewest number of miles, and if we take the interpretation that she walks on days that are multiples of three, then the answer is 36. However, if we take the interpretation that she can choose the starting day to minimize the number of walks, then it's 4. But since 4 seems too low, and math problems usually don't allow such arbitrary starting days unless specified, I think the intended interpretation is that she walks on days that are multiples of three, leading to 9 walks, 36 miles.

But let me check with another approach. Let's assume that "every third day" means that she walks on days that are spaced three days apart, starting from some day. To find the fewest number of walks, we need to find the starting day that results in the fewest number of walks in February. As calculated earlier, the minimum is 1 walk. But maybe there's a constraint that she must walk at least once? Or maybe the problem assumes that she walks every third day starting from the first day of February.

Alternatively, maybe the problem is from a specific source where "every third day" is defined as every third day starting from day 1, and the answer is 40. However, since the problem says "fewest," it's possible that the answer is 36.

Wait, let me think again. If the problem is from a math competition, like the AMC 8 or similar, they usually specify if the starting day is the first day. Since it's not specified, but asks for the fewest number of miles, it's possible that the answer is 36. But I need to verify with an example.

Suppose February has 28 days. If she walks every third day, starting on day 1: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28: 10 days.

If she walks every third day starting on day 3: 3, 6, 9, 12, 15, 18, 21, 24, 27: 9 days.

If she walks every third day starting on day 2: 2, 5, 8, 11, 14, 17, 20, 23, 26: 9 days.

So, depending on the starting day, the number of walks can be 9 or 10. Therefore, the fewest number of walks is 9, leading to 9 * 4 = 36 miles. Therefore, the answer is 36.

But earlier, I found that starting on day 26, she walks once. But that seems to contradict. Why is that?

Because if you start on day 26, and walk every third day, you walk on day 26, then day 29 (which is outside). So only one walk. But why is that different from starting on day 3, which gives 9 walks? Because depending on the starting day, the number of walks varies. So, if the problem allows choosing the starting day, then the answer is 4. But if the problem assumes that she starts on day 1, then it's 40. If it assumes that she starts on day 3 or day 2, it's 36.

But since the problem says "fewest number of miles," it's likely that we are to assume that she can choose the starting day to minimize the number of walks. However, in most math problems, unless specified, "every third day" usually means starting from the first day. But since the problem asks for the fewest, it's possible that the answer is 36. But I need to check if there's a standard way.

Wait, let's think about the definition of "every third day." In common language, if someone says they do something every third day, it usually means that they do it on days that are three days apart, starting from some day. However, the starting day is not specified, so it's ambiguous. But if you want the fewest number of times, you would choose the starting day that results in the fewest number of walks.

But in that case, the answer would be 1 walk, 4 miles. But that seems too low. Maybe there's a constraint that she must walk at least once? Or maybe the problem assumes that she walks every third day starting from the first day of February.

Alternatively, maybe the problem is referring to the number of days in February, and "every third day" means that she walks on the third day of the month, then the sixth, etc. In that case, it's 9 walks, 36 miles.

Given that ambiguity, but considering that in most math problems, "every third day" is interpreted as every third day starting from day 1, leading to 10 walks in February (28 days), which would be 40 miles. However, since the problem asks for the fewest number of miles, and if we consider that she can choose the starting day, then the answer could be lower.

But to resolve this, let's think about the possible answers. If the answer is 36, that would be if she walks on days 3, 6, 9,..., which is 9 walks. If the answer is 40, that's 10 walks. If the answer is 4, that's 1 walk.

But since the problem is about February, which has 28 or 29 days, and if we consider that "every third day" means every third day of the month, then it's 9 or 10 walks. But if it's every third day starting from day 1, then it's 10 walks.

However, since the problem says "fewest number of miles," and if we take the interpretation that she can choose the starting day, then the answer is 4. But that seems too low. Maybe the problem assumes that she walks every third day, meaning that she walks on days that are multiples of three, i.e., 3, 6, 9,..., which would be 9 walks.

Alternatively, maybe the problem is from a source where "every third day" is defined as every third day starting from day 1, and the answer is 40.

Since I need to choose, and given that in many math problems, "every third day" is interpreted as starting from day 1, but since the problem asks for the fewest, maybe the answer is 36.

Wait, let me check with the total number of days. If she walks every third day starting from day 1, then in 28 days, she walks on days 1, 4, 7, 10, 13, 16, 19, 22, 25, 28: 10 days. If she walks every third day starting from day 3, she walks on days 3, 6, 9, 12, 15, 18, 21, 24, 27: 9 days. So, if the problem allows choosing the starting day, then 9 is fewer than 10. So, the fewest number of miles would be 9 * 4 = 36.

Therefore, the answer is 36.

But why would the problem mention February? Because if it's about the fewest number of miles, it depends on the number of days in February. If February has 28 days, then starting on day 3 gives 9 walks. If it has 29 days, starting on day 3 gives 9 walks as well (days 3, 6, ..., 27, 30). But 30 is beyond 29, so still 9 walks. So, regardless of leap year, it's 9 walks. Hence, 36 miles.

Therefore, I think the intended answer is 36.

But earlier, I thought of starting on day 26, which gives 1 walk, but that seems to be a different interpretation. If the problem allows choosing the starting day, then 1 walk is possible. But maybe the problem assumes that she walks every third day, meaning that she walks on days that are spaced three days apart, but she must walk at least once. However, if she starts on day 26, she walks once, which is allowed.

But why would the problem ask for the fewest number of miles if it's so low? Maybe there's a misunderstanding. Let me think again.

If Suzanne walks every third day, it might mean that she walks on days that are three days apart, but she must walk on at least one day. However, if she starts on day 26, she walks once. But maybe the problem assumes that she walks every third day starting from the first day of February, so she walks on day 1, 4, 7,..., which is 10 times.

Since the problem is ambiguous, but given that in most cases, "every third day" is interpreted as starting from the first day, and the answer is 40, but since the problem asks for the fewest, and if we consider starting from day 3, it's 36.

Alternatively, maybe the problem is from a specific source where "every third day" means every third day of the month, i.e., days 3, 6, 9,..., leading to 9 walks.

Given that, and since 36 is a common answer for such problems, I think the intended answer is 36.

But to confirm, let's think about the total number of days. If she walks every third day, starting from day 1, the number of walks is ceil(28 / 3) = 10. If she walks every third day starting from day 3, it's floor(28 / 3) = 9. So, depending on the starting day, it can be 9 or 10. Hence, the fewest is 9, leading to 36 miles.

Therefore, I think the answer is 36.

**Final Answer**
The fewest number of miles Suzanne can walk in February is \boxed{36}.

To determine the **fewest number of miles** Suzanne can walk in February, we need to carefully interpret the phrase **"walks four miles every third day."**

---

### Step 1: Understanding the Phrase

The phrase **"every third day"** can be interpreted in a few ways:

- **Interpretation A:** She walks on days that are **multiples of three** (i.e., days 3, 6, 9, ...).
- **Interpretation B:** She walks on days that are **spaced three days apart**, starting from some day (e.g., day 1, 4, 7, ...).
- **Interpretation C:** She can choose the **starting day** to minimize the number of walks.

Since the problem asks for the **fewest number of miles**, we must consider **Interpretation C**, where Suzanne can choose the starting day to minimize the number of walks.

