Explanation:
Okay, let's try to figure out how many students take physics. There are 360 people in total. Some take calculus, physics, chemistry, or none. Let me jot down the given information first.

First, 15 students take all three subjects: calculus, physics, and chemistry. Then, 15 don't take any of them. So, the number of students taking at least one of the subjects is 360 - 15 = 345.

Next, 180 take calculus. So, the total number of students in calculus is 180. Also, twice as many students take chemistry as take physics. Let me note that as C = 2P, where C is the number taking chemistry and P is the number taking physics.

Then, there are some overlaps: 75 take both calculus and chemistry, 75 take both physics and chemistry, and only 30 take both physics and calculus. Wait, the problem says "Only 30 take both physics and calculus." So, the intersection of physics and calculus is 30. But earlier, it said 15 take all three. So, does that mean that the number taking both physics and calculus is 30, which includes those who also take chemistry? So, the 30 includes the 15 who take all three? Let me confirm.

Yes, when they say "Only 30 take both physics and calculus," that probably means that the intersection of physics and calculus is 30, which includes those who might also be taking chemistry. But since 15 take all three, that would mean that the number taking both physics and calculus but not chemistry is 30 - 15 = 15? Wait, but maybe the problem is using "only" to mean that there are no other overlaps? Wait, no, the wording is a bit confusing here. Let me check again.

The problem says: "Only 30 take both physics and calculus." Hmm. The word "only" here might be indicating that the number of students taking both physics and calculus is 30, and there are no students taking all three? But that contradicts the first statement that 15 take all three. So, maybe the problem is using "only" to mean that the number taking both physics and calculus, excluding those taking all three, is 30? Wait, that might not be standard. Let me think.

In standard set theory, when they say "the number taking both physics and calculus is 30," that includes those taking all three. But if they say "only 30 take both physics and calculus," that might mean that the intersection of physics and calculus is 30, and there are no students taking all three? But that contradicts the first statement. Therefore, maybe the problem is using "only" in a different way. Wait, maybe the problem is saying that the number taking both physics and calculus, but not chemistry, is 30? That would make sense. Let me check the original problem again.

Original problem: "Only 30 take both physics and calculus." So, maybe "only" here is meant to indicate that there are no students taking all three? But that conflicts with the first statement. Wait, the first statement says 15 take all three. So, perhaps the problem is using "only" to mean that the number taking both physics and calculus is 30, and that there are no students taking all three? But that contradicts the first statement. Therefore, maybe the problem is using "only" in the sense that the number taking both physics and calculus, excluding those taking all three, is 30? That is, the intersection of physics and calculus is 30 + 15 = 45? Wait, but the problem says "Only 30 take both physics and calculus," so maybe the total number taking both physics and calculus is 30, which includes those taking all three? But then how does that reconcile with the 15 taking all three?

This is confusing. Let me try to parse the problem again:

"15 take calculus, physics, and chemistry, and 15 don't take any of them. 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics?"

So, the key is that "Only 30 take both physics and calculus." So, the number of students taking both physics and calculus is 30. However, there are 15 students taking all three. Therefore, the number of students taking both physics and calculus but not chemistry is 30 - 15 = 15. Is that correct? Because the total number taking both physics and calculus is 30, which includes those taking all three. Therefore, the overlap between physics and calculus is 30, which includes the 15 who take all three. Therefore, the number taking both physics and calculus but not chemistry is 30 - 15 = 15.

Similarly, the number taking both calculus and chemistry is 75. Since 15 take all three, the number taking both calculus and chemistry but not physics is 75 - 15 = 60. Similarly, the number taking both physics and chemistry is 75, so the number taking both physics and chemistry but not calculus is 75 - 15 = 60.

So, with that in mind, let me try to use the principle of inclusion-exclusion for three sets. Let me denote:

Let C be the number taking calculus, P the number taking physics, Ch the number taking chemistry.

We know that:

Total students taking at least one subject: 345 (since 360 - 15 = 345)

We have:

C = 180

Ch = 2P (since twice as many take chemistry as take physics)

Also, the number taking both calculus and chemistry is 75, which includes those taking all three. Similarly, the number taking both physics and chemistry is 75, and the number taking both physics and calculus is 30, which includes those taking all three.

So, using inclusion-exclusion formula for three sets:

Total = C + P + Ch - (number taking two subjects) + (number taking all three)

Wait, the formula is:

|C ∪ P ∪ Ch| = |C| + |P| + |Ch| - |C ∩ P| - |C ∩ Ch| - |P ∩ Ch| + |C ∩ P ∩ Ch|

So, substituting the known values:

345 = 180 + P + Ch - 30 - 75 - 75 + 15

Let me compute that step by step.

First, compute the sum of individual sets: 180 + P + Ch

Then subtract the pairwise intersections: -30 -75 -75 = - (30 + 75 + 75) = -180

Then add back the triple intersection: +15

So overall:

345 = 180 + P + Ch - 180 + 15

Simplify:

345 = (180 - 180) + P + Ch + 15

Which is:

345 = 0 + P + Ch + 15

Therefore:

P + Ch = 345 - 15 = 330

But we also know that Ch = 2P. Therefore:

P + 2P = 3P = 330 => P = 110

Therefore, the number of students taking physics is 110?

Wait, but let me check if this is consistent with the other numbers. Let me verify.

So, if P = 110, then Ch = 220.

Now, let's check if the inclusion-exclusion formula holds. Let me recompute:

Total = 180 + 110 + 220 - 30 - 75 - 75 + 15

Compute step by step:

180 + 110 = 290; 290 + 220 = 510

Then subtract 30 + 75 + 75 = 180: 510 - 180 = 330

Then add 15: 330 + 15 = 345. Which matches the total number of students taking at least one subject. So that seems correct.

But let me check if the numbers for the overlaps make sense. For example, the number taking both calculus and chemistry is 75. Which includes those taking all three. So, the number taking both calculus and chemistry is 75. Similarly, the number taking both physics and chemistry is 75. The number taking both physics and calculus is 30.

But let me check if the numbers for individual subjects are consistent. For example, the number taking calculus is 180. Let's break down the calculus takers:

Students taking calculus can be divided into:

- Those taking only calculus

- Those taking calculus and physics but not chemistry

- Those taking calculus and chemistry but not physics

- Those taking all three

We know that the number taking all three is 15.

Number taking both calculus and physics is 30, which includes those taking all three. Therefore, the number taking calculus and physics but not chemistry is 30 - 15 = 15.

Number taking both calculus and chemistry is 75, which includes those taking all three. Therefore, the number taking calculus and chemistry but not physics is 75 - 15 = 60.

Therefore, total number taking calculus is:

Only calculus + (calculus and physics only) + (calculus and chemistry only) + all three.

So, let me denote:

Only calculus = ?

We have:

Total calculus = 180 = Only calculus + 15 (calculus and physics only) + 60 (calculus and chemistry only) + 15 (all three)

So:

Only calculus = 180 - 15 - 60 - 15 = 180 - 90 = 90

So, 90 students take only calculus.

Similarly, let's check for physics. The number taking physics is 110. Let's break it down:

Physics takers include:

- Only physics

- Physics and calculus only

- Physics and chemistry only

- All three

We know:

Physics and calculus only is 15 (from earlier, since total physics and calculus is 30, minus all three)

Physics and chemistry only is 75 - 15 = 60 (since total physics and chemistry is 75, minus all three)

All three is 15

Therefore, total physics takers:

Only physics + 15 + 60 + 15 = Only physics + 90

But total physics is 110, so Only physics = 110 - 90 = 20

Similarly, for chemistry, which is 220. Let's check:

Chemistry takers include:

- Only chemistry

- Chemistry and calculus only

- Chemistry and physics only

- All three

We have:

Chemistry and calculus only is 60 (from earlier)

Chemistry and physics only is 60 (from earlier)

All three is 15

Therefore, total chemistry takers:

Only chemistry + 60 + 60 + 15 = Only chemistry + 135

But total chemistry is 220, so Only chemistry = 220 - 135 = 85

Now, let's check the total number of students taking at least one subject:

Only calculus (90) + Only physics (20) + Only chemistry (85) + (calculus and physics only) (15) + (calculus and chemistry only) (60) + (physics and chemistry only) (60) + all three (15)

Adding these up:

90 + 20 = 110; 110 + 85 = 195; 195 + 15 = 210; 210 + 60 = 270; 270 + 60 = 330; 330 + 15 = 345. Which matches the total of 345. So that seems consistent.

Therefore, the number of students taking physics is 110. But let me check if there's another way to get this, or if I made any wrong assumptions.

Wait, the problem says "Only 30 take both physics and calculus." So, does that mean that the number taking both physics and calculus is 30, which includes those taking all three? Which is what I assumed. So, if that's the case, then the calculation is correct.

Alternatively, if "Only 30 take both physics and calculus" meant that the number taking both physics and calculus but not chemistry is 30, then the total number taking both physics and calculus would be 30 + 15 = 45. But in that case, the inclusion-exclusion formula would have different numbers. Let me check that possibility.

Suppose that "Only 30 take both physics and calculus" means that the number taking both physics and calculus but not chemistry is 30. Then, the total number taking both physics and calculus (including those taking all three) would be 30 + 15 = 45. Then, in the inclusion-exclusion formula, the pairwise intersections would be:

|C ∩ P| = 45, |C ∩ Ch| = 75, |P ∩ Ch| = 75. Then, the formula would be:

Total = 180 + P + Ch - 45 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = P + Ch + 15 = 345 => P + Ch = 330. Which is the same as before. So even if "Only 30" refers to the exclusive overlap, the result is the same. Wait, how?

Wait, if |C ∩ P| is 45, then in the inclusion-exclusion formula, we subtract 45, 75, 75. But then adding back 15. So, 180 + P + Ch - 45 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = same as before. Therefore, regardless of whether |C ∩ P| is 30 or 45, the result is same? Wait, no. Wait, if |C ∩ P| is 45, then the formula would be:

Total = 180 + P + Ch - 45 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = P + Ch + 15 = 345 => P + Ch = 330. But if |C ∩ P| is 30, then:

Total = 180 + P + Ch - 30 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = same thing. Wait, so regardless of the value of |C ∩ P|, as long as the other numbers are fixed, the result is same? That seems odd. Wait, no. Wait, if |C ∩ P| is 30, then the formula is:

Total = 180 + P + Ch - 30 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = P + Ch + 15 = 345 => P + Ch = 330.

If |C ∩ P| is 45, then:

Total = 180 + P + Ch - 45 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = same thing. Wait, so regardless of the value of |C ∩ P|, as long as the other numbers are fixed, the result is same? That seems contradictory. How can that be?

Wait, no. Wait, the problem says that 75 take both calculus and chemistry, and 75 take both physics and chemistry. Also, 15 take all three. So, the number taking both calculus and chemistry is 75, which includes the 15 taking all three. Similarly, the number taking both physics and chemistry is 75, which includes the 15 taking all three. Therefore, the number taking both calculus and chemistry but not physics is 75 - 15 = 60, and the number taking both physics and chemistry but not calculus is 75 - 15 = 60.

But the number taking both calculus and physics is given as 30 (if we take the problem's wording as "Only 30 take both physics and calculus"), which includes the 15 taking all three. Therefore, the number taking both calculus and physics but not chemistry is 30 - 15 = 15.

But if the problem had said that the number taking both calculus and physics is 30, not including those taking all three, then the total number taking both calculus and physics would be 30 + 15 = 45. However, in that case, the inclusion-exclusion formula would still result in P + Ch = 330. Therefore, regardless of whether the 30 includes the triple overlap or not, the result is same? That seems strange. Wait, but in reality, the value of |C ∩ P| is fixed by the problem's statement. If the problem says "Only 30 take both physics and calculus," then that should mean that |C ∩ P| is 30, which includes the triple overlap. Therefore, the calculation is correct.

But why does the inclusion-exclusion formula give the same result regardless? Because when we subtract |C ∩ P|, |C ∩ Ch|, |P ∩ Ch|, and add back |C ∩ P ∩ Ch|, the total is adjusted accordingly. However, in this case, since we have the triple overlap already accounted for in the pairwise intersections, the formula still ends up with P + Ch = 330.

Therefore, regardless of the interpretation of "Only 30 take both physics and calculus," the result is same? Wait, no. If the problem had said that the number taking both physics and calculus is 30, not including those taking all three, then |C ∩ P| would be 30 + 15 = 45. Then, the inclusion-exclusion formula would be:

Total = 180 + P + Ch - 45 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = same as before. Therefore, even if |C ∩ P| is different, the formula still gives P + Ch = 330. Therefore, the value of P + Ch is fixed regardless of the value of |C ∩ P|? That seems contradictory. Wait, but in reality, the value of |C ∩ P| is fixed by the problem. If the problem says "Only 30 take both physics and calculus," then that should mean that |C ∩ P| is 30, which includes the triple overlap. Therefore, the calculation is correct.

