Explanation:
Okay, let me try to figure out this sequence problem. So, the sequence starts with 0, and then each subsequent term is alternately adding and multiplying by each successive integer starting from 1. Let me try to write down the terms step by step to understand the pattern.

First term: 0.

Then the second term is produced by adding 1 to the first term. So, 0 + 1 = 1. That makes sense. The third term is produced by multiplying the second term by 1. Wait, so after adding, we multiply? Let me check. The problem says "alternately adding and multiplying by each successive integer beginning with 1." So, starting from the first term, the operations alternate between adding and multiplying? Or is it that each term is generated by either adding or multiplying, alternating between the two operations, and each time using the next integer?

Let me parse the problem again. "Subsequent terms are produced by alternately adding and multiplying by each successive integer beginning with 1." For instance, the second term is produced by adding 1 to the first term; the third term is produced by multiplying the second term by 1; the fourth term is produced by adding 2 to the third term; and so on.

So, operation alternates between add and multiply, starting with adding. The numbers to add or multiply by are successive integers starting with 1. So:

Term 1: 0

Term 2: Term1 + 1 = 0 + 1 = 1

Term3: Term2 * 1 = 1 * 1 = 1

Term4: Term3 + 2 = 1 + 2 = 3

Term5: Term4 * 2 = 3 * 2 = 6

Term6: Term5 + 3 = 6 + 3 = 9

Term7: Term6 * 3 = 9 * 3 = 27

Term8: Term7 + 4 = 27 + 4 = 31

Term9: Term8 * 4 = 31 * 4 = 124

Term10: Term9 + 5 = 124 + 5 = 129

Term11: Term10 * 5 = 129 * 5 = 645

Wait, so according to this, the terms go 0, 1, 1, 3, 6, 9, 27, 31, 124, 129, 645,...

So, the question is asking for the first term that is greater than 125. Looking at the terms I just calculated, term 10 is 129, which is greater than 125. But let me check if I did all the steps correctly.

Let me write them down again step by step to make sure I haven't made an error.

Term 1: 0

Term 2: Term1 + 1 = 0 + 1 = 1

Term 3: Term2 * 1 = 1 * 1 = 1

Term 4: Term3 + 2 = 1 + 2 = 3

Term 5: Term4 * 2 = 3 * 2 = 6

Term 6: Term5 + 3 = 6 + 3 = 9

Term 7: Term6 * 3 = 9 * 3 = 27

Term 8: Term7 + 4 = 27 + 4 = 31

Term 9: Term8 * 4 = 31 * 4 = 124

Term 10: Term9 + 5 = 124 + 5 = 129

Term 11: Term10 * 5 = 129 * 5 = 645

Term 12: Term11 + 6 = 645 + 6 = 651

Term 13: Term12 * 6 = 651 * 6 = 3906

So, according to this, term 10 is 129, which is the first term greater than 125. However, let me check whether the operations are indeed alternating starting with add, then multiply, then add, then multiply, etc. So the operations for term 2 is add, term3 multiply, term4 add, term5 multiply, term6 add, term7 multiply, term8 add, term9 multiply, term10 add, term11 multiply, term12 add, term13 multiply, etc. So yes, the operations alternate between add and multiply, starting with add for term2. So for term n, where n ≥ 2, if n is even, is it add or multiply? Let me check.

Term2: add 1 (operation is add, since starting with add)

Term3: multiply by 1

Term4: add 2

Term5: multiply by 2

Term6: add 3

Term7: multiply by 3

Term8: add 4

Term9: multiply by 4

Term10: add 5

Term11: multiply by 5

So, for term number k, the operation depends on whether k is even or odd? Wait, term2 is add, term3 is multiply, term4 is add, term5 is multiply... So for term numbers starting from 2:

If term number is even (term2, term4, term6, ...), then the operation is add?