---

### Step 2: Determine the Number of Days in February

February has **28 days** in a non-leap year. If it's a leap year, it has **29 days**. However, since the problem does not specify, and we are to find the **fewest** number of miles, we consider the **shortest possible February** — **28 days**.

---

### Step 3: Minimize the Number of Walks

If Suzanne walks **every third day**, starting from a chosen day, the number of walks depends on how the days align. To **minimize** the number of walks, we want to **start as late as possible** in the month, so that the next walk would fall **after** the end of February.

Let’s test this:

- If she starts on **day 26**, the next walk would be on **day 29**, which is **after** February (28 days). So she walks **only once**.
- If she starts on **day 27**, the next walk is on **day 30**, also after February — only **one walk**.
- If she starts on **day 28**, the next walk is on **day 31** — again, only **one walk**.

So, **starting on day 26, 27, or 28** results in **only one walk** in February.

---

### Step 4: Calculate the Fewest Miles

If Suzanne walks **only once**, and each walk is **4 miles**, then the **fewest number of miles** she can walk in February is:

$$
1 \text{ walk} \times 4 \text{ miles} = \boxed{4} \text{ miles}
$$

---

### Step 5: Re-evaluating the Interpretation

However, this interpretation assumes that Suzanne can **choose the starting day** to minimize the number of walks. If the problem assumes that she walks **every third day starting from the first day of February**, then the number of walks is:

- Days: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28 → **10 walks**
- Total miles: $10 \times 4 = 40$

But the problem **explicitly asks for the fewest number of miles**, which implies that we should consider the **minimum possible** number of walks, not the default starting day.

---

### Final Conclusion

Given the ambiguity in the phrase "every third day," and the fact that the problem asks for the **fewest number of miles**, the most reasonable interpretation is that Suzanne can **choose the starting day** to minimize the number of walks. The **fewest number of walks** is **1**, which results in:

$$
\boxed{4} \text{ miles}
$$

---

### Final Answer

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so Suzanne walks four miles every third day. The question is asking for the fewest number of miles she can walk in February. Hmm, let me think about how to approach this.

First, I need to figure out how many days are in February. Wait, February can have either 28 or 29 days depending on whether it's a leap year. But the problem doesn't specify if it's a leap year or not. Hmm, maybe I need to consider both possibilities? Or maybe there's a standard assumption here? Let me check. Usually, unless specified, February is considered to have 28 days. But since the question is about the fewest number of miles, maybe I should consider the shortest possible February, which is 28 days? Wait, but if it's a leap year, it's 29 days. But if we want the fewest number of miles, maybe we need to consider the minimum number of days? Wait, no. Wait, if February has 28 days, then she walks fewer times compared to 29 days? Wait, no. Wait, if she walks every third day, then the number of times she walks depends on how many days there are. Let me think.

So, if February has 28 days, how many times does she walk? Let me figure that out. Every third day means she walks on day 1, day 4, day 7, day 10, etc. So, starting from day 1, every third day. So, the number of times she walks is the number of days in February divided by 3, rounded up or down? Wait, let me think. For example, in 28 days, how many multiples of 3 are there?

Wait, if she walks every third day, starting on day 1, then the days she walks are 1, 4, 7, 10, 13, 16, 19, 22, 25, 28. Let me count that. Starting from 1, adding 3 each time. So, 1 + 3*(n-1) ≤ 28. Let me solve for n. 3*(n-1) ≤ 27. So n-1 ≤ 9, so n ≤ 10. So there are 10 days in February where she walks. Therefore, 10 times 4 miles is 40 miles. But wait, if February has 28 days, then the last day she walks is day 28. But if February had 29 days, then day 29 would be the next day after day 28. So, if February has 29 days, then the days she walks would be 1, 4, 7,..., 28, and then 31? Wait, 28 + 3 = 31, but February only has 29 days. So, the last day she walks would be day 25, then day 28, and then day 31 is beyond February. Wait, so in 29 days, how many times does she walk?

Let me calculate. Starting from day 1, adding 3 each time. Let's list them:

1, 4, 7, 10, 13, 16, 19, 22, 25, 28. Then the next one would be 31, which is beyond 29. So even in 29 days, she still walks 10 times. Wait, so whether it's 28 or 29 days, she walks 10 times? Wait, but if February has 29 days, then day 29 is the last day. So, does she walk on day 29? Let me check. If she walks every third day, starting on day 1, then the days are 1, 4, 7,..., 28, and then 31. But 31 is beyond 29, so she doesn't walk on day 29. Therefore, even in a leap year, she still walks 10 times. Therefore, regardless of whether February has 28 or 29 days, she walks 10 times. Therefore, the total miles would be 10 * 4 = 40 miles. But wait, the question says "the fewest number of miles she can walk in February." So, if February has 28 days, she walks 10 times. If it has 29 days, she still walks 10 times. Therefore, the fewest number of miles is 40? But wait, maybe there's a way to have fewer walks? Wait, maybe if she starts walking on a different day? Wait, the problem says "every third day." Does that mean starting from day 1, or can she choose the starting day? Wait, the problem says "Suzanne walks four miles every third day." So, does that mean that she walks on days that are multiples of three? Or does it mean that she walks every third day, starting from some day?

Wait, the problem is a bit ambiguous. Let me think. If it's "every third day," that could be interpreted as every third day starting from the first day. So, for example, if she starts on day 1, then day 1, day 4, day 7, etc. But if she starts on day 2, then day 2, day 5, day 8, etc. But the problem doesn't specify when she starts. So, maybe depending on when she starts, the number of walks can be different? Therefore, to find the fewest number of miles, we need to find the starting day that minimizes the number of walks in February.

Ah! That's probably the key here. The problem says "every third day," but doesn't specify the starting day. Therefore, depending on when she starts, she might walk fewer times. Therefore, to find the fewest number of miles, we need to find the minimal number of walks possible in February, given that she walks every third day, starting on some day.

Therefore, the problem is not about how many times she walks if she starts on day 1, but rather, what's the minimal number of walks possible in February, depending on her starting day.

Therefore, we need to figure out the minimal number of days she can walk in February, given that she walks every third day, starting on some day. Then multiply that by 4 to get the fewest number of miles.

So, how do we find the minimal number of walks?

Let me think. Let's consider February has 28 days. Let me think about how to schedule her walks so that she walks as few times as possible. Since she walks every third day, the number of walks depends on how the days align.

If she starts on day 1, then she walks on days 1, 4, 7,..., 28. As before, that's 10 days.

If she starts on day 2, then she walks on days 2, 5, 8,..., 29. But February only has 28 days, so the last day would be day 26 (since 2 + 3*(n-1) ≤ 28). Let me calculate:

Starting on day 2: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29. But 29 is beyond 28, so only 9 days. Wait, so starting on day 2, she walks 9 times? Wait, let me count:

Day 2, 5, 8, 11, 14, 17, 20, 23, 26. That's 9 days. Then the next would be 29, which is outside. So 9 walks. So that's fewer than 10. So that's better for minimizing.

Similarly, if she starts on day 3: days 3, 6, 9,..., 27, 30. So 3, 6, 9, 12, 15, 18, 21, 24, 27. Then next is 30, which is beyond. So that's 9 days as well. So 9 walks.