But in this case, the problem gives us the number of students taking both calculus and chemistry (75), both physics and chemistry (75), and both physics and calculus (30). Therefore, the inclusion-exclusion formula is:

Total = C + P + Ch - (C∩P + C∩Ch + P∩Ch) + C∩P∩Ch

Which is 345 = 180 + P + Ch - (30 + 75 + 75) + 15

Which is 345 = 180 + P + Ch - 180 + 15 => 345 = P + Ch + 15 => P + Ch = 330.

Since Ch = 2P, then P = 110. Therefore, the answer is 110.

But let me check once again with the numbers. If P = 110, Ch = 220. Then, the number taking only physics is 20, only chemistry is 85, only calculus is 90. Then, the numbers for overlaps:

Calculus and physics only: 15

Calculus and chemistry only: 60

Physics and chemistry only: 60

All three: 15

Total students taking at least one subject: 90 + 20 + 85 + 15 + 60 + 60 + 15 = 90 + 20 = 110; 110 + 85 = 195; 195 + 15 = 210; 210 + 60 = 270; 270 + 60 = 330; 330 + 15 = 345. Correct.

Also, the number taking calculus is 90 (only) + 15 (calculus and physics) + 60 (calculus and chemistry) + 15 (all three) = 90 + 15 + 60 + 15 = 180. Correct.

Number taking physics is 20 (only) + 15 (calculus and physics) + 60 (physics and chemistry) + 15 (all three) = 20 + 15 + 60 + 15 = 110. Correct.

Number taking chemistry is 85 (only) + 60 (calculus and chemistry) + 60 (physics and chemistry) + 15 (all three) = 85 + 60 + 60 + 15 = 220. Correct.

Therefore, all numbers check out. Therefore, the number of students taking physics is 110.

But let me check if there's another way someone might approach this problem and get a different answer. For example, if someone misreads the "Only 30 take both physics and calculus" as meaning that the number taking both physics and calculus is 30, not including those taking all three. Then, they would have |C ∩ P| = 30, and |C ∩ P ∩ Ch| = 15. Therefore, the number taking both physics and calculus but not chemistry is 30. Then, the total number taking both physics and calculus is 30 + 15 = 45. Then, in the inclusion-exclusion formula, they would have:

Total = 180 + P + Ch - 45 - 75 - 75 + 15 = same as before. So, same result. Therefore, regardless of interpretation, the answer is same.

Therefore, the answer is 110.

**Final Answer**
The number of students who take physics is \boxed{110}.

To determine how many students take **physics**, we are given a set of constraints involving three subjects: **calculus (C)**, **physics (P)**, and **chemistry (Ch)**. The total number of students is 360, and 15 students do **not** take any of the three subjects. This means that the number of students taking **at least one** of the subjects is:

$$
360 - 15 = 345
$$

We are also given the following:

- 15 students take **all three** subjects: $ |C \cap P \cap Ch| = 15 $
- 180 students take **calculus**: $ |C| = 180 $
- Twice as many students take **chemistry** as take **physics**: $ |Ch| = 2|P| $
- 75 students take **both calculus and chemistry**: $ |C \cap Ch| = 75 $
- 75 students take **both physics and chemistry**: $ |P \cap Ch| = 75 $
- **Only 30** students take **both physics and calculus**: This means $ |C \cap P| = 30 $, which includes those who also take chemistry.

---

### Step 1: Use the Inclusion-Exclusion Principle

The formula for the number of students taking **at least one** of the three subjects is:

$$
|C \cup P \cup Ch| = |C| + |P| + |Ch| - |C \cap P| - |C \cap Ch| - |P \cap Ch| + |C \cap P \cap Ch|
$$

Substitute the known values:

$$
345 = 180 + P + 2P - 30 - 75 - 75 + 15
$$

Simplify the right-hand side:

$$
345 = 180 + 3P - 180 + 15 = 3P + 15
$$

Solve for $ P $:

$$
3P = 345 - 15 = 330 \Rightarrow P = \frac{330}{3} = 110
$$

---

### Step 2: Verify the Result

Let’s verify that the total number of students taking at least one subject is indeed 345, using the calculated values:

- $ |C| = 180 $
- $ |P| = 110 $
- $ |Ch| = 2P = 220 $
- $ |C \cap P| = 30 $
- $ |C \cap Ch| = 75 $
- $ |P \cap Ch| = 75 $
- $ |C \cap P \cap Ch| = 15 $

Now apply the inclusion-exclusion formula again:

$$
|C \cup P \cup Ch| = 180 + 110 + 220 - 30 - 75 - 75 + 15 = 345
$$

This confirms the calculation is consistent.

---

### Final Answer

$$
\boxed{110}
$$

[Assistant]
Explanation:
Okay, let's try to figure out how many students take physics. There are 360 people in total. Some take calculus, physics, chemistry, or none. Let me jot down the given information first.

First, 15 students take all three subjects: calculus, physics, and chemistry. Then, 15 don't take any of them. So, the number of students taking at least one of the subjects is 360 - 15 = 345.

Next, 180 take calculus. So, the total number of students in calculus is 180. Also, twice as many students take chemistry as take physics. Let me note that as C = 2P, where C is the number taking chemistry and P is the number taking physics.

Then, there are some overlaps: 75 take both calculus and chemistry, 75 take both physics and chemistry, and only 30 take both physics and calculus. Wait, the problem says "Only 30 take both physics and calculus." So, the intersection of physics and calculus is 30. But earlier, it said 15 take all three. So, does that mean that the number taking both physics and calculus is 30, which includes those who also take chemistry? So, the 30 includes the 15 who take all three? Let me confirm.

Yes, when they say "Only 30 take both physics and calculus," that probably means that the intersection of physics and calculus is 30, which includes those who might also be taking chemistry. But since 15 take all three, that would mean that the number taking both physics and calculus but not chemistry is 30 - 15 = 15? Wait, but maybe the problem is using "only" to mean that there are no other overlaps? Wait, no, the wording is a bit confusing here. Let me check again.

The problem says: "Only 30 take both physics and calculus." Hmm. The word "only" here might be indicating that the number of students taking both physics and calculus is 30, and there are no students taking all three? But that contradicts the first statement that 15 take all three. So, maybe the problem is using "only" to mean that the number taking both physics and calculus, excluding those taking all three, is 30? Wait, that might not be standard. Let me think.

In standard set theory, when they say "the number taking both physics and calculus is 30," that includes those taking all three. But if they say "only 30 take both physics and calculus," that might mean that the intersection of physics and calculus is 30, and there are no students taking all three? But that contradicts the first statement. Therefore, maybe the problem is using "only" in a different way. Wait, maybe the problem is saying that the number taking both physics and calculus, but not chemistry, is 30? That would make sense. Let me check the original problem again.

Original problem: "Only 30 take both physics and calculus." So, maybe "only" here is meant to indicate that there are no students taking all three? But that conflicts with the first statement. Wait, the first statement says 15 take all three. So, perhaps the problem is using "only" to mean that the number taking both physics and calculus is 30, and that there are no students taking all three? But that contradicts the first statement. Therefore, maybe the problem is using "only" in the sense that the number taking both physics and calculus, excluding those taking all three, is 30? That is, the intersection of physics and calculus is 30 + 15 = 45? Wait, but the problem says "Only 30 take both physics and calculus," so maybe the total number taking both physics and calculus is 30, which includes those taking all three? But then how does that reconcile with the 15 taking all three?

This is confusing. Let me try to parse the problem again:

"15 take calculus, physics, and chemistry, and 15 don't take any of them. 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics?"

So, the key is that "Only 30 take both physics and calculus." So, the number of students taking both physics and calculus is 30. However, there are 15 students taking all three. Therefore, the number of students taking both physics and calculus but not chemistry is 30 - 15 = 15. Is that correct? Because the total number taking both physics and calculus is 30, which includes those taking all three. Therefore, the overlap between physics and calculus is 30, which includes the 15 who take all three. Therefore, the number taking both physics and calculus but not chemistry is 30 - 15 = 15.

Similarly, the number taking both calculus and chemistry is 75. Since 15 take all three, the number taking both calculus and chemistry but not physics is 75 - 15 = 60. Similarly, the number taking both physics and chemistry is 75, so the number taking both physics and chemistry but not calculus is 75 - 15 = 60.

So, with that in mind, let me try to use the principle of inclusion-exclusion for three sets. Let me denote:

Let C be the number taking calculus, P the number taking physics, Ch the number taking chemistry.

We know that:

Total students taking at least one subject: 345 (since 360 - 15 = 345)

We have:

C = 180

Ch = 2P (since twice as many take chemistry as take physics)

Also, the number taking both calculus and chemistry is 75, which includes those taking all three. Similarly, the number taking both physics and chemistry is 75, and the number taking both physics and calculus is 30, which includes those taking all three.

So, using inclusion-exclusion formula for three sets:

Total = C + P + Ch - (number taking two subjects) + (number taking all three)

Wait, the formula is:

|C ∪ P ∪ Ch| = |C| + |P| + |Ch| - |C ∩ P| - |C ∩ Ch| - |P ∩ Ch| + |C ∩ P ∩ Ch|

So, substituting the known values:

345 = 180 + P + Ch - 30 - 75 - 75 + 15

Let me compute that step by step.

First, compute the sum of individual sets: 180 + P + Ch

Then subtract the pairwise intersections: -30 -75 -75 = - (30 + 75 + 75) = -180

Then add back the triple intersection: +15

So overall:

345 = 180 + P + Ch - 180 + 15

Simplify:

345 = (180 - 180) + P + Ch + 15

Which is:

345 = 0 + P + Ch + 15

Therefore:

P + Ch = 345 - 15 = 330

But we also know that Ch = 2P. Therefore:

P + 2P = 3P = 330 => P = 110

Therefore, the number of students taking physics is 110?

Wait, but let me check if this is consistent with the other numbers. Let me verify.

So, if P = 110, then Ch = 220.

Now, let's check if the inclusion-exclusion formula holds. Let me recompute:

Total = 180 + 110 + 220 - 30 - 75 - 75 + 15

Compute step by step:

180 + 110 = 290; 290 + 220 = 510

Then subtract 30 + 75 + 75 = 180: 510 - 180 = 330

Then add 15: 330 + 15 = 345. Which matches the total number of students taking at least one subject. So that seems correct.

But let me check if the numbers for the overlaps make sense. For example, the number taking both calculus and chemistry is 75. Which includes those taking all three. So, the number taking both calculus and chemistry is 75. Similarly, the number taking both physics and chemistry is 75. The number taking both physics and calculus is 30.

But let me check if the numbers for individual subjects are consistent. For example, the number taking calculus is 180. Let's break down the calculus takers:

Students taking calculus can be divided into:

- Those taking only calculus

- Those taking calculus and physics but not chemistry

- Those taking calculus and chemistry but not physics

- Those taking all three

We know that the number taking all three is 15.

Number taking both calculus and physics is 30, which includes those taking all three. Therefore, the number taking calculus and physics but not chemistry is 30 - 15 = 15.

Number taking both calculus and chemistry is 75, which includes those taking all three. Therefore, the number taking calculus and chemistry but not physics is 75 - 15 = 60.

Therefore, total number taking calculus is:

Only calculus + (calculus and physics only) + (calculus and chemistry only) + all three.

So, let me denote:

Only calculus = ?

We have:

Total calculus = 180 = Only calculus + 15 (calculus and physics only) + 60 (calculus and chemistry only) + 15 (all three)

So:

Only calculus = 180 - 15 - 60 - 15 = 180 - 90 = 90

So, 90 students take only calculus.

Similarly, let's check for physics. The number taking physics is 110. Let's break it down:

Physics takers include:

- Only physics

- Physics and calculus only

- Physics and chemistry only

- All three

We know:

Physics and calculus only is 15 (from earlier, since total physics and calculus is 30, minus all three)

Physics and chemistry only is 75 - 15 = 60 (since total physics and chemistry is 75, minus all three)

All three is 15

Therefore, total physics takers:

Only physics + 15 + 60 + 15 = Only physics + 90

But total physics is 110, so Only physics = 110 - 90 = 20

Similarly, for chemistry, which is 220. Let's check:

Chemistry takers include:

- Only chemistry

- Chemistry and calculus only

- Chemistry and physics only

- All three

We have:

Chemistry and calculus only is 60 (from earlier)

Chemistry and physics only is 60 (from earlier)

All three is 15

Therefore, total chemistry takers:

Only chemistry + 60 + 60 + 15 = Only chemistry + 135

But total chemistry is 220, so Only chemistry = 220 - 135 = 85

Now, let's check the total number of students taking at least one subject:

Only calculus (90) + Only physics (20) + Only chemistry (85) + (calculus and physics only) (15) + (calculus and chemistry only) (60) + (physics and chemistry only) (60) + all three (15)

Adding these up:

90 + 20 = 110; 110 + 85 = 195; 195 + 15 = 210; 210 + 60 = 270; 270 + 60 = 330; 330 + 15 = 345. Which matches the total of 345. So that seems consistent.