Wait term2 is add, term4 is add, term6 is add, term8 is add, term10 is add, etc. So for term numbers that are even (starting from term2), it's adding. Then term numbers that are odd (term3, term5, term7, etc.) are multiplying. But note that the number to add or multiply by is the integer starting from 1, increasing by 1 each time. So for term2, we add 1; term3, multiply by 1; term4, add 2; term5, multiply by 2; term6, add 3; term7, multiply by 3; term8, add 4; term9, multiply by 4; term10, add 5; term11, multiply by 5; term12, add 6; etc. So the number to add or multiply by is (n-1)/2 when n is odd? Wait, maybe not. Let's think.

Looking at term2: add 1, term3: multiply by 1, term4: add 2, term5: multiply by 2, term6: add 3, term7: multiply by 3, term8: add 4, term9: multiply by 4, term10: add 5, term11: multiply by 5, term12: add 6, term13: multiply by 6,...

So for term number k ≥ 2, if k is even, then we add (k/2) to the previous term. If k is odd, then we multiply the previous term by ( (k-1)/2 ). Wait let me check:

For term2: k=2, even, so add 1. k/2 = 1. Correct.

term3: k=3, odd, multiply by (3-1)/2 = 1. Correct.

term4: k=4, even, add 4/2=2. Correct.

term5: k=5, odd, multiply by (5-1)/2=2. Correct.

term6: k=6, even, add 6/2=3. Correct. Yes, so generalizing:

For term k (starting from k=2):

If k is even, then term_k = term_{k-1} + (k/2)

If k is odd, then term_k = term_{k-1} * ((k-1)/2)

But let's check that with term3:

k=3, odd, term3 = term2 * ((3-1)/2) = term2 * 1 = 1*1=1. Correct.

term5: k=5, odd, term5 = term4 * ((5-1)/2)= term4 * 2 = 3*2=6. Correct.

term7: term6 * 3 = 9*3=27. Correct. So that formula seems correct.

Therefore, the sequence can be generated by the following rules:

Term1 = 0

For k ≥ 2:

If k is even, term_k = term_{k-1} + (k/2)

If k is odd, term_k = term_{k-1} * ((k-1)/2)

Alternatively, another way to think is that for each operation, starting with adding 1, then multiplying by 1, then adding 2, multiplying by 2, adding 3, multiplying by 3, etc. So alternately adding and multiplying, with the number increasing by 1 each time. So the operation alternates between add and multiply, and the number to add/multiply alternates between 1, 1, 2, 2, 3, 3, 4, 4,...

Therefore, the sequence is generated as follows:

Start with 0.

Then for each step, alternate between adding and multiplying, with the number to add/multiply being 1, then 1, then 2, then 2, then 3, then 3, etc. So for step 1 (to get term2), add 1; step2 (to get term3), multiply by 1; step3 (to get term4), add 2; step4 (to get term5), multiply by 2; step5 (to get term6), add 3; step6 (to get term7), multiply by 3; and so on. So each pair of steps (add and multiply) uses the same number, increasing by 1 each pair. So that's another way to think about it.

So given that, let me make sure that my previous calculations were correct.

Term1: 0

Term2: Term1 + 1 = 0 + 1 = 1

Term3: Term2 * 1 = 1 * 1 = 1

Term4: Term3 + 2 = 1 + 2 = 3

Term5: Term4 * 2 = 3 * 2 = 6

Term6: Term5 + 3 = 6 + 3 = 9

Term7: Term6 * 3 = 9 * 3 = 27

Term8: Term7 + 4 = 27 + 4 = 31

Term9: Term8 * 4 = 31 * 4 = 124

Term10: Term9 + 5 = 124 + 5 = 129

Term11: Term10 * 5 = 129 * 5 = 645

Term12: Term11 + 6 = 645 + 6 = 651

Term13: Term12 * 6 = 651 * 6 = 3906

Term14: Term13 + 7 = 3906 + 7 = 3913

Term15: Term14 * 7 = 3913 * 7 = 27391

So, the terms are: 0, 1, 1, 3, 6, 9, 27, 31, 124, 129, 645, 651, 3906, 3913, 27391,...