If she starts on day 4: days 4, 7, 10,..., 28, 31. So 4, 7, 10, 13, 16, 19, 22, 25, 28. Then next is 31. So that's 9 days. Wait, same as before.

Wait, so starting on day 1 gives 10 walks, starting on day 2, 3, 4 gives 9 walks. What about starting on day 5?

Starting on day 5: 5, 8, 11, 14, 17, 20, 23, 26, 29. But 29 is beyond 28, so 8 days? Wait, 5, 8, 11, 14, 17, 20, 23, 26. That's 8 days. Wait, so starting on day 5, she walks 8 times? Wait, that seems less. Wait, let me check again.

Starting on day 5:

First walk: day 5

Second: 5 + 3 = 8

Third: 11

Fourth: 14

Fifth: 17

Sixth: 20

Seventh: 23

Eighth: 26

Ninth: 29 (which is beyond 28). So only 8 walks? Wait, that's 8 walks? Wait, but 5 + 3*(n-1) ≤ 28. Let me solve for n:

3(n - 1) ≤ 23

n - 1 ≤ 23/3 ≈ 7.666...

So n - 1 = 7, so n = 8. So yes, 8 walks. So starting on day 5, she walks 8 times. That's even fewer. So that's better.

Similarly, starting on day 6: days 6, 9, 12, 15, 18, 21, 24, 27, 30. So 6, 9, 12, 15, 18, 21, 24, 27. Next is 30, which is beyond. So 8 walks. Same as starting on day 5.

Starting on day 7: days 7, 10, 13, 16, 19, 22, 25, 28, 31. So 7, 10, 13, 16, 19, 22, 25, 28. That's 8 walks. Next is 31, which is beyond. So 8 walks.

Starting on day 8: days 8, 11, 14, 17, 20, 23, 26, 29. So 8, 11, 14, 17, 20, 23, 26. Next is 29, which is beyond. So 7 walks? Wait, let me count:

Day 8, 11, 14, 17, 20, 23, 26. That's 7 days. Wait, so starting on day 8, she walks 7 times? Wait, that's even fewer. Wait, let me check the math again.

If starting on day 8, then the nth day is 8 + 3(n - 1). Let me solve 8 + 3(n - 1) ≤ 28.

So 3(n - 1) ≤ 20

n - 1 ≤ 20/3 ≈ 6.666...

So n - 1 = 6, so n = 7. Therefore, 7 walks. So that's 7 walks. So that's even fewer. So starting on day 8, she walks 7 times. So that's better.

Similarly, starting on day 9: days 9, 12, 15, 18, 21, 24, 27, 30. So 9, 12, 15, 18, 21, 24, 27. Next is 30. So 7 walks. Same as starting on day 8.

Starting on day 10: days 10, 13, 16, 19, 22, 25, 28, 31. So 10, 13, 16, 19, 22, 25, 28. That's 7 walks. Next is 31, which is beyond. So 7 walks.

Starting on day 11: days 11, 14, 17, 20, 23, 26, 29. So 11, 14, 17, 20, 23, 26. Next is 29, which is beyond. So 6 walks. Wait, that's even fewer. Let me check:

Starting on day 11: 11 + 3(n - 1) ≤ 28.

So 3(n - 1) ≤ 17

n - 1 ≤ 5.666...

So n - 1 = 5, so n = 6. Therefore, 6 walks. So starting on day 11, she walks 6 times. That's even fewer.

Similarly, starting on day 12: days 12, 15, 18, 21, 24, 27, 30. So 12, 15, 18, 21, 24, 27. Next is 30. So 6 walks. Same as starting on day 11.

Starting on day 13: days 13, 16, 19, 22, 25, 28, 31. So 13, 16, 19, 22, 25, 28. Next is 31. So 6 walks.

Starting on day 14: days 14, 17, 20, 23, 26, 29. So 14, 17, 20, 23, 26. Next is 29. So 5 walks. Let me check:

14 + 3(n - 1) ≤ 28

3(n - 1) ≤ 14

n - 1 ≤ 4.666...

So n - 1 = 4, n = 5. So 5 walks. So starting on day 14, she walks 5 times. That's even fewer.

Similarly, starting on day 15: days 15, 18, 21, 24, 27, 30. So 15, 18, 21, 24, 27. Next is 30. So 5 walks.

Starting on day 16: days 16, 19, 22, 25, 28, 31. So 16, 19, 22, 25, 28. Next is 31. So 5 walks.

Starting on day 17: days 17, 20, 23, 26, 29. So 17, 20, 23, 26. Next is 29. So 4 walks. Let me check:

17 + 3(n - 1) ≤ 28

3(n - 1) ≤ 11

n - 1 ≤ 3.666...

So n - 1 = 3, n = 4. So 4 walks. So starting on day 17, she walks 4 times.

Starting on day 18: days 18, 21, 24, 27, 30. So 18, 21, 24, 27. Next is 30. So 4 walks.

Starting on day 19: days 19, 22, 25, 28, 31. So 19, 22, 25, 28. Next is 31. So 4 walks.

Starting on day 20: days 20, 23, 26, 29. So 20, 23, 26. Next is 29. So 3 walks. Let me check:

20 + 3(n - 1) ≤ 28

3(n - 1) ≤ 8

n - 1 ≤ 2.666...

So n - 1 = 2, n = 3. So 3 walks. So starting on day 20, she walks 3 times.

Starting on day 21: days 21, 24, 27, 30. So 21, 24, 27. Next is 30. So 3 walks.

Starting on day 22: days 22, 25, 28, 31. So 22, 25, 28. Next is 31. So 3 walks.

Starting on day 23: days 23, 26, 29. So 23, 26. Next is 29. So 2 walks.

Starting on day 24: days 24, 27, 30. So 24, 27. Next is 30. So 2 walks.

Starting on day 25: days 25, 28, 31. So 25, 28. Next is 31. So 2 walks.

Starting on day 26: days 26, 29. So 26. Next is 29. So 1 walk.

Starting on day 27: days 27, 30. So 27. Next is 30. So 1 walk.

Starting on day 28: days 28, 31. So 28. Next is 31. So 1 walk.

Starting on day 29: but February only has 28 days, so starting on day 29 is not possible.

So, depending on the starting day, the number of walks varies from 1 to 10. Therefore, the minimal number of walks is 1? Wait, but if she starts on day 26, she walks on day 26 and 29. But day 29 is beyond February, so only 1 walk. Similarly, starting on day 27, she walks on day 27. Starting on day 28, she walks on day 28. So, if she starts on day 26, she walks once. But is that possible? Wait, the problem says "every third day." So if she starts on day 26, then the next walk would be day 26 + 3 = day 29, which is outside of February. Therefore, she only walks once. So, in that case, the fewest number of miles would be 4 miles. But that seems too low. Is there a constraint that she must walk every third day starting from the first day? Or can she choose any starting day?

Wait, the problem says "Suzanne walks four miles every third day." So, does that mean that she walks on days that are every third day, but not necessarily starting from day 1? So, if she can choose when to start, then she can minimize the number of walks. Therefore, the minimal number of walks is 1, so 4 miles. But that seems counterintuitive. Maybe there's a misunderstanding here.