Therefore, the number of students taking physics is 110. But let me check if there's another way to get this, or if I made any wrong assumptions.

Wait, the problem says "Only 30 take both physics and calculus." So, does that mean that the number taking both physics and calculus is 30, which includes those taking all three? Which is what I assumed. So, if that's the case, then the calculation is correct.

Alternatively, if "Only 30 take both physics and calculus" meant that the number taking both physics and calculus but not chemistry is 30, then the total number taking both physics and calculus would be 30 + 15 = 45. But in that case, the inclusion-exclusion formula would have different numbers. Let me check that possibility.

Suppose that "Only 30 take both physics and calculus" means that the number taking both physics and calculus but not chemistry is 30. Then, the total number taking both physics and calculus (including those taking all three) would be 30 + 15 = 45. Then, in the inclusion-exclusion formula, the pairwise intersections would be:

|C ∩ P| = 45, |C ∩ Ch| = 75, |P ∩ Ch| = 75. Then, the formula would be:

Total = 180 + P + Ch - 45 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = P + Ch + 15 = 345 => P + Ch = 330. Which is the same as before. So even if "Only 30" refers to the exclusive overlap, the result is the same. Wait, how?

Wait, if |C ∩ P| is 45, then in the inclusion-exclusion formula, we subtract 45, 75, 75. But then adding back 15. So, 180 + P + Ch - 45 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = same as before. Therefore, regardless of whether |C ∩ P| is 30 or 45, the result is same? Wait, no. Wait, if |C ∩ P| is 45, then the formula would be:

Total = 180 + P + Ch - 45 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = P + Ch + 15 = 345 => P + Ch = 330. But if |C ∩ P| is 30, then:

Total = 180 + P + Ch - 30 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = same thing. Wait, so regardless of the value of |C ∩ P|, as long as the other numbers are fixed, the result is same? That seems odd. Wait, no. Wait, if |C ∩ P| is 30, then the formula is:

Total = 180 + P + Ch - 30 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = P + Ch + 15 = 345 => P + Ch = 330.

If |C ∩ P| is 45, then:

Total = 180 + P + Ch - 45 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = same thing. Wait, so regardless of the value of |C ∩ P|, as long as the other numbers are fixed, the result is same? That seems contradictory. How can that be?

Wait, no. Wait, the problem says that 75 take both calculus and chemistry, and 75 take both physics and chemistry. Also, 15 take all three. So, the number taking both calculus and chemistry is 75, which includes the 15 taking all three. Similarly, the number taking both physics and chemistry is 75, which includes the 15 taking all three. Therefore, the number taking both calculus and chemistry but not physics is 75 - 15 = 60, and the number taking both physics and chemistry but not calculus is 75 - 15 = 60.

But the number taking both calculus and physics is given as 30 (if we take the problem's wording as "Only 30 take both physics and calculus"), which includes the 15 taking all three. Therefore, the number taking both calculus and physics but not chemistry is 30 - 15 = 15.

But if the problem had said that the number taking both calculus and physics is 30, not including those taking all three, then the total number taking both calculus and physics would be 30 + 15 = 45. However, in that case, the inclusion-exclusion formula would still result in P + Ch = 330. Therefore, regardless of whether the 30 includes the triple overlap or not, the result is same? That seems strange. Wait, but in reality, the value of |C ∩ P| is fixed by the problem's statement. If the problem says "Only 30 take both physics and calculus," then that should mean that |C ∩ P| is 30, which includes the triple overlap. Therefore, the calculation is correct.

But why does the inclusion-exclusion formula give the same result regardless? Because when we subtract |C ∩ P|, |C ∩ Ch|, |P ∩ Ch|, and add back |C ∩ P ∩ Ch|, the total is adjusted accordingly. However, in this case, since we have the triple overlap already accounted for in the pairwise intersections, the formula still ends up with P + Ch = 330.

Therefore, regardless of the interpretation of "Only 30 take both physics and calculus," the result is same? Wait, no. If the problem had said that the number taking both physics and calculus is 30, not including those taking all three, then |C ∩ P| would be 30 + 15 = 45. Then, the inclusion-exclusion formula would be:

Total = 180 + P + Ch - 45 - 75 - 75 + 15 = 180 + P + Ch - 180 + 15 = same as before. Therefore, even if |C ∩ P| is different, the formula still gives P + Ch = 330. Therefore, the value of P + Ch is fixed regardless of the value of |C ∩ P|? That seems contradictory. Wait, but in reality, the value of |C ∩ P| is fixed by the problem. If the problem says "Only 30 take both physics and calculus," then that should mean that |C ∩ P| is 30, which includes the triple overlap. Therefore, the calculation is correct.

But in this case, the problem gives us the number of students taking both calculus and chemistry (75), both physics and chemistry (75), and both physics and calculus (30). Therefore, the inclusion-exclusion formula is:

Total = C + P + Ch - (C∩P + C∩Ch + P∩Ch) + C∩P∩Ch

Which is 345 = 180 + P + Ch - (30 + 75 + 75) + 15

Which is 345 = 180 + P + Ch - 180 + 15 => 345 = P + Ch + 15 => P + Ch = 330.

Since Ch = 2P, then P = 110. Therefore, the answer is 110.

But let me check once again with the numbers. If P = 110, Ch = 220. Then, the number taking only physics is 20, only chemistry is 85, only calculus is 90. Then, the numbers for overlaps:

Calculus and physics only: 15

Calculus and chemistry only: 60

Physics and chemistry only: 60

All three: 15

Total students taking at least one subject: 90 + 20 + 85 + 15 + 60 + 60 + 15 = 90 + 20 = 110; 110 + 85 = 195; 195 + 15 = 210; 210 + 60 = 270; 270 + 60 = 330; 330 + 15 = 345. Correct.

Also, the number taking calculus is 90 (only) + 15 (calculus and physics) + 60 (calculus and chemistry) + 15 (all three) = 90 + 15 + 60 + 15 = 180. Correct.

Number taking physics is 20 (only) + 15 (calculus and physics) + 60 (physics and chemistry) + 15 (all three) = 20 + 15 + 60 + 15 = 110. Correct.

Number taking chemistry is 85 (only) + 60 (calculus and chemistry) + 60 (physics and chemistry) + 15 (all three) = 85 + 60 + 60 + 15 = 220. Correct.

Therefore, all numbers check out. Therefore, the number of students taking physics is 110.

But let me check if there's another way someone might approach this problem and get a different answer. For example, if someone misreads the "Only 30 take both physics and calculus" as meaning that the number taking both physics and calculus is 30, not including those taking all three. Then, they would have |C ∩ P| = 30, and |C ∩ P ∩ Ch| = 15. Therefore, the number taking both physics and calculus but not chemistry is 30. Then, the total number taking both physics and calculus is 30 + 15 = 45. Then, in the inclusion-exclusion formula, they would have:

Total = 180 + P + Ch - 45 - 75 - 75 + 15 = same as before. So, same result. Therefore, regardless of interpretation, the answer is same.

Therefore, the answer is 110.

**Final Answer**
The number of students who take physics is \boxed{110}.

To determine how many students take **physics**, we are given a set of constraints involving three subjects: **calculus (C)**, **physics (P)**, and **chemistry (Ch)**. The total number of students is 360, and 15 students do **not** take any of the three subjects. This means that the number of students taking **at least one** of the subjects is:

$$
360 - 15 = 345
$$

We are also given the following:

- 15 students take **all three** subjects: $ |C \cap P \cap Ch| = 15 $
- 180 students take **calculus**: $ |C| = 180 $
- Twice as many students take **chemistry** as take **physics**: $ |Ch| = 2|P| $
- 75 students take **both calculus and chemistry**: $ |C \cap Ch| = 75 $
- 75 students take **both physics and chemistry**: $ |P \cap Ch| = 75 $
- **Only 30** students take **both physics and calculus**: This means $ |C \cap P| = 30 $, which includes those who also take chemistry.

---

### Step 1: Use the Inclusion-Exclusion Principle

The formula for the number of students taking **at least one** of the three subjects is:

$$
|C \cup P \cup Ch| = |C| + |P| + |Ch| - |C \cap P| - |C \cap Ch| - |P \cap Ch| + |C \cap P \cap Ch|
$$

Substitute the known values:

$$
345 = 180 + P + 2P - 30 - 75 - 75 + 15
$$

Simplify the right-hand side:

$$
345 = 180 + 3P - 180 + 15 = 3P + 15
$$

Solve for $ P $:

$$
3P = 345 - 15 = 330 \Rightarrow P = \frac{330}{3} = 110
$$

---

### Step 2: Verify the Result

Let’s verify that the total number of students taking at least one subject is indeed 345, using the calculated values:

- $ |C| = 180 $
- $ |P| = 110 $
- $ |Ch| = 2P = 220 $
- $ |C \cap P| = 30 $
- $ |C \cap Ch| = 75 $
- $ |P \cap Ch| = 75 $
- $ |C \cap P \cap Ch| = 15 $

Now apply the inclusion-exclusion formula again:

$$
|C \cup P \cup Ch| = 180 + 110 + 220 - 30 - 75 - 75 + 15 = 345
$$

This confirms the calculation is consistent.

---

### Final Answer

$$
\boxed{110}
$$

Solution:
Let $x$ be the number of students taking physics, so the number in chemistry is $2x$.  There are 15 students taking all three, and 30 students in both physics and calculus, meaning there are $30 - 15 = 15$ students in just physics and calculus.  Similarly there are $60$ students in just chemistry and calculus, and $60$ in physics and chemistry.  Since there are $x$ students in physics and $15 + 15 + 60 = 90$ students taking physics along with other classes, $x - 90$ students are just taking physics.  Similarly, there are $2x - 135$ students taking just chemistry and $90$ students taking just calculus.  Knowing that there are 15 students not taking any of them, the sum of these eight categories is 360, the total number of people at the school:   \[
(x - 90) + (2x - 135) + 90 + 60 + 15 + 60 + 15 + 15 = 360.
\] We solve for $x$ and find that the number of physics students is $x = \boxed{110}$.


Explanation:
Okay, so Sam has this 2-gallon bucket of milk, and he needs to take it from the barn to his house, which is 3 miles away. But there's a leak in the bucket. The problem says that for each mile he walks, there's 2/3 as much milk in the bucket as at the beginning of the mile. Hmm, I need to figure out how much milk is left when he gets home. Let me think about this step by step.

First, let me make sure I understand the problem correctly. The bucket starts with 2 gallons. For each mile he walks, the amount of milk decreases to 2/3 of what it was at the beginning of that mile. So, if he walks one mile, then after that mile, the milk is 2/3 of the original amount. Then, if he walks another mile, it's 2/3 of the amount at the start of the second mile, which would be 2/3 of the previous amount? Wait, is that how it works?

Wait, the wording says: "For each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, does that mean that after each mile, the milk is 2/3 of what it was at the beginning of that mile? So, if he starts with 2 gallons, after the first mile, it's 2*(2/3). Then, after the second mile, it's (2*(2/3))*(2/3) = 2*(2/3)^2. Then after the third mile, it's 2*(2/3)^3. So, since he walks 3 miles, the total milk left would be 2*(2/3)^3?

Wait, but let me check if that's the correct interpretation. The problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, maybe it's not that after each mile, the amount is multiplied by 2/3, but rather that during each mile, the amount of milk decreases such that at the end of the mile, it's 2/3 of what it was at the beginning of the mile. So, yes, that would mean that each mile, the amount is multiplied by 2/3. Therefore, after each mile, the amount is 2/3 of the previous amount. Therefore, after 3 miles, it would be 2*(2/3)^3.

But let me think again. Suppose he starts with 2 gallons. Let's break it down mile by mile.

First mile: starts with 2 gallons. After walking the first mile, the milk is 2/3 of the beginning of the mile. So, 2*(2/3) = 4/3 gallons.

Second mile: starts with 4/3 gallons. After walking the second mile, it's 2/3 of that, so (4/3)*(2/3) = 8/9 gallons.

Third mile: starts with 8/9 gallons. After walking the third mile, it's 2/3 of that, so (8/9)*(2/3) = 16/27 gallons.

Therefore, when he gets home, there's 16/27 gallons left. Let me check if that's correct.

Alternatively, maybe the problem is that the leak is constant, so the amount of milk decreases continuously? But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, it's not a continuous leak, but rather, after each mile, the amount is 2/3 of what it was at the beginning of that mile. So, the decrease happens at the end of each mile. Therefore, the calculation above is correct.

So, 2*(2/3)^3 = 2*(8/27) = 16/27. Which is approximately 0.59259... gallons. So, 16/27 gallons is the answer. Let me check if there's another way to interpret the problem.

Wait, another way: maybe the leak is such that for each mile walked, the milk decreases by 2/3? But that would be different. If it's decreasing by 2/3, then after the first mile, it would be 2 - (2/3)*2 = 2 - 4/3 = 2/3. But the problem says "there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that would mean that the amount is multiplied by 2/3, not that it's decreased by 2/3. So, the first interpretation is correct.