So looking for the first term greater than 125. Term10 is 129, which is greater than 125. Term9 is 124, which is less than 125. Therefore, term10 is the first term greater than 125, so the answer is 129?

Wait, but let me check again. Maybe I miscalculated some terms. Let me verify term by term again:

Term1: 0

Term2: 0 + 1 = 1

Term3: 1 * 1 = 1

Term4: 1 + 2 = 3

Term5: 3 * 2 = 6

Term6: 6 + 3 = 9

Term7: 9 * 3 = 27

Term8: 27 + 4 = 31

Term9: 31 * 4 = 124

Term10: 124 + 5 = 129

Yes, that's correct. Term10 is 129. Term9 is 124. So 129 is the first term exceeding 125. Therefore, the answer is 129. However, let me check if the problem says "the first term that is greater than 125", so 129 is the first such term. So maybe the answer is 129.

Wait, but maybe I need to check the problem statement again. The problem says "the first term that is greater than 125". So 129 is greater than 125, and the previous term was 124, which is less than 125. Therefore, term10 is 129. But let me check the problem statement once again to make sure that I didn't misinterpret the operation.

Original problem statement:

"In the sequence 0, 1, 1, 3, 6, 9, 27, ..., the first term is 0. Subsequent terms are produced by alternately adding and multiplying by each successive integer beginning with 1. For instance, the second term is produced by adding 1 to the first term; the third term is produced by multiplying the second term by 1; the fourth term is produced by adding 2 to the third term; and so on. What is the value of the first term that is greater than 125?"

Yes, that seems to match with my calculations. Let me check if there is a term between term9 and term10 that could be greater than 125? But term9 is 124, term10 is 129. So no. So term10 is 129. Therefore, the answer is 129?

But wait, let me check if I have the correct operations for term numbers. Let me think again. Maybe I miscounted the term numbers? Let me list them again with term numbers:

Term1: 0

Term2: 0 + 1 = 1

Term3: 1 * 1 = 1

Term4: 1 + 2 = 3

Term5: 3 * 2 = 6

Term6: 6 + 3 = 9

Term7: 9 * 3 = 27

Term8: 27 + 4 = 31

Term9: 31 * 4 = 124

Term10: 124 + 5 = 129

Yes, so term numbers 1 through 10 are as above. So term10 is 129. Therefore, the first term greater than 125 is term10, which is 129.

But wait, let me check term 11: 129 * 5 = 645. So term11 is 645. Term10 is 129, which is the first term exceeding 125. Therefore, the answer is 129.

But wait, maybe there's a mistake here. Let me check if the operations alternate starting with adding, then multiplying, then adding, etc. So starting from term1, the operations to get term2 is add, term3 is multiply, term4 add, term5 multiply, term6 add, term7 multiply, term8 add, term9 multiply, term10 add, term11 multiply, etc. Therefore, for term numbers:

Term2: add 1

Term3: multiply by 1

Term4: add 2

Term5: multiply by 2

Term6: add 3

Term7: multiply by 3

Term8: add 4

Term9: multiply by 4

Term10: add 5

Term11: multiply by 5

Term12: add 6

So, each time, the operation alternates, and the number to add/multiply is increasing by 1 every two terms. So, first add 1, multiply by 1, add 2, multiply by 2, add 3, multiply by 3, etc. Therefore, the numbers to add/multiply are 1,1,2,2,3,3,4,4,5,5,6,6,...

So that seems to be consistent with the operations.

So, given that, term10 is adding 5 to term9, which was 124, so 124 + 5 = 129. Term11 is multiplying term10 by 5, so 129 * 5 = 645. Therefore, term10 is 129, which is the first term exceeding 125. Therefore, the answer should be 129.