Wait, maybe "every third day" is meant to be interpreted as every third day starting from the first day of February. So, if she starts on day 1, then walks every third day. But the problem doesn't specify. However, in common language, "every third day" usually means that she walks on days that are spaced three days apart, but the starting point could be any day. However, if the problem is from a math competition or something, maybe there's a standard interpretation.

Alternatively, maybe the problem is referring to walking every third day, meaning that she walks on days 3, 6, 9, etc. But that would be every third day starting from day 3. But that's not clear.

Wait, let me check the exact wording: "Suzanne walks four miles every third day." So, the phrase "every third day" is a bit ambiguous. In some contexts, it could mean that she walks on days that are multiples of three, i.e., days 3, 6, 9, etc. In other contexts, it could mean that she walks on days that are spaced three days apart, starting from some day. For example, if she walks on day 1, then day 4, day 7, etc. So, depending on the interpretation, the answer could vary.

But given that the problem is asking for the fewest number of miles, and if we take the interpretation that she can choose the starting day to minimize the number of walks, then the answer could be as low as 4 miles (if she walks only once). However, that seems too low. Maybe there's a different interpretation.

Alternatively, maybe "every third day" is meant to be that she walks on the third day of each week, but that's not necessarily related to February.

Alternatively, maybe the problem is referring to walking every third day of the month, meaning that she walks on the 3rd, 6th, 9th, etc., days of February. In that case, the number of walks would be floor(28 / 3) = 9 (since 3*9=27, 3*10=30 which is beyond 28). Therefore, 9 walks, 9*4=36 miles. But if February has 29 days, then floor(29/3)=9 (since 3*9=27, 3*10=30). So still 9 walks. But then the fewest number of miles would be 36. However, this is assuming that she walks on the 3rd, 6th, 9th, etc., days, which is a different interpretation.

But the problem says "every third day," not "every third day of the month." So, if it's every third day, starting from day 1, then it's days 1, 4, 7,... as before. But if she can choose the starting day, then she can minimize the number of walks.

But maybe the problem assumes that she starts walking on the first day of February. In that case, the number of walks is 10 for 28 days, as calculated earlier. But then the answer would be 40 miles. However, the problem says "the fewest number of miles she can walk in February," implying that depending on the starting day, the number of walks can vary, so we need to find the minimum.

But if we take the interpretation that "every third day" means that she walks on days that are multiples of three, i.e., days 3, 6, 9,..., then the number of walks is floor(28/3)=9, so 9*4=36. But if she can choose the starting day, then she can have fewer walks. For example, if she starts on day 1, walks on days 1,4,7,... which is different from multiples of three.

This is confusing. Maybe I need to check standard interpretations of "every third day."

In common usage, "every third day" usually means that the interval between walks is three days. So, if she walks on day 1, then the next walk is day 4 (three days later), then day 7, etc. So, the starting day is not specified, but the interval is three days. Therefore, depending on when she starts, the number of walks can vary. Therefore, to find the fewest number of miles, we need to find the starting day that results in the fewest number of walks in February.

Earlier, I found that if she starts on day 26, she walks once. But that seems odd. Let me verify with an example. Suppose February has 28 days. If she starts on day 26, then she walks on day 26. The next walk would be day 26 + 3 = day 29, which is beyond February. Therefore, only one walk. So, 4 miles. But is that acceptable? If she walks every third day, starting on day 26, then yes, she walks once. But maybe the problem assumes that she walks on at least one day? Or maybe the problem assumes that she walks every third day starting from the first day of February? The problem doesn't specify, so it's ambiguous.

However, given that the problem asks for the fewest number of miles, and if we take the interpretation that she can choose the starting day, then the answer could be as low as 4 miles. But that seems too low. Maybe there's a different way to interpret "every third day."

Alternatively, maybe "every third day" means that she walks on days that are the third day of each week, i.e., every Wednesday. But that would be a fixed day of the week, not dependent on the month. However, February can have 28 or 29 days, which is a multiple of 7 or not. For example, 28 days is exactly 4 weeks, so if she walks every Wednesday, she would walk 4 times in February (if 28 days) or 5 times (if 29 days). But the problem doesn't mention weeks or days of the week, so this is probably not the intended interpretation.

Alternatively, maybe the problem is from a math competition where "every third day" is meant to be every third day starting from day 1, and the answer is 40 miles. However, given that the problem says "the fewest number of miles," it's possible that the answer is lower if we consider different starting days.

But maybe there's a constraint that she must walk on at least one day? Or maybe the problem assumes that she walks every third day starting from the first day of February. In that case, the answer would be 10 walks, 40 miles. However, since the problem says "fewest," maybe we need to consider that she could have fewer walks if she starts later.

But if we take the starting day as day 26, then she walks once. But maybe the problem assumes that she walks every third day without skipping days? Wait, no. If she walks every third day, starting on day 26, then she walks on day 26, and then the next would be day 29, which is outside. So, only one walk. But maybe the problem assumes that she walks every third day, meaning that she walks on days 3, 6, 9,..., which would be different.

Alternatively, maybe the problem is in the style of the SAT or similar, where "every third day" is meant to be every third day of the month, i.e., days 3, 6, 9, etc. In that case, for February with 28 days, the number of walks is floor(28 / 3) = 9, so 9 * 4 = 36 miles. But if February has 29 days, it's still floor(29 / 3) = 9, so same answer. But the problem says "fewest number of miles," so maybe if February has 28 days, it's 9 walks, but if it has 29 days, it's still 9. So 36 miles. But then why mention February? Maybe the problem is from a specific context where February is considered to have 28 days, and the answer is 36. But I need to check.

Alternatively, maybe the problem is from a math competition where "every third day" is meant to be every third day starting from day 1, and the answer is 40. But then why ask for the fewest? Maybe there's a different way.

Wait, let me think again. If the problem is from a math competition, maybe the answer is 36. Let me check with both interpretations.

First interpretation: every third day starting from day 1. For February with 28 days, number of walks is ceil(28 / 3) = 10? Wait, no. Wait, if you start on day 1, then the number of walks is floor((28 - 1)/3) + 1 = floor(27/3) + 1 = 9 + 1 = 10. So 10 walks. So 10 * 4 = 40.

Second interpretation: every third day meaning days that are multiples of 3, i.e., days 3, 6, 9,... So number of walks is floor(28 / 3) = 9. So 9 * 4 = 36.

Third interpretation: choosing the starting day to minimize the number of walks, which could be as low as 1 walk, 4 miles.

But which interpretation is correct? Since the problem says "every third day," without specifying starting day, but asking for the fewest number of miles, it's likely that the intended interpretation is the second one, where she walks on days that are multiples of three, i.e., days 3, 6, 9,..., leading to 9 walks in February, hence 36 miles. However, if the problem is from a source that considers "every third day" as every third day starting from day 1, then it's 40 miles. But since the problem asks for the fewest number of miles, maybe the answer is 36. But I need to verify.

Alternatively, maybe the problem is from a math competition where "every third day" is meant to be every third day starting from day 1, and the answer is 40. However, if you can choose the starting day, then the answer could be lower. But maybe there's a standard way to interpret "every third day" in math problems.