Alternatively, maybe the problem is that the rate of leakage is such that over each mile, the milk is reduced to 2/3 of the original. So, if you think of it as a geometric sequence where each term is 2/3 of the previous term. So, starting with 2 gallons, after 1 mile: 2*(2/3), after 2 miles: 2*(2/3)^2, after 3 miles: 2*(2/3)^3. Therefore, 16/27 gallons. So, that seems consistent.

Alternatively, maybe the problem is trying to trick me? Let me think again. Suppose that the leak is such that for each mile, the amount of milk is 2/3 of what it was at the beginning of the mile. So, if he walks one mile, then the amount is 2/3 of the original. Then, if he walks another mile, does the 2/3 apply to the original amount or to the current amount? The wording says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, "at the beginning of the mile" – so for each mile, the amount at the end of the mile is 2/3 of the amount at the beginning of that mile. Therefore, each mile, the amount is multiplied by 2/3. Therefore, after three miles, it's 2*(2/3)^3. So, 16/27. So, that seems correct.

Alternatively, maybe the problem is that the leak is proportional to the distance? But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that seems to indicate that the amount is multiplied by 2/3 per mile. So, the answer is 16/27.

But let me check with another approach. Let's think of it as a geometric progression. The initial amount is 2 gallons. After each mile, the amount is multiplied by 2/3. So, after 1 mile: 2*(2/3). After 2 miles: 2*(2/3)^2. After 3 miles: 2*(2/3)^3. So, 2*(8/27) = 16/27. So, that's the same as before.

Alternatively, maybe the problem is that the leak is such that the milk is decreasing continuously, and the rate is such that after each mile, it's 2/3. But that would be similar to the same result. Wait, if it's a continuous leak, then the amount would be decreasing exponentially. But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that seems to indicate that after each mile, regardless of the continuous process, the amount is 2/3 of the beginning. So, maybe it's a discrete process, not continuous. So, the answer is 16/27.

Alternatively, maybe the problem is that the leak is such that for each mile, the amount lost is 2/3 of the original amount? But that would be different. For example, if he walks one mile, he loses 2/3 of the original amount, so remaining is 2 - (2/3)*2 = 2 - 4/3 = 2/3. Then, for the second mile, he loses 2/3 of the original amount again? But that would be 2/3 of 2 gallons, so 4/3 gallons, but he only has 2/3 gallons left. That doesn't make sense. So, that interpretation is probably wrong.

Alternatively, maybe the problem is that the rate of leakage is such that the amount of milk decreases by a factor of 2/3 per mile. Which is the same as multiplying by 2/3 each mile. So, that's the same as the first interpretation.

Therefore, I think the answer is 16/27 gallons. Let me check with another way. Let me think of each mile as a step.

First mile: starts with 2 gallons. After walking 1 mile, milk is 2*(2/3) = 4/3.

Second mile: starts with 4/3 gallons. After walking 1 mile, milk is (4/3)*(2/3) = 8/9.

Third mile: starts with 8/9 gallons. After walking 1 mile, milk is (8/9)*(2/3) = 16/27.

Yes, that's the same result. So, 16/27 gallons. Therefore, the answer is 16/27. Let me check if that reduces or if there's a different way to write it. 16 and 27 have no common factors, so 16/27 is the simplified fraction.

Alternatively, maybe the problem is expecting a different answer? Let me think again. Suppose that the leak is such that for each mile, the amount of milk is 2/3 of what it was at the beginning of the mile. So, if he walks 3 miles, then the total amount is 2*(2/3)^3. Which is 16/27. So, that's the same as before.

Alternatively, maybe the problem is that the leak is proportional to the distance, so that the amount of milk lost per mile is proportional to the current amount? But that would be a continuous process, and the rate would be such that the amount decreases exponentially. However, the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that seems to indicate that after each mile, regardless of the process, the amount is 2/3 of the beginning. Therefore, it's a discrete process, not continuous. Therefore, the answer is 16/27.

Alternatively, maybe the problem is that the leak is such that the milk is decreasing by 1/3 each mile? Wait, if it's decreasing by 1/3, then after each mile, it's 2/3 of the original. So, that's the same as the problem says. So, that's the same as before.

Therefore, I think the answer is 16/27. Let me check with decimal to see if it makes sense. 2 gallons is 2.0. After first mile: 2*(2/3) ≈ 1.333... After second mile: 1.333...*(2/3) ≈ 0.888... After third mile: 0.888...*(2/3) ≈ 0.592... Which is 16/27 ≈ 0.59259... So, that seems reasonable.

Alternatively, if the problem had said that the milk is decreasing by 2/3 each mile, meaning losing 2/3 of the current amount, then it would be different. For example, if you lose 2/3 of the current amount each mile, then after first mile, you have 2 - (2/3)*2 = 2 - 4/3 = 2/3. Then, second mile, you lose 2/3 of 2/3, which is 4/9, so remaining is 2/3 - 4/9 = 2/9. Third mile, lose 2/3 of 2/9 = 4/27, remaining is 2/9 - 4/27 = 2/27. But that would be a different answer. However, the problem says "there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that wording suggests that the amount at the end of the mile is 2/3 of the beginning of the mile, not that you lose 2/3 of the current amount. Therefore, the first interpretation is correct.

Therefore, I think the answer is 16/27 gallons. Therefore, the answer is \boxed{\dfrac{16}{27}}.

**Final Answer**
The amount of milk in the bucket when Sam gets home is \boxed{\dfrac{16}{27}} gallons.

To determine how much milk remains in Sam's bucket when he reaches his house, we analyze the situation step by step.

---

### **Understanding the Problem**

Sam starts with a **2-gallon bucket of milk** and walks **3 miles** to his house. However, the bucket has a **leak** such that **after each mile**, the amount of milk in the bucket is **2/3 of what it was at the beginning of that mile**.

This means that the amount of milk **decreases multiplicatively** by a factor of $ \frac{2}{3} $ after each mile. So, the amount of milk after each mile is:

$$
\text{Milk after } n \text{ miles} = 2 \times \left( \frac{2}{3} \right)^n
$$

---

### **Step-by-Step Calculation**

Let’s compute the amount of milk after each mile:

- **After 1 mile**:
  $$
  2 \times \frac{2}{3} = \frac{4}{3}
  $$

- **After 2 miles**:
  $$
  \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}
  $$

- **After 3 miles**:
  $$
  \frac{8}{9} \times \frac{2}{3} = \frac{16}{27}
  $$

---

### **Final Answer**

Thus, when Sam reaches his house after walking 3 miles, the amount of milk remaining in the bucket is:

$$
\boxed{\dfrac{16}{27}}
$$

[Assistant]
Explanation:
Okay, so Sam has this 2-gallon bucket of milk, and he needs to take it from the barn to his house, which is 3 miles away. But there's a leak in the bucket. The problem says that for each mile he walks, there's 2/3 as much milk in the bucket as at the beginning of the mile. Hmm, I need to figure out how much milk is left when he gets home. Let me think about this step by step.

First, let me make sure I understand the problem correctly. The bucket starts with 2 gallons. For each mile he walks, the amount of milk decreases to 2/3 of what it was at the beginning of that mile. So, if he walks one mile, then after that mile, the milk is 2/3 of the original amount. Then, if he walks another mile, it's 2/3 of the amount at the start of the second mile, which would be 2/3 of the previous amount? Wait, is that how it works?

Wait, the wording says: "For each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, does that mean that after each mile, the milk is 2/3 of what it was at the beginning of that mile? So, if he starts with 2 gallons, after the first mile, it's 2*(2/3). Then, after the second mile, it's (2*(2/3))*(2/3) = 2*(2/3)^2. Then after the third mile, it's 2*(2/3)^3. So, since he walks 3 miles, the total milk left would be 2*(2/3)^3?

Wait, but let me check if that's the correct interpretation. The problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, maybe it's not that after each mile, the amount is multiplied by 2/3, but rather that during each mile, the amount of milk decreases such that at the end of the mile, it's 2/3 of what it was at the beginning of the mile. So, yes, that would mean that each mile, the amount is multiplied by 2/3. Therefore, after each mile, the amount is 2/3 of the previous amount. Therefore, after 3 miles, it would be 2*(2/3)^3.

But let me think again. Suppose he starts with 2 gallons. Let's break it down mile by mile.

First mile: starts with 2 gallons. After walking the first mile, the milk is 2/3 of the beginning of the mile. So, 2*(2/3) = 4/3 gallons.

Second mile: starts with 4/3 gallons. After walking the second mile, it's 2/3 of that, so (4/3)*(2/3) = 8/9 gallons.

Third mile: starts with 8/9 gallons. After walking the third mile, it's 2/3 of that, so (8/9)*(2/3) = 16/27 gallons.

Therefore, when he gets home, there's 16/27 gallons left. Let me check if that's correct.

Alternatively, maybe the problem is that the leak is constant, so the amount of milk decreases continuously? But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, it's not a continuous leak, but rather, after each mile, the amount is 2/3 of what it was at the beginning of that mile. So, the decrease happens at the end of each mile. Therefore, the calculation above is correct.

So, 2*(2/3)^3 = 2*(8/27) = 16/27. Which is approximately 0.59259... gallons. So, 16/27 gallons is the answer. Let me check if there's another way to interpret the problem.

Wait, another way: maybe the leak is such that for each mile walked, the milk decreases by 2/3? But that would be different. If it's decreasing by 2/3, then after the first mile, it would be 2 - (2/3)*2 = 2 - 4/3 = 2/3. But the problem says "there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that would mean that the amount is multiplied by 2/3, not that it's decreased by 2/3. So, the first interpretation is correct.

Alternatively, maybe the problem is that the rate of leakage is such that over each mile, the milk is reduced to 2/3 of the original. So, if you think of it as a geometric sequence where each term is 2/3 of the previous term. So, starting with 2 gallons, after 1 mile: 2*(2/3), after 2 miles: 2*(2/3)^2, after 3 miles: 2*(2/3)^3. Therefore, 16/27 gallons. So, that seems consistent.

Alternatively, maybe the problem is trying to trick me? Let me think again. Suppose that the leak is such that for each mile, the amount of milk is 2/3 of what it was at the beginning of the mile. So, if he walks one mile, then the amount is 2/3 of the original. Then, if he walks another mile, does the 2/3 apply to the original amount or to the current amount? The wording says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, "at the beginning of the mile" – so for each mile, the amount at the end of the mile is 2/3 of the amount at the beginning of that mile. Therefore, each mile, the amount is multiplied by 2/3. Therefore, after three miles, it's 2*(2/3)^3. So, 16/27. So, that seems correct.

Alternatively, maybe the problem is that the leak is proportional to the distance? But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that seems to indicate that the amount is multiplied by 2/3 per mile. So, the answer is 16/27.

But let me check with another approach. Let's think of it as a geometric progression. The initial amount is 2 gallons. After each mile, the amount is multiplied by 2/3. So, after 1 mile: 2*(2/3). After 2 miles: 2*(2/3)^2. After 3 miles: 2*(2/3)^3. So, 2*(8/27) = 16/27. So, that's the same as before.

Alternatively, maybe the problem is that the leak is such that the milk is decreasing continuously, and the rate is such that after each mile, it's 2/3. But that would be similar to the same result. Wait, if it's a continuous leak, then the amount would be decreasing exponentially. But the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that seems to indicate that after each mile, regardless of the continuous process, the amount is 2/3 of the beginning. So, maybe it's a discrete process, not continuous. So, the answer is 16/27.

Alternatively, maybe the problem is that the leak is such that for each mile, the amount lost is 2/3 of the original amount? But that would be different. For example, if he walks one mile, he loses 2/3 of the original amount, so remaining is 2 - (2/3)*2 = 2 - 4/3 = 2/3. Then, for the second mile, he loses 2/3 of the original amount again? But that would be 2/3 of 2 gallons, so 4/3 gallons, but he only has 2/3 gallons left. That doesn't make sense. So, that interpretation is probably wrong.

Alternatively, maybe the problem is that the rate of leakage is such that the amount of milk decreases by a factor of 2/3 per mile. Which is the same as multiplying by 2/3 each mile. So, that's the same as the first interpretation.

Therefore, I think the answer is 16/27 gallons. Let me check with another way. Let me think of each mile as a step.

First mile: starts with 2 gallons. After walking 1 mile, milk is 2*(2/3) = 4/3.

Second mile: starts with 4/3 gallons. After walking 1 mile, milk is (4/3)*(2/3) = 8/9.

Third mile: starts with 8/9 gallons. After walking 1 mile, milk is (8/9)*(2/3) = 16/27.

Yes, that's the same result. So, 16/27 gallons. Therefore, the answer is 16/27. Let me check if that reduces or if there's a different way to write it. 16 and 27 have no common factors, so 16/27 is the simplified fraction.