But wait, let me check if maybe there is a term before term10 that is greater than 125? Let's check term7: 27, term8: 31, term9: 124, term10: 129. So term9 is 124, which is less than 125, term10 is 129, which is greater. So yes, 129 is the first term greater than 125.

But just to make sure, let me re-express the sequence:

Term1: 0

Term2: 0 + 1 = 1

Term3: 1 × 1 = 1

Term4: 1 + 2 = 3

Term5: 3 × 2 = 6

Term6: 6 + 3 = 9

Term7: 9 × 3 = 27

Term8: 27 + 4 = 31

Term9: 31 × 4 = 124

Term10: 124 + 5 = 129

Term11: 129 × 5 = 645

Term12: 645 + 6 = 651

So, yes, term10 is 129. Therefore, the answer is 129. Wait, but maybe the problem counts the first term greater than 125 as term10, but maybe I need to check if there is a term between term9 and term10? But no, there are no terms in between. Each term is generated one after another. So term9 is 124, term10 is 129, so term10 is the first term after 124 that is greater than 125.

Therefore, I think the answer is 129. However, let me check once again, maybe my initial calculations were wrong. Let me verify term8 and term9 again.

Term7: 27

Term8: Term7 + 4 = 27 + 4 = 31. Correct.

Term9: Term8 × 4 = 31 × 4 = 124. Correct.

Term10: Term9 + 5 = 124 + 5 = 129. Correct.

Yes, that's right. So term10 is 129, which is greater than 125. Therefore, the answer is 129. However, I feel like maybe the problem expects a different answer, so let me check if I misread the question.

Wait, the problem says "the first term that is greater than 125." So 129 is greater than 125, and there is no term between term9 and term10. So, yes, 129 is the answer.

But let me check if there is a different interpretation of the problem. For example, maybe the operations are alternately adding and multiplying, but using the successive integers starting from 1, meaning that for term2, you add 1; term3, you multiply by 2; term4, you add 3; term5, you multiply by 4; etc. But that would be different.

Wait, the problem says: "alternately adding and multiplying by each successive integer beginning with 1."

The example says:

The second term is produced by adding 1 to the first term; the third term is produced by multiplying the second term by 1; the fourth term is produced by adding 2 to the third term; and so on.

Therefore, the numbers to add/multiply are 1, 1, 2, 2, 3, 3,... So for term2: add 1; term3: multiply by 1; term4: add 2; term5: multiply by 2; term6: add 3; term7: multiply by 3; etc. So the numbers are 1, 1, 2, 2, 3, 3, 4, 4,... So my original interpretation was correct.

Therefore, term10 is 129. So the answer is 129.

But let me check another way. Maybe I can model this sequence with a formula or find a pattern.

Looking at the sequence:

Term1: 0

Term2: 1

Term3: 1

Term4: 3

Term5: 6

Term6: 9

Term7: 27

Term8: 31

Term9: 124

Term10: 129

Term11: 645

Term12: 651

Term13: 3906

Term14: 3913

Term15: 27391

So looking at this, after term7, which is 27, term8 is adding 4, term9 is multiplying by 4, term10 is adding 5, term11 multiplying by 5, etc.

So, if we consider that every pair of operations (add and multiply) with the same number, like for number 1: add 1 then multiply by 1; number 2: add 2 then multiply by 2; etc. However, starting from term2, which is adding 1, then term3 is multiplying by 1. Then term4 is adding 2, term5 is multiplying by 2. So for each number n ≥1, we have an add operation with n and then a multiply operation with n. Except for n=1, since the first add is with n=1, and the first multiply is with n=1. Then for n=2, add 2 and multiply by 2, etc.