Looking up similar problems, usually, "every third day" is interpreted as every third day starting from the first day. For example, if someone says "I walk every third day," it usually means they walk on day 1, day 4, day 7, etc. So, in that case, the number of walks would be 10 for 28 days, leading to 40 miles. However, since the problem asks for the fewest number of miles, maybe there's a different way.

Wait, but if you can choose the starting day, then you can have fewer walks. For example, if she starts on day 2, then walks on days 2, 5, 8,..., which in 28 days would be 9 walks. If she starts on day 3, same thing. If she starts on day 4, same. If she starts on day 5, 8 walks. If she starts on day 6, 8 walks. If she starts on day 7, 8 walks. If she starts on day 8, 7 walks. If she starts on day 9, 7 walks. If she starts on day 10, 7 walks. If she starts on day 11, 6 walks. If she starts on day 12, 6 walks. If she starts on day 13, 6 walks. If she starts on day 14, 5 walks. If she starts on day 15, 5 walks. If she starts on day 16, 5 walks. If she starts on day 17, 4 walks. If she starts on day 18, 4 walks. If she starts on day 19, 4 walks. If she starts on day 20, 3 walks. If she starts on day 21, 3 walks. If she starts on day 22, 3 walks. If she starts on day 23, 2 walks. If she starts on day 24, 2 walks. If she starts on day 25, 2 walks. If she starts on day 26, 1 walk. If she starts on day 27, 1 walk. If she starts on day 28, 1 walk.

So, the minimal number of walks is 1, leading to 4 miles. But is this acceptable? If the problem allows Suzanne to choose when to start walking, then yes. But maybe the problem assumes that she starts walking on the first day of February. In that case, the answer would be 10 walks, 40 miles. But since the problem says "fewest number of miles," it's possible that the answer is 4 miles. However, this seems too low, and maybe there's a misunderstanding.

Alternatively, maybe the problem is referring to walking every third day of the month, meaning that she walks on the 3rd, 6th, 9th, etc., days of February. In that case, for February with 28 days, the number of walks is floor(28 / 3) = 9, so 9 * 4 = 36 miles. This is a common interpretation in math problems where "every third day" refers to the day number being a multiple of three.

Given that ambiguity, but considering that the problem asks for the fewest number of miles, and if we take the interpretation that she walks on days that are multiples of three, then the answer is 36. However, if we take the interpretation that she can choose the starting day to minimize the number of walks, then it's 4. But since 4 seems too low, and math problems usually don't allow such arbitrary starting days unless specified, I think the intended interpretation is that she walks on days that are multiples of three, leading to 9 walks, 36 miles.

But let me check with another approach. Let's assume that "every third day" means that she walks on days that are spaced three days apart, starting from some day. To find the fewest number of walks, we need to find the starting day that results in the fewest number of walks in February. As calculated earlier, the minimum is 1 walk. But maybe there's a constraint that she must walk at least once? Or maybe the problem assumes that she walks every third day starting from the first day of February.

Alternatively, maybe the problem is from a specific source where "every third day" is defined as every third day starting from day 1, and the answer is 40. However, since the problem says "fewest," it's possible that the answer is 36.

Wait, let me think again. If the problem is from a math competition, like the AMC 8 or similar, they usually specify if the starting day is the first day. Since it's not specified, but asks for the fewest number of miles, it's possible that the answer is 36. But I need to verify with an example.

Suppose February has 28 days. If she walks every third day, starting on day 1: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28: 10 days.

If she walks every third day starting on day 3: 3, 6, 9, 12, 15, 18, 21, 24, 27: 9 days.

If she walks every third day starting on day 2: 2, 5, 8, 11, 14, 17, 20, 23, 26: 9 days.

So, depending on the starting day, the number of walks can be 9 or 10. Therefore, the fewest number of walks is 9, leading to 9 * 4 = 36 miles. Therefore, the answer is 36.

But earlier, I found that starting on day 26, she walks once. But that seems to contradict. Why is that?

Because if you start on day 26, and walk every third day, you walk on day 26, then day 29 (which is outside). So only one walk. But why is that different from starting on day 3, which gives 9 walks? Because depending on the starting day, the number of walks varies. So, if the problem allows choosing the starting day, then the answer is 4. But if the problem assumes that she starts on day 1, then it's 40. If it assumes that she starts on day 3 or day 2, it's 36.

But since the problem says "fewest number of miles," it's likely that we are to assume that she can choose the starting day to minimize the number of walks. However, in most math problems, unless specified, "every third day" usually means starting from the first day. But since the problem asks for the fewest, it's possible that the answer is 36. But I need to check if there's a standard way.

Wait, let's think about the definition of "every third day." In common language, if someone says they do something every third day, it usually means that they do it on days that are three days apart, starting from some day. However, the starting day is not specified, so it's ambiguous. But if you want the fewest number of times, you would choose the starting day that results in the fewest number of walks.

But in that case, the answer would be 1 walk, 4 miles. But that seems too low. Maybe there's a constraint that she must walk at least once? Or maybe the problem assumes that she walks every third day starting from the first day of February.

Alternatively, maybe the problem is referring to the number of days in February, and "every third day" means that she walks on the third day of the month, then the sixth, etc. In that case, it's 9 walks, 36 miles.

Given that ambiguity, but considering that in most math problems, "every third day" is interpreted as every third day starting from day 1, leading to 10 walks in February (28 days), which would be 40 miles. However, since the problem asks for the fewest number of miles, and if we consider that she can choose the starting day, then the answer could be lower.

But to resolve this, let's think about the possible answers. If the answer is 36, that would be if she walks on days 3, 6, 9,..., which is 9 walks. If the answer is 40, that's 10 walks. If the answer is 4, that's 1 walk.

But since the problem is about February, which has 28 or 29 days, and if we consider that "every third day" means every third day of the month, then it's 9 or 10 walks. But if it's every third day starting from day 1, then it's 10 walks.

However, since the problem says "fewest number of miles," and if we take the interpretation that she can choose the starting day, then the answer is 4. But that seems too low. Maybe the problem assumes that she walks every third day, meaning that she walks on days that are multiples of three, i.e., 3, 6, 9,..., which would be 9 walks.

Alternatively, maybe the problem is from a source where "every third day" is defined as every third day starting from day 1, and the answer is 40.

Since I need to choose, and given that in many math problems, "every third day" is interpreted as starting from day 1, but since the problem asks for the fewest, maybe the answer is 36.

Wait, let me check with the total number of days. If she walks every third day starting from day 1, then in 28 days, she walks on days 1, 4, 7, 10, 13, 16, 19, 22, 25, 28: 10 days. If she walks every third day starting from day 3, she walks on days 3, 6, 9, 12, 15, 18, 21, 24, 27: 9 days. So, if the problem allows choosing the starting day, then 9 is fewer than 10. So, the fewest number of miles would be 9 * 4 = 36.

Therefore, the answer is 36.

But why would the problem mention February? Because if it's about the fewest number of miles, it depends on the number of days in February. If February has 28 days, then starting on day 3 gives 9 walks. If it has 29 days, starting on day 3 gives 9 walks as well (days 3, 6, ..., 27, 30). But 30 is beyond 29, so still 9 walks. So, regardless of leap year, it's 9 walks. Hence, 36 miles.

Therefore, I think the intended answer is 36.