Alternatively, maybe the problem is expecting a different answer? Let me think again. Suppose that the leak is such that for each mile, the amount of milk is 2/3 of what it was at the beginning of the mile. So, if he walks 3 miles, then the total amount is 2*(2/3)^3. Which is 16/27. So, that's the same as before.

Alternatively, maybe the problem is that the leak is proportional to the distance, so that the amount of milk lost per mile is proportional to the current amount? But that would be a continuous process, and the rate would be such that the amount decreases exponentially. However, the problem says "for each mile he walks, there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that seems to indicate that after each mile, regardless of the process, the amount is 2/3 of the beginning. Therefore, it's a discrete process, not continuous. Therefore, the answer is 16/27.

Alternatively, maybe the problem is that the leak is such that the milk is decreasing by 1/3 each mile? Wait, if it's decreasing by 1/3, then after each mile, it's 2/3 of the original. So, that's the same as the problem says. So, that's the same as before.

Therefore, I think the answer is 16/27. Let me check with decimal to see if it makes sense. 2 gallons is 2.0. After first mile: 2*(2/3) ≈ 1.333... After second mile: 1.333...*(2/3) ≈ 0.888... After third mile: 0.888...*(2/3) ≈ 0.592... Which is 16/27 ≈ 0.59259... So, that seems reasonable.

Alternatively, if the problem had said that the milk is decreasing by 2/3 each mile, meaning losing 2/3 of the current amount, then it would be different. For example, if you lose 2/3 of the current amount each mile, then after first mile, you have 2 - (2/3)*2 = 2 - 4/3 = 2/3. Then, second mile, you lose 2/3 of 2/3, which is 4/9, so remaining is 2/3 - 4/9 = 2/9. Third mile, lose 2/3 of 2/9 = 4/27, remaining is 2/9 - 4/27 = 2/27. But that would be a different answer. However, the problem says "there is 2/3 as much milk in the bucket as at the beginning of the mile." So, that wording suggests that the amount at the end of the mile is 2/3 of the beginning of the mile, not that you lose 2/3 of the current amount. Therefore, the first interpretation is correct.

Therefore, I think the answer is 16/27 gallons. Therefore, the answer is \boxed{\dfrac{16}{27}}.

**Final Answer**
The amount of milk in the bucket when Sam gets home is \boxed{\dfrac{16}{27}} gallons.

To determine how much milk remains in Sam's bucket when he reaches his house, we analyze the situation step by step.

---

### **Understanding the Problem**

Sam starts with a **2-gallon bucket of milk** and walks **3 miles** to his house. However, the bucket has a **leak** such that **after each mile**, the amount of milk in the bucket is **2/3 of what it was at the beginning of that mile**.

This means that the amount of milk **decreases multiplicatively** by a factor of $ \frac{2}{3} $ after each mile. So, the amount of milk after each mile is:

$$
\text{Milk after } n \text{ miles} = 2 \times \left( \frac{2}{3} \right)^n
$$

---

### **Step-by-Step Calculation**

Let’s compute the amount of milk after each mile:

- **After 1 mile**:
  $$
  2 \times \frac{2}{3} = \frac{4}{3}
  $$

- **After 2 miles**:
  $$
  \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}
  $$

- **After 3 miles**:
  $$
  \frac{8}{9} \times \frac{2}{3} = \frac{16}{27}
  $$

---

### **Final Answer**

Thus, when Sam reaches his house after walking 3 miles, the amount of milk remaining in the bucket is:

$$
\boxed{\dfrac{16}{27}}
$$

Solution: At the end of the first mile, there will be $\frac{2}{3}$ of the initial milk in the bucket. Each additional mile multiplies this amount by $\frac{2}{3}$. Thus, when he arrives at home at the end of the third mile, there will be $\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \left(\frac{2}{3}\right)^{3}$ as much milk in the bucket. Since he had 2 gallons initially, the amount in the bucket when he gets home is $2 \cdot \left(\frac{2}{3}\right)^{3}$. Because $\left(\frac{a}{b}\right)^{n} = \frac{a^{n}}{b^{n}}$, this expression is equivalent to $2 \cdot \frac{2^{3}}{3^{3}}$. Because $n^{a} \cdot n^{b} = n^{a+b}$, this equals $\frac{2^{4}}{3^{3}}$. Multiplying the exponents out, we get $\boxed{\frac{16}{27}}$ gallons.


Explanation:
Okay, so I need to evaluate the sum S = 1/(1 + ω) + 1/(1 + ω²) + ... + 1/(1 + ω¹⁹⁹⁷), where ω is a 1997th root of unity, not equal to 1. Hmm, let me think about how to approach this.

First, I remember that roots of unity have some nice properties. Since ω¹⁹⁹⁷ = 1 and ω ≠ 1, ω is a primitive 1997th root of unity. But maybe I don't need to worry about primitive roots here. The key thing is that ω is a root of the equation x¹⁹⁹⁷ - 1 = 0, so it satisfies that equation. Also, the sum of all the roots of xⁿ - 1 = 0 is 0, right? Because the coefficient of xⁿ⁻¹ is 0. So, the sum of all the roots ωᵏ for k = 0 to 1996 is 0. But here, our sum is from k=1 to 1997? Wait, the original sum is from 1/(1 + ω) up to 1/(1 + ω¹⁹⁹⁷). Wait, but ω¹⁹⁹⁷ is 1, so the last term is 1/(1 + 1) = 1/2. But the problem says ω ≠ 1, so maybe the sum is from k=1 to 1996? Wait, let me check the original problem again.

The problem says: "If ω¹⁹⁹⁷ = 1 and ω ≠ 1, then evaluate [sum from 1/(1 + ω) + ... + 1/(1 + ω¹⁹⁹⁷)]." So, the exponents go from 1 to 1997. But ω¹⁹⁹⁷ is 1, so the last term is 1/(1 + 1) = 1/2. However, the original roots of unity are ω⁰, ω¹, ..., ω¹⁹⁹⁶. So ω¹⁹⁹⁷ is the same as ω⁰, which is 1. Therefore, the sum is from k=1 to 1997, but since ω¹⁹⁹⁷ = ω⁰, so the terms are 1/(1 + ω), 1/(1 + ω²), ..., 1/(1 + ω¹⁹⁹⁶), and then 1/(1 + ω¹⁹⁹⁷) = 1/(1 + 1). Therefore, the sum is over all the roots except ω⁰ = 1, but including ω⁰? Wait, no. Wait, the original sum is from k=1 to 1997. But since ω¹⁹⁹⁷ = 1, so the terms are 1/(1 + ω), 1/(1 + ω²), ..., 1/(1 + ω¹⁹⁹⁶), and then 1/(1 + ω¹⁹⁹⁷) = 1/(1 + 1). Therefore, the sum is over all the roots except ω⁰? Wait, no. The exponents go from 1 to 1997, but since ω¹⁹⁹⁷ is 1, which is ω⁰. So the sum is over all the roots except ω⁰? Wait, no. Let me think again.

If we have exponents from 1 to 1997, but since ω¹⁹⁹⁷ = 1, then the exponents 1 to 1996 are distinct roots, and then the 1997th term is 1/(1 + 1). Therefore, the sum is over all the roots except ω⁰, but including ω⁰? Wait, no. The original roots are ω⁰, ω¹, ..., ω¹⁹⁹⁶. So if we take exponents from 1 to 1997, that's equivalent to exponents from 1 to 1996 and then 1997, which is 1. Therefore, the sum is over all roots except ω⁰, but then adding 1/(1 + 1). Wait, but ω⁰ is 1, so the term 1/(1 + ω⁰) would be 1/2. But in our sum, we have 1/(1 + ω¹⁹⁹⁷) = 1/(1 + 1). Therefore, the sum S is equal to the sum from k=1 to 1996 of 1/(1 + ωᵏ) plus 1/2. But maybe there's a way to relate this sum to something else?

Alternatively, maybe I can consider pairing terms or using some symmetry. Let me recall that for roots of unity, sometimes you can use the fact that 1 + ωᵏ is related to some other expression. Also, maybe consider multiplying numerator and denominator by something?

Alternatively, perhaps consider the sum S = sum_{k=1}^{1997} 1/(1 + ωᵏ). But since ω¹⁹⁹⁷ = 1, and ω ≠ 1, so the roots are ω, ω², ..., ω¹⁹⁹⁶, and 1. Wait, but the sum is from k=1 to 1997, so that includes 1/(1 + 1) as the last term. So maybe S = sum_{k=1}^{1996} 1/(1 + ωᵏ) + 1/2. But how do I compute that?

Alternatively, maybe consider that for each root ωᵏ, we can pair it with another root? For example, if we have ω and ω^{n - k} or something? Wait, but 1997 is a prime number? Wait, 1997 is a prime? Let me check. 1997 divided by... Well, 1997 is actually a prime number. Yes, I think so. Let me confirm. Well, 1997 is a prime. So, the multiplicative group modulo 1997 is cyclic of order 1996. But maybe that's not directly helpful here.

Alternatively, maybe consider that the sum over k=1 to n-1 of 1/(1 + ωᵏ) where ω is a primitive nth root of unity. Wait, but here n is 1997, and the sum is from k=1 to n, which includes the term 1/(1 + 1). Maybe I can relate this to some known formula.

Alternatively, think about the sum S = sum_{k=1}^{n} 1/(1 + ωᵏ), where ω is a primitive nth root of unity. Wait, but in our case, n is 1997, and the sum is from k=1 to n. However, since ωⁿ = 1, so the term for k = n is 1/(1 + 1) = 1/2. So maybe I can find a general formula for this sum?

Alternatively, consider that 1/(1 + ωᵏ) can be written as (1 - ωᵏ)/(1 - ω²ᵏ). Wait, no. Wait, 1/(1 + x) = (1 - x)/(1 - x²). So if I let x = ωᵏ, then 1/(1 + ωᵏ) = (1 - ωᵏ)/(1 - ω²ᵏ). But I don't know if that helps. Maybe?

Alternatively, consider that sum_{k=1}^{n} 1/(1 + ωᵏ) = sum_{k=1}^{n} [1/(1 + ωᵏ)]. Let me think about complex analysis or residues? Maybe not necessary here.

Alternatively, consider that for each term 1/(1 + ωᵏ), if I multiply numerator and denominator by (1 - ωᵏ), then I get (1 - ωᵏ)/(1 - ω²ᵏ). But then, if I sum over k, maybe?

But maybe there's another approach. Let me think about the sum S = sum_{k=1}^{n} 1/(1 + ωᵏ), where ω is a primitive nth root of unity. Let me consider that n is odd? Since 1997 is odd. So n is odd. Let me check for small n. Let me take n = 3. Let me test with n = 3. Let ω be a primitive 3rd root of unity. Then the sum would be 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + 1). Let me compute that. Since ω² = ω⁻¹, and 1 + ω + ω² = 0. So 1 + ω = -ω², and 1 + ω² = -ω. Therefore, 1/(1 + ω) = 1/(-ω²) = -ω, and 1/(1 + ω²) = -ω. Then adding them gives -ω - ω = -2ω. Then adding 1/2 gives -2ω + 1/2. Wait, but this seems messy. Wait, but maybe I made a mistake here. Wait, let me compute again.

Wait, for n = 3, the sum is from k=1 to 3. So:

Term 1: 1/(1 + ω)

Term 2: 1/(1 + ω²)

Term 3: 1/(1 + 1) = 1/2

But since ω² = ω⁻¹, and 1 + ω + ω² = 0 => 1 + ω = -ω², so 1/(1 + ω) = 1/(-ω²) = -ω^{-2} = -ω (since ω³ = 1, so ω^{-2} = ω^{1} because -2 ≡ 1 mod 3). Similarly, 1/(1 + ω²) = -ω^{-1} = -ω². Therefore, the sum is -ω - ω² + 1/2. But since ω + ω² = -1, so -ω - ω² = 1. Therefore, total sum is 1 + 1/2 = 3/2. Wait, that's interesting. So for n = 3, the sum is 3/2. Hmm. Let me check if that's correct.

Alternatively, compute numerically. Let me take ω = e^(2πi/3) = -1/2 + i√3/2. Then 1 + ω = 1 - 1/2 + i√3/2 = 1/2 + i√3/2. The reciprocal is 1/(1/2 + i√3/2) = (1/2 - i√3/2)/[(1/2)^2 + (√3/2)^2] = (1/2 - i√3/2)/1 = 1/2 - i√3/2. Similarly, 1/(1 + ω²) would be the same as 1/(1 + ω⁻¹) = 1/(1 + conjugate(ω)) = conjugate(1/(1 + ω)) = 1/2 + i√3/2. Therefore, adding them gives (1/2 - i√3/2) + (1/2 + i√3/2) = 1. Then adding 1/2 gives 1 + 1/2 = 3/2. So yes, that works. So for n = 3, the sum is 3/2. Interesting. So maybe for general odd n, the sum is n/2? Wait, but n = 3 gives 3/2. Let me check for n = 5. Maybe?