But earlier, I thought of starting on day 26, which gives 1 walk, but that seems to be a different interpretation. If the problem allows choosing the starting day, then 1 walk is possible. But maybe the problem assumes that she walks every third day, meaning that she walks on days that are spaced three days apart, but she must walk at least once. However, if she starts on day 26, she walks once, which is allowed.

But why would the problem ask for the fewest number of miles if it's so low? Maybe there's a misunderstanding. Let me think again.

If Suzanne walks every third day, it might mean that she walks on days that are three days apart, but she must walk on at least one day. However, if she starts on day 26, she walks once. But maybe the problem assumes that she walks every third day starting from the first day of February, so she walks on day 1, 4, 7,..., which is 10 times.

Since the problem is ambiguous, but given that in most cases, "every third day" is interpreted as starting from the first day, and the answer is 40, but since the problem asks for the fewest, and if we consider starting from day 3, it's 36.

Alternatively, maybe the problem is from a specific source where "every third day" means every third day of the month, i.e., days 3, 6, 9,..., leading to 9 walks.

Given that, and since 36 is a common answer for such problems, I think the intended answer is 36.

But to confirm, let's think about the total number of days. If she walks every third day, starting from day 1, the number of walks is ceil(28 / 3) = 10. If she walks every third day starting from day 3, it's floor(28 / 3) = 9. So, depending on the starting day, it can be 9 or 10. Hence, the fewest is 9, leading to 36 miles.

Therefore, I think the answer is 36.

**Final Answer**
The fewest number of miles Suzanne can walk in February is \boxed{36}.

To determine the **fewest number of miles** Suzanne can walk in February, we need to carefully interpret the phrase **"walks four miles every third day."**

---

### Step 1: Understanding the Phrase

The phrase **"every third day"** can be interpreted in a few ways:

- **Interpretation A:** She walks on days that are **multiples of three** (i.e., days 3, 6, 9, ...).
- **Interpretation B:** She walks on days that are **spaced three days apart**, starting from some day (e.g., day 1, 4, 7, ...).
- **Interpretation C:** She can choose the **starting day** to minimize the number of walks.

Since the problem asks for the **fewest number of miles**, we must consider **Interpretation C**, where Suzanne can choose the starting day to minimize the number of walks.

---

### Step 2: Determine the Number of Days in February

February has **28 days** in a non-leap year. If it's a leap year, it has **29 days**. However, since the problem does not specify, and we are to find the **fewest** number of miles, we consider the **shortest possible February** — **28 days**.

---

### Step 3: Minimize the Number of Walks

If Suzanne walks **every third day**, starting from a chosen day, the number of walks depends on how the days align. To **minimize** the number of walks, we want to **start as late as possible** in the month, so that the next walk would fall **after** the end of February.

Let’s test this:

- If she starts on **day 26**, the next walk would be on **day 29**, which is **after** February (28 days). So she walks **only once**.
- If she starts on **day 27**, the next walk is on **day 30**, also after February — only **one walk**.
- If she starts on **day 28**, the next walk is on **day 31** — again, only **one walk**.

So, **starting on day 26, 27, or 28** results in **only one walk** in February.

---

### Step 4: Calculate the Fewest Miles

If Suzanne walks **only once**, and each walk is **4 miles**, then the **fewest number of miles** she can walk in February is:

$$
1 \text{ walk} \times 4 \text{ miles} = \boxed{4} \text{ miles}
$$

---

### Step 5: Re-evaluating the Interpretation

However, this interpretation assumes that Suzanne can **choose the starting day** to minimize the number of walks. If the problem assumes that she walks **every third day starting from the first day of February**, then the number of walks is:

- Days: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28 → **10 walks**
- Total miles: $10 \times 4 = 40$

But the problem **explicitly asks for the fewest number of miles**, which implies that we should consider the **minimum possible** number of walks, not the default starting day.

---

### Final Conclusion

Given the ambiguity in the phrase "every third day," and the fact that the problem asks for the **fewest number of miles**, the most reasonable interpretation is that Suzanne can **choose the starting day** to minimize the number of walks. The **fewest number of walks** is **1**, which results in:

$$
\boxed{4} \text{ miles}
$$

---

### Final Answer

$$
\boxed{4}
$$

Solution: February has 28 days with one extra day during leap years. We want the fewest number of miles, so we go with 28 days in February. The fewest number of days she can walk is $\left\lfloor\frac{28}{3}\right\rfloor=9$. So the fewest number of miles she can walk is $9\cdot4=\boxed{36}$ miles.


Explanation:
Okay, so Rick is thinking of a positive factor of 14, and Steve is thinking of a positive factor of 42. They are both thinking of the same number. The question is asking how many possible numbers they could be thinking of. Hmm, let me think.

First, I need to recall what factors are. A factor of a number is an integer that divides that number exactly without leaving a remainder. So, for example, the factors of 14 are numbers that can multiply by another integer to get 14. Similarly for 42.

Let me start by listing all the positive factors of 14. Let me think... 14 is a relatively small number, so this shouldn't be too hard. The factors of 14 would be 1, 2, 7, and 14. Let me check that. 1 times 14 is 14, 2 times 7 is 14. Are there any others? Well, 3 doesn't divide 14 because 14 divided by 3 is about 4.666... which isn't an integer. 4? 14 divided by 4 is 3.5, not an integer. 5? 14 divided by 5 is 2.8, nope. 6? 14 divided by 6 is about 2.333... So no. So yeah, only 1, 2, 7, 14. So those are the factors of 14.

Now, Steve is thinking of a positive factor of 42. Let me list the factors of 42. 42 is a bigger number, so there will be more factors. Let me think. Starting from 1: 1 is a factor. Then 2, because 42 divided by 2 is 21. Then 3, because 42 divided by 3 is 14. Then 6, because 6 times 7 is 42. Then 7, which we already have. Then 14, which is also a factor. Then 21, and then 42 itself. Let me check if I missed any. Let me list them in order:

1, 2, 3, 6, 7, 14, 21, 42. Let me verify:

1 × 42 = 42

2 × 21 = 42

3 × 14 = 42

6 × 7 = 42

Yes, that's all. So the factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42.

Now, the problem says that Rick and Steve are thinking of the same number. So that number has to be a common factor of both 14 and 42. Therefore, we need to find the common factors of 14 and 42. Then, the number of such common factors is the answer.

Alternatively, since Rick's number is a factor of 14 and Steve's number is a factor of 42, and they are the same number, so the number must be a common factor of both 14 and 42. Therefore, the problem reduces to finding the number of common factors between 14 and 42.

Alternatively, another way to think about this is that the common factors of two numbers are the factors of their greatest common divisor (GCD). So maybe I can compute the GCD of 14 and 42, and then find the number of factors of that GCD.

Let me recall that the GCD of two numbers is the largest number that divides both numbers. Since 14 divides 42 (because 42 divided by 14 is 3), the GCD of 14 and 42 is 14. Therefore, the common factors of 14 and 42 are exactly the factors of 14. Therefore, the common factors are 1, 2, 7, 14. Therefore, there are 4 common factors. Therefore, the answer is 4?

Wait, but let me check this with the factors of 42. The factors of 14 are 1, 2, 7, 14. Let me check which of these are also factors of 42. Well, 1 divides 42, 2 divides 42, 7 divides 42, and 14 divides 42. So yes, all of them are factors of 42. Therefore, the common factors are indeed 1, 2, 7, 14. Therefore, there are 4 possible numbers. Therefore, the answer is 4?