But before that, let me think. If the answer is n/2, then for n = 1997, the answer would be 1997/2. But wait, in the case of n = 3, the sum is 3/2. However, in our problem, the sum is from k=1 to n, which includes the term 1/(1 + 1) = 1/2. But if n is odd, then the number of terms is n. However, if we consider the sum from k=1 to n-1 of 1/(1 + ωᵏ) plus 1/2. But in the case of n = 3, the sum from k=1 to 2 is 1/(1 + ω) + 1/(1 + ω²) = 1, and adding 1/2 gives 3/2. So maybe in general, the sum from k=1 to n of 1/(1 + ωᵏ) is n/2? But let me check for n = 1. Wait, n = 1 would be trivial, but ω¹ = 1, but the problem states ω ≠ 1. So n must be at least 2. Let me check n = 5.

Let me try n = 5. Let ω be a primitive 5th root of unity. Then the sum S = sum_{k=1}^5 1/(1 + ωᵏ). Let me compute this. First, note that 1 + ω + ω² + ω³ + ω⁴ = 0. Let me try to compute the sum. Maybe there's a pattern here. Alternatively, maybe use the same approach as with n = 3. Let me consider pairing terms. For example, note that ω^k and ω^{n - k} are complex conjugates. So 1/(1 + ω^k) + 1/(1 + ω^{n - k}) would be 2 * Re[1/(1 + ω^k)]. But maybe this is too vague.

Alternatively, consider that for each term 1/(1 + ω^k), multiply numerator and denominator by (1 - ω^k):

1/(1 + ω^k) = (1 - ω^k)/(1 - ω^{2k}).

But then, if I sum over k from 1 to n, then:

Sum_{k=1}^n [ (1 - ω^k) / (1 - ω^{2k}) ].

But note that ω^{2k} is another root of unity. However, since n is odd, 2k mod n is not equal to 0 unless k is multiple of n/2, but since n is odd, there is no such k. Therefore, the denominator 1 - ω^{2k} is not zero. However, I don't know if this helps. Let me think.

Alternatively, consider that the sum S = sum_{k=1}^n 1/(1 + ω^k). Let me consider multiplying both sides by something. Let me think about the sum over all roots. Let me consider that for each root ω^k, except ω^0 = 1, we have terms. But maybe consider the sum over all k=0 to n-1 of 1/(1 + ω^k). Then subtract the term for k=0, which is 1/(1 + 1) = 1/2. But in our problem, the sum is from k=1 to n, which includes k=n, which is 1/(1 + 1). However, since ω^n = 1, so the term for k=n is same as k=0. Therefore, the sum from k=1 to n is same as sum from k=0 to n-1 of 1/(1 + ω^k) + 1/(1 + 1) - 1/(1 + 1) ??? Wait, no. Wait, if we have sum from k=1 to n, that is sum from k=1 to n-1 plus 1/(1 + 1). But sum from k=0 to n-1 is 1/(1 + 1) + sum from k=1 to n-1. Therefore, sum from k=1 to n is sum from k=1 to n-1 + 1/(1 + 1) = [sum from k=0 to n-1 - 1/(1 + 1)] + 1/(1 + 1) = sum from k=0 to n-1. Therefore, the sum from k=1 to n is equal to the sum from k=0 to n-1. Therefore, S = sum_{k=0}^{n-1} 1/(1 + ω^k). Therefore, if I can compute this sum, then I can get the answer.

So, let me focus on computing sum_{k=0}^{n-1} 1/(1 + ω^k). Let me denote this sum as T. So T = sum_{k=0}^{n-1} 1/(1 + ω^k). How can I compute T?

I remember that sometimes sums over roots of unity can be evaluated by considering the derivative of the polynomial or using some other trick. Let me think. Let me consider the polynomial f(x) = xⁿ - 1, which has roots ω^k for k = 0, 1, ..., n-1. Then, perhaps I can relate the sum T to the derivative of f(x) or something else.

Alternatively, consider that 1/(1 + ω^k) can be written as 1/(1 - (-ω^k)). So, if I think of the sum as sum_{k=0}^{n-1} 1/(1 - (-ω^k)). There's a formula for sums of 1/(1 - a^k) over roots of unity? Maybe not directly. Alternatively, consider that sum_{k=0}^{n-1} 1/(1 + ω^k) = sum_{k=0}^{n-1} 1/(1 - (-ω^k)).

There is a formula for sum_{k=0}^{n-1} 1/(a - ω^k) = n / (aⁿ - 1) if a ≠ 1? Wait, no, that formula is for sum_{k=0}^{n-1} 1/(a - ω^k). Let me recall. If f(x) = xⁿ - 1, then f'(x) = n x^{n-1}, and f'(ω^k) = n ω^{k(n - 1)}. Also, the sum_{k=0}^{n-1} 1/(a - ω^k) = f'(a)/f(a). Wait, is that true?

Yes! Because if you have a polynomial f(x) with roots r_k, then sum_{k} 1/(a - r_k) = f'(a)/f(a). Let me check. For example, if f(x) = (x - r_1)(x - r_2)...(x - r_n), then f'(x) = sum_{i=1}^n product_{j≠i} (x - r_j). Therefore, f'(a)/f(a) = sum_{i=1}^n 1/(a - r_i). Yes! So that formula holds. Therefore, in our case, f(x) = xⁿ - 1, so f'(x) = n x^{n - 1}. Therefore, sum_{k=0}^{n - 1} 1/(a - ω^k) = f'(a)/f(a) = n a^{n - 1}/(aⁿ - 1). Therefore, if I set a = -1, then sum_{k=0}^{n - 1} 1/(-1 - ω^k) = n (-1)^{n - 1}/[(-1)^n - 1]. Therefore, sum_{k=0}^{n - 1} 1/(1 + ω^k) = sum_{k=0}^{n - 1} 1/(-1 - ω^k) * (-1) = - sum_{k=0}^{n - 1} 1/(-1 - ω^k) = - [n (-1)^{n - 1}/[(-1)^n - 1]].

Let me compute that. Let me write it step by step.

Let me denote a = -1. Then:

sum_{k=0}^{n - 1} 1/(a - ω^k) = f'(a)/f(a) = n a^{n - 1}/(aⁿ - 1).

Therefore, sum_{k=0}^{n - 1} 1/(a - ω^k) = n a^{n - 1}/(aⁿ - 1).

But we want sum_{k=0}^{n - 1} 1/(1 + ω^k) = sum_{k=0}^{n - 1} 1/( - (-1 - ω^k) ) = - sum_{k=0}^{n - 1} 1/( -1 - ω^k ) = - [n (-1)^{n - 1}/( (-1)^n - 1 ) ].

Therefore, simplifying:

sum_{k=0}^{n - 1} 1/(1 + ω^k) = - [n (-1)^{n - 1}/( (-1)^n - 1 ) ].

Let me compute the denominator: (-1)^n - 1. If n is odd, then (-1)^n = -1, so denominator is -1 - 1 = -2. If n is even, (-1)^n = 1, so denominator is 1 - 1 = 0. But in our problem, n = 1997, which is odd. Therefore, denominator is -2. Then numerator is - [n (-1)^{n - 1} ].

Since n is odd, n - 1 is even, so (-1)^{n - 1} = 1. Therefore, numerator is - [n * 1 ] = -n. Therefore, the entire expression is (-n)/(-2) = n/2. Therefore, sum_{k=0}^{n - 1} 1/(1 + ω^k) = n/2. Therefore, since our original sum S is equal to this sum T, which is n/2. Therefore, the answer is 1997/2. Wait, but let me check with n = 3. For n = 3, we had 3/2, which matches. For n = 5, it would be 5/2. Let me check with n = 5.

But let me think again. Wait, in our problem, the sum S is from k=1 to n, which we showed is equal to sum_{k=0}^{n - 1} 1/(1 + ω^k) = n/2. Therefore, the answer is n/2. Therefore, for n = 1997, the answer is 1997/2. But let me check with n = 3 again. For n = 3, the sum is 3/2, which is correct. Therefore, seems like the answer is n/2. Therefore, the answer is 1997/2. But let me check with another example. Let me take n = 1. Wait, n = 1, but ω ≠ 1, so n must be at least 2. Let me take n = 2. Wait, n = 2, but ω² = 1, so ω = -1. Then the sum from k=1 to 2 of 1/(1 + ω^k). For k=1: 1/(1 + (-1)) = 1/0, which is undefined. But n = 2 is even, and our formula would have denominator (-1)^n - 1 = 1 - 1 = 0, which is why we had to exclude even n? Wait, but in our problem, n is 1997, which is odd, so the formula works. For even n, the denominator becomes zero, which suggests that the formula is not valid for even n. But in our case, n is odd, so it's okay.

Therefore, the conclusion is that the sum S = n/2. Therefore, for n = 1997, the answer is 1997/2. Therefore, the answer is \boxed{\dfrac{1997}{2}}.

But let me check again with n = 3. If I compute sum_{k=0}^2 1/(1 + ω^k) = 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + (-ω) + (-ω²) = 1/2 + (- (ω + ω²)) = 1/2 + 1 = 3/2. Which is indeed 3/2. So that works. Therefore, the formula holds. Therefore, the answer is 1997/2.

But let me think again why the formula works. Because when we use the formula sum_{k=0}^{n - 1} 1/(a - ω^k) = f'(a)/f(a). Then, by setting a = -1, we get the sum involving 1/( -1 - ω^k ), and then we adjust the sign to get the sum involving 1/(1 + ω^k). Then, for odd n, we get n/2. Therefore, the answer is n/2. Therefore, the answer is 1997/2. Therefore, I think that is correct.

But let me check another way. Suppose I consider that for each term 1/(1 + ω^k), and note that 1 + ω^k = 0 when ω^k = -1. But since n is odd, -1 is not a root of xⁿ - 1 = 0, because (-1)^n = -1 ≠ 1. Therefore, none of the terms 1 + ω^k is zero, so all denominators are non-zero. Therefore, the sum is well-defined.

Alternatively, think about pairing terms. For example, for each k, pair 1/(1 + ω^k) and 1/(1 + ω^{n - k}). Let me see. Let me take n = 1997, which is odd. Let me consider k and n - k. Note that ω^{n - k} = ω^{-k} because ω^n = 1. Therefore, 1/(1 + ω^k) + 1/(1 + ω^{-k}) = [1/(1 + ω^k) + 1/(1 + ω^{-k})]. Let me compute this:

Let me write it as [ (1 + ω^{-k}) + (1 + ω^k) ] / [ (1 + ω^k)(1 + ω^{-k}) ] = [2 + ω^k + ω^{-k}]/[1 + ω^k + ω^{-k} + 1] = [2 + 2 Re(ω^k)] / [2 + 2 Re(ω^k)] = 1. Wait, that's interesting. Wait, let me check:

Wait, numerator: 1 + ω^{-k} + 1 + ω^k = 2 + ω^k + ω^{-k}

Denominator: (1 + ω^k)(1 + ω^{-k}) = 1 + ω^k + ω^{-k} + ω^k * ω^{-k} = 1 + ω^k + ω^{-k} + 1 = 2 + ω^k + ω^{-k}

Therefore, the sum of the two terms is [2 + ω^k + ω^{-k}]/[2 + ω^k + ω^{-k}] = 1. Therefore, each pair sums to 1. Therefore, if we can pair all terms except for one term, then the total sum would be (number of pairs) * 1 + remaining term. However, since n is odd, the number of terms from k=1 to n-1 is even? Wait, n is 1997, so n - 1 = 1996, which is even. Therefore, the number of terms from k=1 to n-1 is even, so we can pair them all. Therefore, the sum from k=1 to n-1 of 1/(1 + ω^k) is equal to (n - 1)/2 * 1 = (n - 1)/2. Then, adding the term for k = n, which is 1/(1 + 1) = 1/2, gives total sum S = (n - 1)/2 + 1/2 = n/2. Therefore, this confirms the result. Therefore, the answer is n/2, which for n = 1997 is 1997/2. Therefore, the answer is \boxed{\dfrac{1997}{2}}.

This method seems more straightforward. By pairing terms, we can see that each pair sums to 1, and since there are (n - 1)/2 pairs, adding the last term 1/2 gives n/2. Therefore, this confirms the result. Therefore, the answer is 1997/2.

**Final Answer**
\boxed{\dfrac{1997}{2}}

To evaluate the sum

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} + \dots + \frac{1}{1 + \omega^{1997}},
$$

we are given that $\omega^{1997} = 1$ and $\omega \neq 1$. This tells us that $\omega$ is a **primitive 1997th root of unity**, and the full set of 1997th roots of unity are $\omega^0, \omega^1, \omega^2, \dots, \omega^{1996}$, with $\omega^{1997} = \omega^0 = 1$.

---

### Step 1: Understanding the Sum

The sum in question is:

$$
S = \sum_{k=1}^{1997} \frac{1}{1 + \omega^k}.
$$

Note that $\omega^{1997} = 1$, so the last term is:

$$
\frac{1}{1 + \omega^{1997}} = \frac{1}{1 + 1} = \frac{1}{2}.
$$

Thus, the sum can be rewritten as:

$$
S = \sum_{k=1}^{1996} \frac{1}{1 + \omega^k} + \frac{1}{2}.
$$

---

### Step 2: Pairing Terms

We now consider the sum over all $k = 1$ to $1996$, and pair each term $\frac{1}{1 + \omega^k}$ with $\frac{1}{1 + \omega^{-k}}$, where $\omega^{-k} = \omega^{1997 - k}$ (since $\omega^{1997} = 1$).

Let’s compute the sum of a pair:

$$
\frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{-k}} = \frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{1997 - k}}.
$$

Let’s simplify this expression:

$$
\frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{-k}} = \frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{-k}}.
$$

Multiply numerator and denominator of the second term by $\omega^k$:

$$
\frac{1}{1 + \omega^k} + \frac{\omega^k}{\omega^k + 1} = \frac{1 + \omega^k}{1 + \omega^k} = 1.
$$

So, **each pair of terms sums to 1**.

---

### Step 3: Counting the Pairs

There are $1996$ terms from $k = 1$ to $1996$, and since $1996$ is even, we can pair them all up. That gives us:

$$
\frac{1996}{2} = 998 \text{ pairs}.
$$

Each pair contributes 1, so the sum of the first $1996$ terms is:

$$
998 \times 1 = 998.
$$

Adding the last term $\frac{1}{2}$, we get:

$$
S = 998 + \frac{1}{2} = \frac{1996}{2} + \frac{1}{2} = \frac{1997}{2}.
$$

---

### Final Answer

$$
\boxed{\dfrac{1997}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to evaluate the sum S = 1/(1 + ω) + 1/(1 + ω²) + ... + 1/(1 + ω¹⁹⁹⁷), where ω is a 1997th root of unity, not equal to 1. Hmm, let me think about how to approach this.

First, I remember that roots of unity have some nice properties. Since ω¹⁹⁹⁷ = 1 and ω ≠ 1, ω is a primitive 1997th root of unity. But maybe I don't need to worry about primitive roots here. The key thing is that ω is a root of the equation x¹⁹⁹⁷ - 1 = 0, so it satisfies that equation. Also, the sum of all the roots of xⁿ - 1 = 0 is 0, right? Because the coefficient of xⁿ⁻¹ is 0. So, the sum of all the roots ωᵏ for k = 0 to 1996 is 0. But here, our sum is from k=1 to 1997? Wait, the original sum is from 1/(1 + ω) up to 1/(1 + ω¹⁹⁹⁷). Wait, but ω¹⁹⁹⁷ is 1, so the last term is 1/(1 + 1) = 1/2. But the problem says ω ≠ 1, so maybe the sum is from k=1 to 1996? Wait, let me check the original problem again.

The problem says: "If ω¹⁹⁹⁷ = 1 and ω ≠ 1, then evaluate [sum from 1/(1 + ω) + ... + 1/(1 + ω¹⁹⁹⁷)]." So, the exponents go from 1 to 1997. But ω¹⁹⁹⁷ is 1, so the last term is 1/(1 + 1) = 1/2. However, the original roots of unity are ω⁰, ω¹, ..., ω¹⁹⁹⁶. So ω¹⁹⁹⁷ is the same as ω⁰, which is 1. Therefore, the sum is from k=1 to 1997, but since ω¹⁹⁹⁷ = ω⁰, so the terms are 1/(1 + ω), 1/(1 + ω²), ..., 1/(1 + ω¹⁹⁹⁶), and then 1/(1 + ω¹⁹⁹⁷) = 1/(1 + 1). Therefore, the sum is over all the roots except ω⁰ = 1, but including ω⁰? Wait, no. Wait, the original sum is from k=1 to 1997. But since ω¹⁹⁹⁷ = 1, so the terms are 1/(1 + ω), 1/(1 + ω²), ..., 1/(1 + ω¹⁹⁹⁶), and then 1/(1 + ω¹⁹⁹⁷) = 1/(1 + 1). Therefore, the sum is over all the roots except ω⁰? Wait, no. The exponents go from 1 to 1997, but since ω¹⁹⁹⁷ is 1, which is ω⁰. So the sum is over all the roots except ω⁰? Wait, no. Let me think again.

If we have exponents from 1 to 1997, but since ω¹⁹⁹⁷ = 1, then the exponents 1 to 1996 are distinct roots, and then the 1997th term is 1/(1 + 1). Therefore, the sum is over all the roots except ω⁰, but including ω⁰? Wait, no. The original roots are ω⁰, ω¹, ..., ω¹⁹⁹⁶. So if we take exponents from 1 to 1997, that's equivalent to exponents from 1 to 1996 and then 1997, which is 1. Therefore, the sum is over all roots except ω⁰, but then adding 1/(1 + 1). Wait, but ω⁰ is 1, so the term 1/(1 + ω⁰) would be 1/2. But in our sum, we have 1/(1 + ω¹⁹⁹⁷) = 1/(1 + 1). Therefore, the sum S is equal to the sum from k=1 to 1996 of 1/(1 + ωᵏ) plus 1/2. But maybe there's a way to relate this sum to something else?

Alternatively, maybe I can consider pairing terms or using some symmetry. Let me recall that for roots of unity, sometimes you can use the fact that 1 + ωᵏ is related to some other expression. Also, maybe consider multiplying numerator and denominator by something?

Alternatively, perhaps consider the sum S = sum_{k=1}^{1997} 1/(1 + ωᵏ). But since ω¹⁹⁹⁷ = 1, and ω ≠ 1, so the roots are ω, ω², ..., ω¹⁹⁹⁶, and 1. Wait, but the sum is from k=1 to 1997, so that includes 1/(1 + 1) as the last term. So maybe S = sum_{k=1}^{1996} 1/(1 + ωᵏ) + 1/2. But how do I compute that?

Alternatively, maybe consider that for each root ωᵏ, we can pair it with another root? For example, if we have ω and ω^{n - k} or something? Wait, but 1997 is a prime number? Wait, 1997 is a prime? Let me check. 1997 divided by... Well, 1997 is actually a prime number. Yes, I think so. Let me confirm. Well, 1997 is a prime. So, the multiplicative group modulo 1997 is cyclic of order 1996. But maybe that's not directly helpful here.

Alternatively, maybe consider that the sum over k=1 to n-1 of 1/(1 + ωᵏ) where ω is a primitive nth root of unity. Wait, but here n is 1997, and the sum is from k=1 to n, which includes the term 1/(1 + 1). Maybe I can relate this to some known formula.

Alternatively, think about the sum S = sum_{k=1}^{n} 1/(1 + ωᵏ), where ω is a primitive nth root of unity. Wait, but in our case, n is 1997, and the sum is from k=1 to n. However, since ωⁿ = 1, so the term for k = n is 1/(1 + 1) = 1/2. So maybe I can find a general formula for this sum?

Alternatively, consider that 1/(1 + ωᵏ) can be written as (1 - ωᵏ)/(1 - ω²ᵏ). Wait, no. Wait, 1/(1 + x) = (1 - x)/(1 - x²). So if I let x = ωᵏ, then 1/(1 + ωᵏ) = (1 - ωᵏ)/(1 - ω²ᵏ). But I don't know if that helps. Maybe?

Alternatively, consider that sum_{k=1}^{n} 1/(1 + ωᵏ) = sum_{k=1}^{n} [1/(1 + ωᵏ)]. Let me think about complex analysis or residues? Maybe not necessary here.

Alternatively, consider that for each term 1/(1 + ωᵏ), if I multiply numerator and denominator by (1 - ωᵏ), then I get (1 - ωᵏ)/(1 - ω²ᵏ). But then, if I sum over k, maybe?

But maybe there's another approach. Let me think about the sum S = sum_{k=1}^{n} 1/(1 + ωᵏ), where ω is a primitive nth root of unity. Let me consider that n is odd? Since 1997 is odd. So n is odd. Let me check for small n. Let me take n = 3. Let me test with n = 3. Let ω be a primitive 3rd root of unity. Then the sum would be 1/(1 + ω) + 1/(1 + ω²) + 1/(1 + 1). Let me compute that. Since ω² = ω⁻¹, and 1 + ω + ω² = 0. So 1 + ω = -ω², and 1 + ω² = -ω. Therefore, 1/(1 + ω) = 1/(-ω²) = -ω, and 1/(1 + ω²) = -ω. Then adding them gives -ω - ω = -2ω. Then adding 1/2 gives -2ω + 1/2. Wait, but this seems messy. Wait, but maybe I made a mistake here. Wait, let me compute again.

Wait, for n = 3, the sum is from k=1 to 3. So:

Term 1: 1/(1 + ω)

Term 2: 1/(1 + ω²)

Term 3: 1/(1 + 1) = 1/2

But since ω² = ω⁻¹, and 1 + ω + ω² = 0 => 1 + ω = -ω², so 1/(1 + ω) = 1/(-ω²) = -ω^{-2} = -ω (since ω³ = 1, so ω^{-2} = ω^{1} because -2 ≡ 1 mod 3). Similarly, 1/(1 + ω²) = -ω^{-1} = -ω². Therefore, the sum is -ω - ω² + 1/2. But since ω + ω² = -1, so -ω - ω² = 1. Therefore, total sum is 1 + 1/2 = 3/2. Wait, that's interesting. So for n = 3, the sum is 3/2. Hmm. Let me check if that's correct.

Alternatively, compute numerically. Let me take ω = e^(2πi/3) = -1/2 + i√3/2. Then 1 + ω = 1 - 1/2 + i√3/2 = 1/2 + i√3/2. The reciprocal is 1/(1/2 + i√3/2) = (1/2 - i√3/2)/[(1/2)^2 + (√3/2)^2] = (1/2 - i√3/2)/1 = 1/2 - i√3/2. Similarly, 1/(1 + ω²) would be the same as 1/(1 + ω⁻¹) = 1/(1 + conjugate(ω)) = conjugate(1/(1 + ω)) = 1/2 + i√3/2. Therefore, adding them gives (1/2 - i√3/2) + (1/2 + i√3/2) = 1. Then adding 1/2 gives 1 + 1/2 = 3/2. So yes, that works. So for n = 3, the sum is 3/2. Interesting. So maybe for general odd n, the sum is n/2? Wait, but n = 3 gives 3/2. Let me check for n = 5. Maybe?

But before that, let me think. If the answer is n/2, then for n = 1997, the answer would be 1997/2. But wait, in the case of n = 3, the sum is 3/2. However, in our problem, the sum is from k=1 to n, which includes the term 1/(1 + 1) = 1/2. But if n is odd, then the number of terms is n. However, if we consider the sum from k=1 to n-1 of 1/(1 + ωᵏ) plus 1/2. But in the case of n = 3, the sum from k=1 to 2 is 1/(1 + ω) + 1/(1 + ω²) = 1, and adding 1/2 gives 3/2. So maybe in general, the sum from k=1 to n of 1/(1 + ωᵏ) is n/2? But let me check for n = 1. Wait, n = 1 would be trivial, but ω¹ = 1, but the problem states ω ≠ 1. So n must be at least 2. Let me check n = 5.

Let me try n = 5. Let ω be a primitive 5th root of unity. Then the sum S = sum_{k=1}^5 1/(1 + ωᵏ). Let me compute this. First, note that 1 + ω + ω² + ω³ + ω⁴ = 0. Let me try to compute the sum. Maybe there's a pattern here. Alternatively, maybe use the same approach as with n = 3. Let me consider pairing terms. For example, note that ω^k and ω^{n - k} are complex conjugates. So 1/(1 + ω^k) + 1/(1 + ω^{n - k}) would be 2 * Re[1/(1 + ω^k)]. But maybe this is too vague.

Alternatively, consider that for each term 1/(1 + ω^k), multiply numerator and denominator by (1 - ω^k):

1/(1 + ω^k) = (1 - ω^k)/(1 - ω^{2k}).

But then, if I sum over k from 1 to n, then:

Sum_{k=1}^n [ (1 - ω^k) / (1 - ω^{2k}) ].

But note that ω^{2k} is another root of unity. However, since n is odd, 2k mod n is not equal to 0 unless k is multiple of n/2, but since n is odd, there is no such k. Therefore, the denominator 1 - ω^{2k} is not zero. However, I don't know if this helps. Let me think.