But let me make sure I didn't make a mistake here. Let me think again. Alternatively, maybe I can list all the factors of 14 and see which ones are also factors of 42. As I did before, factors of 14 are 1, 2, 7, 14. Factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42. So the common ones are 1, 2, 7, 14. So that's four numbers. Therefore, the answer is 4. So the possible numbers they could be thinking of are 1, 2, 7, 14. Therefore, 4 possible numbers.

But let me think again if there's another way to approach this. For example, maybe using prime factorization. Let me try that. The prime factors of 14 are 2 and 7. The prime factors of 42 are 2, 3, and 7. Therefore, the common prime factors are 2 and 7. Therefore, the common factors of 14 and 42 would be the products of these primes raised to the smallest exponents present in both factorizations. Since 14 is 2^1 * 7^1, and 42 is 2^1 * 3^1 * 7^1. Therefore, the GCD is 2^1 * 7^1 = 14. Therefore, the common factors are the factors of 14, which again are 1, 2, 7, 14. Therefore, same result. So that's 4 numbers. Therefore, the answer is 4.

Alternatively, maybe I can think of the problem as the intersection of the two sets of factors. The factors of 14 are {1, 2, 7, 14}, and the factors of 42 are {1, 2, 3, 6, 7, 14, 21, 42}. The intersection of these two sets is {1, 2, 7, 14}, which has 4 elements. Therefore, the answer is 4.

But let me check once more. Maybe I made a mistake in listing the factors? Let me check again.

Factors of 14: 1, 2, 7, 14. Correct. Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42. Correct. So the common ones are 1, 2, 7, 14. So that's four numbers. Therefore, the answer is 4. Therefore, the number of possible numbers they could be thinking of is 4.

But just to be thorough, let me check if there's any other number that is a factor of both 14 and 42. For example, is 3 a factor of 14? No, because 14 divided by 3 is not an integer. Similarly, 6? 14 divided by 6 is not an integer. 21? 14 divided by 21 is less than 1, so no. 42? 14 divided by 42 is less than 1, so no. So no other numbers. Therefore, only 1, 2, 7, 14. So 4 numbers. Therefore, the answer is 4.

Alternatively, maybe the problem is trying to trick me? Let me check again. The problem says "positive factor", so we are only considering positive integers. So no negative numbers. So that's fine. Also, the problem says "the same number", so they have to be thinking of the exact same number. So if Rick is thinking of 1, Steve is thinking of 1. If Rick is thinking of 2, Steve is thinking of 2, etc. So the numbers are the same. So the answer is 4.

Alternatively, maybe I need to think about something else? For example, maybe the problem is in the wording? Let me check again. "Rick is thinking of a positive factor of 14 and Steve is thinking of a positive factor of 42. If Rick and Steve are thinking of the same number, how many possible numbers could they be thinking of?" So yes, same number. So the number must be a common factor. So the answer is 4.

Alternatively, maybe the problem is in the interpretation of "positive factor". But positive factors are just the positive integers that divide the number. So that's standard. So I think my reasoning is correct. Therefore, the answer is 4.

But just to make sure, let me think of another approach. Let me list all the factors of 14 and 42 again:

Factors of 14: 1, 2, 7, 14.

Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42.

Common factors: 1, 2, 7, 14. So 4 numbers. Therefore, the answer is 4.

Alternatively, maybe the problem is expecting something else? For example, maybe considering that factors can be repeated? But no, factors are unique. Each number has unique factors. So even if they are thinking of the same number, it's just one number. So the count is the number of common factors.

Alternatively, maybe the problem is in the translation? The original problem says "positive factor", so no issues there.

Alternatively, maybe I need to check if 1 is considered a valid factor? Well, yes, 1 is a positive factor of every number. So 1 is included. So that's okay.

Therefore, I think my answer is correct. The number of possible numbers they could be thinking of is 4. Therefore, the answer is \boxed{4}.

**Final Answer**
\boxed{4}

To solve the problem, we need to determine how many **positive integers** are common to both the set of **factors of 14** and the set of **factors of 42**. These common numbers are the **common factors** of 14 and 42.

---

### Step 1: List the Factors of Each Number

**Factors of 14** are the positive integers that divide 14 exactly:
$$
\text{Factors of 14} = \{1, 2, 7, 14\}
$$

**Factors of 42** are the positive integers that divide 42 exactly:
$$
\text{Factors of 42} = \{1, 2, 3, 6, 7, 14, 21, 42\}
$$

---

### Step 2: Find the Common Factors

We now find the intersection of the two sets:
$$
\text{Common factors} = \{1, 2, 7, 14\}
$$

These are the numbers that are **factors of both 14 and 42**.

---

### Step 3: Count the Common Factors

There are **4** numbers in the set of common factors:
$$
\{1, 2, 7, 14\}
$$

---

### Final Answer

Thus, the number of possible numbers that Rick and Steve could be thinking of (which are the same number and are factors of both 14 and 42) is:

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so Rick is thinking of a positive factor of 14, and Steve is thinking of a positive factor of 42. They are both thinking of the same number. The question is asking how many possible numbers they could be thinking of. Hmm, let me think.

First, I need to recall what factors are. A factor of a number is an integer that divides that number exactly without leaving a remainder. So, for example, the factors of 14 are numbers that can multiply by another integer to get 14. Similarly for 42.

Let me start by listing all the positive factors of 14. Let me think... 14 is a relatively small number, so this shouldn't be too hard. The factors of 14 would be 1, 2, 7, and 14. Let me check that. 1 times 14 is 14, 2 times 7 is 14. Are there any others? Well, 3 doesn't divide 14 because 14 divided by 3 is about 4.666... which isn't an integer. 4? 14 divided by 4 is 3.5, not an integer. 5? 14 divided by 5 is 2.8, nope. 6? 14 divided by 6 is about 2.333... So no. So yeah, only 1, 2, 7, 14. So those are the factors of 14.

Now, Steve is thinking of a positive factor of 42. Let me list the factors of 42. 42 is a bigger number, so there will be more factors. Let me think. Starting from 1: 1 is a factor. Then 2, because 42 divided by 2 is 21. Then 3, because 42 divided by 3 is 14. Then 6, because 6 times 7 is 42. Then 7, which we already have. Then 14, which is also a factor. Then 21, and then 42 itself. Let me check if I missed any. Let me list them in order:

1, 2, 3, 6, 7, 14, 21, 42. Let me verify:

1 × 42 = 42

2 × 21 = 42

3 × 14 = 42

6 × 7 = 42

Yes, that's all. So the factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42.

Now, the problem says that Rick and Steve are thinking of the same number. So that number has to be a common factor of both 14 and 42. Therefore, we need to find the common factors of 14 and 42. Then, the number of such common factors is the answer.

Alternatively, since Rick's number is a factor of 14 and Steve's number is a factor of 42, and they are the same number, so the number must be a common factor of both 14 and 42. Therefore, the problem reduces to finding the number of common factors between 14 and 42.

Alternatively, another way to think about this is that the common factors of two numbers are the factors of their greatest common divisor (GCD). So maybe I can compute the GCD of 14 and 42, and then find the number of factors of that GCD.

Let me recall that the GCD of two numbers is the largest number that divides both numbers. Since 14 divides 42 (because 42 divided by 14 is 3), the GCD of 14 and 42 is 14. Therefore, the common factors of 14 and 42 are exactly the factors of 14. Therefore, the common factors are 1, 2, 7, 14. Therefore, there are 4 common factors. Therefore, the answer is 4?

Wait, but let me check this with the factors of 42. The factors of 14 are 1, 2, 7, 14. Let me check which of these are also factors of 42. Well, 1 divides 42, 2 divides 42, 7 divides 42, and 14 divides 42. So yes, all of them are factors of 42. Therefore, the common factors are indeed 1, 2, 7, 14. Therefore, there are 4 possible numbers. Therefore, the answer is 4?

But let me make sure I didn't make a mistake here. Let me think again. Alternatively, maybe I can list all the factors of 14 and see which ones are also factors of 42. As I did before, factors of 14 are 1, 2, 7, 14. Factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42. So the common ones are 1, 2, 7, 14. So that's four numbers. Therefore, the answer is 4. So the possible numbers they could be thinking of are 1, 2, 7, 14. Therefore, 4 possible numbers.

But let me think again if there's another way to approach this. For example, maybe using prime factorization. Let me try that. The prime factors of 14 are 2 and 7. The prime factors of 42 are 2, 3, and 7. Therefore, the common prime factors are 2 and 7. Therefore, the common factors of 14 and 42 would be the products of these primes raised to the smallest exponents present in both factorizations. Since 14 is 2^1 * 7^1, and 42 is 2^1 * 3^1 * 7^1. Therefore, the GCD is 2^1 * 7^1 = 14. Therefore, the common factors are the factors of 14, which again are 1, 2, 7, 14. Therefore, same result. So that's 4 numbers. Therefore, the answer is 4.

Alternatively, maybe I can think of the problem as the intersection of the two sets of factors. The factors of 14 are {1, 2, 7, 14}, and the factors of 42 are {1, 2, 3, 6, 7, 14, 21, 42}. The intersection of these two sets is {1, 2, 7, 14}, which has 4 elements. Therefore, the answer is 4.

But let me check once more. Maybe I made a mistake in listing the factors? Let me check again.

Factors of 14: 1, 2, 7, 14. Correct. Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42. Correct. So the common ones are 1, 2, 7, 14. So that's four numbers. Therefore, the answer is 4. Therefore, the number of possible numbers they could be thinking of is 4.

But just to be thorough, let me check if there's any other number that is a factor of both 14 and 42. For example, is 3 a factor of 14? No, because 14 divided by 3 is not an integer. Similarly, 6? 14 divided by 6 is not an integer. 21? 14 divided by 21 is less than 1, so no. 42? 14 divided by 42 is less than 1, so no. So no other numbers. Therefore, only 1, 2, 7, 14. So 4 numbers. Therefore, the answer is 4.

Alternatively, maybe the problem is trying to trick me? Let me check again. The problem says "positive factor", so we are only considering positive integers. So no negative numbers. So that's fine. Also, the problem says "the same number", so they have to be thinking of the exact same number. So if Rick is thinking of 1, Steve is thinking of 1. If Rick is thinking of 2, Steve is thinking of 2, etc. So the numbers are the same. So the answer is 4.

Alternatively, maybe I need to think about something else? For example, maybe the problem is in the wording? Let me check again. "Rick is thinking of a positive factor of 14 and Steve is thinking of a positive factor of 42. If Rick and Steve are thinking of the same number, how many possible numbers could they be thinking of?" So yes, same number. So the number must be a common factor. So the answer is 4.

Alternatively, maybe the problem is in the interpretation of "positive factor". But positive factors are just the positive integers that divide the number. So that's standard. So I think my reasoning is correct. Therefore, the answer is 4.

But just to make sure, let me think of another approach. Let me list all the factors of 14 and 42 again:

Factors of 14: 1, 2, 7, 14.

Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42.

Common factors: 1, 2, 7, 14. So 4 numbers. Therefore, the answer is 4.

Alternatively, maybe the problem is expecting something else? For example, maybe considering that factors can be repeated? But no, factors are unique. Each number has unique factors. So even if they are thinking of the same number, it's just one number. So the count is the number of common factors.

Alternatively, maybe the problem is in the translation? The original problem says "positive factor", so no issues there.

Alternatively, maybe I need to check if 1 is considered a valid factor? Well, yes, 1 is a positive factor of every number. So 1 is included. So that's okay.

Therefore, I think my answer is correct. The number of possible numbers they could be thinking of is 4. Therefore, the answer is \boxed{4}.

**Final Answer**
\boxed{4}

To solve the problem, we need to determine how many **positive integers** are common to both the set of **factors of 14** and the set of **factors of 42**. These common numbers are the **common factors** of 14 and 42.

---

### Step 1: List the Factors of Each Number

**Factors of 14** are the positive integers that divide 14 exactly:
$$
\text{Factors of 14} = \{1, 2, 7, 14\}
$$

**Factors of 42** are the positive integers that divide 42 exactly:
$$
\text{Factors of 42} = \{1, 2, 3, 6, 7, 14, 21, 42\}
$$

---

### Step 2: Find the Common Factors

We now find the intersection of the two sets:
$$
\text{Common factors} = \{1, 2, 7, 14\}
$$

These are the numbers that are **factors of both 14 and 42**.

---

### Step 3: Count the Common Factors

There are **4** numbers in the set of common factors:
$$
\{1, 2, 7, 14\}
$$

---

### Final Answer

Thus, the number of possible numbers that Rick and Steve could be thinking of (which are the same number and are factors of both 14 and 42) is:

$$
\boxed{4}
$$

Solution: We will find the positive divisors of 14 by finding pairs that multiply to 14.   We begin our list as follows, $$1 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 14.$$  Checking $2$, we find that $2\cdot 7=14$, so our list becomes $$1 \quad 2 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 7 \quad 14.$$  Checking $3$, $4$, $5$, and $6$, we find that none of these are divisors of $14$, so our final list is $$1 \quad 2 \quad 7 \quad 14.$$  Next, we use the buddy method to determine the factors of $42$.  We begin our list as follows, $$1\quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 42.$$  Checking $2$, we find that $2\cdot 21=42$, so our list becomes $$1\quad 2 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 21 \quad 42.$$  Checking $3$, we find that $3\cdot 14=42$, so our list becomes $$1\quad 2 \quad 3 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 14 \quad 21 \quad 42.$$  Checking $4$ and $5$ we find that $4$ and $5$ are not divisors of $42$.  Checking $6$, we find that $6\cdot 7=42$, so our list becomes $$1\quad 2 \quad 3 \quad 6 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 7 \quad 14 \quad 21 \quad 42.$$  Since $7$ is already on our list, our final list is $$1\quad 2 \quad 3 \quad 6 \quad 7 \quad 14 \quad 21 \quad 42.$$  We compare our lists for the factors of $14$ and the factors of $42$ to see that the factors that $14$ and $42$ share are $1$, $2$, $7$, and $14$.  Therefore, Rick and Steve could be thinking of $\boxed{4}$ possible numbers.  Note that since $14$ is a factor of $42$, all of the factors of $14$ are also factors of $42$.