Alternatively, consider that the sum S = sum_{k=1}^n 1/(1 + ω^k). Let me consider multiplying both sides by something. Let me think about the sum over all roots. Let me consider that for each root ω^k, except ω^0 = 1, we have terms. But maybe consider the sum over all k=0 to n-1 of 1/(1 + ω^k). Then subtract the term for k=0, which is 1/(1 + 1) = 1/2. But in our problem, the sum is from k=1 to n, which includes k=n, which is 1/(1 + 1). However, since ω^n = 1, so the term for k=n is same as k=0. Therefore, the sum from k=1 to n is same as sum from k=0 to n-1 of 1/(1 + ω^k) + 1/(1 + 1) - 1/(1 + 1) ??? Wait, no. Wait, if we have sum from k=1 to n, that is sum from k=1 to n-1 plus 1/(1 + 1). But sum from k=0 to n-1 is 1/(1 + 1) + sum from k=1 to n-1. Therefore, sum from k=1 to n is sum from k=1 to n-1 + 1/(1 + 1) = [sum from k=0 to n-1 - 1/(1 + 1)] + 1/(1 + 1) = sum from k=0 to n-1. Therefore, the sum from k=1 to n is equal to the sum from k=0 to n-1. Therefore, S = sum_{k=0}^{n-1} 1/(1 + ω^k). Therefore, if I can compute this sum, then I can get the answer.

So, let me focus on computing sum_{k=0}^{n-1} 1/(1 + ω^k). Let me denote this sum as T. So T = sum_{k=0}^{n-1} 1/(1 + ω^k). How can I compute T?

I remember that sometimes sums over roots of unity can be evaluated by considering the derivative of the polynomial or using some other trick. Let me think. Let me consider the polynomial f(x) = xⁿ - 1, which has roots ω^k for k = 0, 1, ..., n-1. Then, perhaps I can relate the sum T to the derivative of f(x) or something else.

Alternatively, consider that 1/(1 + ω^k) can be written as 1/(1 - (-ω^k)). So, if I think of the sum as sum_{k=0}^{n-1} 1/(1 - (-ω^k)). There's a formula for sums of 1/(1 - a^k) over roots of unity? Maybe not directly. Alternatively, consider that sum_{k=0}^{n-1} 1/(1 + ω^k) = sum_{k=0}^{n-1} 1/(1 - (-ω^k)).

There is a formula for sum_{k=0}^{n-1} 1/(a - ω^k) = n / (aⁿ - 1) if a ≠ 1? Wait, no, that formula is for sum_{k=0}^{n-1} 1/(a - ω^k). Let me recall. If f(x) = xⁿ - 1, then f'(x) = n x^{n-1}, and f'(ω^k) = n ω^{k(n - 1)}. Also, the sum_{k=0}^{n-1} 1/(a - ω^k) = f'(a)/f(a). Wait, is that true?

Yes! Because if you have a polynomial f(x) with roots r_k, then sum_{k} 1/(a - r_k) = f'(a)/f(a). Let me check. For example, if f(x) = (x - r_1)(x - r_2)...(x - r_n), then f'(x) = sum_{i=1}^n product_{j≠i} (x - r_j). Therefore, f'(a)/f(a) = sum_{i=1}^n 1/(a - r_i). Yes! So that formula holds. Therefore, in our case, f(x) = xⁿ - 1, so f'(x) = n x^{n - 1}. Therefore, sum_{k=0}^{n - 1} 1/(a - ω^k) = f'(a)/f(a) = n a^{n - 1}/(aⁿ - 1). Therefore, if I set a = -1, then sum_{k=0}^{n - 1} 1/(-1 - ω^k) = n (-1)^{n - 1}/[(-1)^n - 1]. Therefore, sum_{k=0}^{n - 1} 1/(1 + ω^k) = sum_{k=0}^{n - 1} 1/(-1 - ω^k) * (-1) = - sum_{k=0}^{n - 1} 1/(-1 - ω^k) = - [n (-1)^{n - 1}/[(-1)^n - 1]].

Let me compute that. Let me write it step by step.

Let me denote a = -1. Then:

sum_{k=0}^{n - 1} 1/(a - ω^k) = f'(a)/f(a) = n a^{n - 1}/(aⁿ - 1).

Therefore, sum_{k=0}^{n - 1} 1/(a - ω^k) = n a^{n - 1}/(aⁿ - 1).

But we want sum_{k=0}^{n - 1} 1/(1 + ω^k) = sum_{k=0}^{n - 1} 1/( - (-1 - ω^k) ) = - sum_{k=0}^{n - 1} 1/( -1 - ω^k ) = - [n (-1)^{n - 1}/( (-1)^n - 1 ) ].

Therefore, simplifying:

sum_{k=0}^{n - 1} 1/(1 + ω^k) = - [n (-1)^{n - 1}/( (-1)^n - 1 ) ].

Let me compute the denominator: (-1)^n - 1. If n is odd, then (-1)^n = -1, so denominator is -1 - 1 = -2. If n is even, (-1)^n = 1, so denominator is 1 - 1 = 0. But in our problem, n = 1997, which is odd. Therefore, denominator is -2. Then numerator is - [n (-1)^{n - 1} ].

Since n is odd, n - 1 is even, so (-1)^{n - 1} = 1. Therefore, numerator is - [n * 1 ] = -n. Therefore, the entire expression is (-n)/(-2) = n/2. Therefore, sum_{k=0}^{n - 1} 1/(1 + ω^k) = n/2. Therefore, since our original sum S is equal to this sum T, which is n/2. Therefore, the answer is 1997/2. Wait, but let me check with n = 3. For n = 3, we had 3/2, which matches. For n = 5, it would be 5/2. Let me check with n = 5.

But let me think again. Wait, in our problem, the sum S is from k=1 to n, which we showed is equal to sum_{k=0}^{n - 1} 1/(1 + ω^k) = n/2. Therefore, the answer is n/2. Therefore, for n = 1997, the answer is 1997/2. But let me check with n = 3 again. For n = 3, the sum is 3/2, which is correct. Therefore, seems like the answer is n/2. Therefore, the answer is 1997/2. But let me check with another example. Let me take n = 1. Wait, n = 1, but ω ≠ 1, so n must be at least 2. Let me take n = 2. Wait, n = 2, but ω² = 1, so ω = -1. Then the sum from k=1 to 2 of 1/(1 + ω^k). For k=1: 1/(1 + (-1)) = 1/0, which is undefined. But n = 2 is even, and our formula would have denominator (-1)^n - 1 = 1 - 1 = 0, which is why we had to exclude even n? Wait, but in our problem, n is 1997, which is odd, so the formula works. For even n, the denominator becomes zero, which suggests that the formula is not valid for even n. But in our case, n is odd, so it's okay.

Therefore, the conclusion is that the sum S = n/2. Therefore, for n = 1997, the answer is 1997/2. Therefore, the answer is \boxed{\dfrac{1997}{2}}.

But let me check again with n = 3. If I compute sum_{k=0}^2 1/(1 + ω^k) = 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + (-ω) + (-ω²) = 1/2 + (- (ω + ω²)) = 1/2 + 1 = 3/2. Which is indeed 3/2. So that works. Therefore, the formula holds. Therefore, the answer is 1997/2.

But let me think again why the formula works. Because when we use the formula sum_{k=0}^{n - 1} 1/(a - ω^k) = f'(a)/f(a). Then, by setting a = -1, we get the sum involving 1/( -1 - ω^k ), and then we adjust the sign to get the sum involving 1/(1 + ω^k). Then, for odd n, we get n/2. Therefore, the answer is n/2. Therefore, the answer is 1997/2. Therefore, I think that is correct.

But let me check another way. Suppose I consider that for each term 1/(1 + ω^k), and note that 1 + ω^k = 0 when ω^k = -1. But since n is odd, -1 is not a root of xⁿ - 1 = 0, because (-1)^n = -1 ≠ 1. Therefore, none of the terms 1 + ω^k is zero, so all denominators are non-zero. Therefore, the sum is well-defined.

Alternatively, think about pairing terms. For example, for each k, pair 1/(1 + ω^k) and 1/(1 + ω^{n - k}). Let me see. Let me take n = 1997, which is odd. Let me consider k and n - k. Note that ω^{n - k} = ω^{-k} because ω^n = 1. Therefore, 1/(1 + ω^k) + 1/(1 + ω^{-k}) = [1/(1 + ω^k) + 1/(1 + ω^{-k})]. Let me compute this:

Let me write it as [ (1 + ω^{-k}) + (1 + ω^k) ] / [ (1 + ω^k)(1 + ω^{-k}) ] = [2 + ω^k + ω^{-k}]/[1 + ω^k + ω^{-k} + 1] = [2 + 2 Re(ω^k)] / [2 + 2 Re(ω^k)] = 1. Wait, that's interesting. Wait, let me check:

Wait, numerator: 1 + ω^{-k} + 1 + ω^k = 2 + ω^k + ω^{-k}

Denominator: (1 + ω^k)(1 + ω^{-k}) = 1 + ω^k + ω^{-k} + ω^k * ω^{-k} = 1 + ω^k + ω^{-k} + 1 = 2 + ω^k + ω^{-k}

Therefore, the sum of the two terms is [2 + ω^k + ω^{-k}]/[2 + ω^k + ω^{-k}] = 1. Therefore, each pair sums to 1. Therefore, if we can pair all terms except for one term, then the total sum would be (number of pairs) * 1 + remaining term. However, since n is odd, the number of terms from k=1 to n-1 is even? Wait, n is 1997, so n - 1 = 1996, which is even. Therefore, the number of terms from k=1 to n-1 is even, so we can pair them all. Therefore, the sum from k=1 to n-1 of 1/(1 + ω^k) is equal to (n - 1)/2 * 1 = (n - 1)/2. Then, adding the term for k = n, which is 1/(1 + 1) = 1/2, gives total sum S = (n - 1)/2 + 1/2 = n/2. Therefore, this confirms the result. Therefore, the answer is n/2, which for n = 1997 is 1997/2. Therefore, the answer is \boxed{\dfrac{1997}{2}}.

This method seems more straightforward. By pairing terms, we can see that each pair sums to 1, and since there are (n - 1)/2 pairs, adding the last term 1/2 gives n/2. Therefore, this confirms the result. Therefore, the answer is 1997/2.

**Final Answer**
\boxed{\dfrac{1997}{2}}

To evaluate the sum

$$
\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} + \dots + \frac{1}{1 + \omega^{1997}},
$$

we are given that $\omega^{1997} = 1$ and $\omega \neq 1$. This tells us that $\omega$ is a **primitive 1997th root of unity**, and the full set of 1997th roots of unity are $\omega^0, \omega^1, \omega^2, \dots, \omega^{1996}$, with $\omega^{1997} = \omega^0 = 1$.

---

### Step 1: Understanding the Sum

The sum in question is:

$$
S = \sum_{k=1}^{1997} \frac{1}{1 + \omega^k}.
$$

Note that $\omega^{1997} = 1$, so the last term is:

$$
\frac{1}{1 + \omega^{1997}} = \frac{1}{1 + 1} = \frac{1}{2}.
$$

Thus, the sum can be rewritten as:

$$
S = \sum_{k=1}^{1996} \frac{1}{1 + \omega^k} + \frac{1}{2}.
$$

---

### Step 2: Pairing Terms

We now consider the sum over all $k = 1$ to $1996$, and pair each term $\frac{1}{1 + \omega^k}$ with $\frac{1}{1 + \omega^{-k}}$, where $\omega^{-k} = \omega^{1997 - k}$ (since $\omega^{1997} = 1$).

Let’s compute the sum of a pair:

$$
\frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{-k}} = \frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{1997 - k}}.
$$

Let’s simplify this expression:

$$
\frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{-k}} = \frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{-k}}.
$$

Multiply numerator and denominator of the second term by $\omega^k$:

$$
\frac{1}{1 + \omega^k} + \frac{\omega^k}{\omega^k + 1} = \frac{1 + \omega^k}{1 + \omega^k} = 1.
$$

So, **each pair of terms sums to 1**.

---

### Step 3: Counting the Pairs

There are $1996$ terms from $k = 1$ to $1996$, and since $1996$ is even, we can pair them all up. That gives us:

$$
\frac{1996}{2} = 998 \text{ pairs}.
$$

Each pair contributes 1, so the sum of the first $1996$ terms is:

$$
998 \times 1 = 998.
$$

Adding the last term $\frac{1}{2}$, we get:

$$
S = 998 + \frac{1}{2} = \frac{1996}{2} + \frac{1}{2} = \frac{1997}{2}.
$$

---

### Final Answer

$$
\boxed{\dfrac{1997}{2}}
$$

Solution:
Note that
\begin{align*}
\frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{1997 - k}} &= \frac{1}{1 + \omega^k} + \frac{\omega^k}{\omega^k + \omega^{1997}} \\
&= \frac{1}{1 + \omega^k} + \frac{\omega^k}{\omega^k + 1} \\
&= \frac{1 + \omega^k}{1 + \omega^k} = 1.
\end{align*}Thus, we can pair the terms
\[\frac{1}{1 + \omega}, \ \frac{1}{1 + \omega^2}, \ \dots, \ \frac{1}{1 + \omega^{1995}}, \ \frac{1}{1 + \omega^{1996}}\]into $1996/2 = 998$ pairs, so that the sum of the numbers in each pair is 1.  Also, $\frac{1}{1 + \omega^{1997}} = \frac{1}{2},$ so the sum works out to $998 + \frac{1}{2} = \boxed{\frac{1997}{2}}.$