So, if we think of it as for n ≥1, we do:

Add n to the previous term, then multiply by n. Wait, but in the sequence, term2 is adding 1, term3 is multiplying by 1. Then term4 is adding 2, term5 is multiplying by 2. So for each n, the operations are add n and multiply by n, but for n=1, that's term2 and term3, for n=2, term4 and term5, etc. So the sequence is built by pairs of operations. So if I group them:

Group 1 (n=1): term2 (add 1), term3 (multiply by 1)

Group 2 (n=2): term4 (add 2), term5 (multiply by 2)

Group 3 (n=3): term6 (add 3), term7 (multiply by 3)

Group 4 (n=4): term8 (add 4), term9 (multiply by 4)

Group 5 (n=5): term10 (add 5), term11 (multiply by 5)

Group 6 (n=6): term12 (add 6), term13 (multiply by 6)

and so on.

So each group corresponds to a value of n, starting at 1, and for each group, we perform an add and a multiply operation. So starting from term1 = 0, then:

Group 1 (n=1):

term2 = term1 + 1 = 0 + 1 = 1

term3 = term2 * 1 = 1 * 1 = 1

Group 2 (n=2):

term4 = term3 + 2 = 1 + 2 = 3

term5 = term4 * 2 = 3 * 2 = 6

Group 3 (n=3):

term6 = term5 + 3 = 6 + 3 = 9

term7 = term6 * 3 = 9 * 3 = 27

Group 4 (n=4):

term8 = term7 + 4 = 27 + 4 = 31

term9 = term8 * 4 = 31 * 4 = 124

Group 5 (n=5):

term10 = term9 + 5 = 124 + 5 = 129

term11 = term10 * 5 = 129 * 5 = 645

Group 6 (n=6):

term12 = term11 + 6 = 645 + 6 = 651

term13 = term12 * 6 = 651 * 6 = 3906

...

So, in this grouping, each group n (starting at n=1) performs an add and a multiply. So, for group n, the terms are:

term_{2n} = term_{2n-1} + n

term_{2n+1} = term_{2n} * n

Wait, let me check with n=1:

term2 = term1 + 1 = 0 + 1 = 1 (which is term2)

term3 = term2 * 1 = 1 * 1 = 1 (term3)

Similarly, for n=2:

term4 = term3 + 2 = 1 + 2 = 3 (term4)

term5 = term4 * 2 = 3 * 2 = 6 (term5)

Yes. So for each n ≥ 1, term_{2n} is term_{2n - 1} + n, and term_{2n + 1} is term_{2n} * n. Therefore, this seems to be a way to model it.

Therefore, if we think in groups, each group n consists of two terms:

term_{2n} = term_{2n - 1} + n

term_{2n + 1} = term_{2n} * n

But how can we express term_{2n} and term_{2n +1} in terms of previous terms? It might not be straightforward, but maybe we can find a pattern or formula.

Alternatively, since we need only to find the first term greater than 125, which is term10 (129), and term10 is the result of adding 5 to term9 (which is 124). Term9 is from multiplying term8 by 4. Term8 is adding 4 to term7 (which is 27). Term7 is multiplying term6 by 3, term6 is adding 3 to term5 (6), term5 is multiplying term4 by 2, term4 is adding 2 to term3 (1), term3 is multiplying term2 by 1, term2 is adding 1 to term1 (0). So the sequence is built up step by step. So as calculated earlier, term10 is 129. So unless there's a miscalculation here, which I don't see, 129 should be the correct answer.

Alternatively, maybe the problem is expecting a different answer if there's a different interpretation. Let me check again.

Another possible misinterpretation: Maybe the operations are alternately adding and multiplying, but the numbers to add/multiply are 1, 2, 3, 4,... So for term2: add 1, term3: multiply by 2, term4: add 3, term5: multiply by 4, etc. But according to the problem statement, the example says:

"the second term is produced by adding 1 to the first term; the third term is produced by multiplying the second term by 1; the fourth term is produced by adding 2 to the third term; and so on."

Therefore, the numbers are 1, 1, 2, 2, 3, 3,... So the example explicitly says that term3 is multiplied by 1, which matches with the numbers increasing every two operations, not every operation.

Therefore, the correct interpretation is that for each pair of operations (add and multiply), the number increases by 1. Therefore, my previous calculations hold.

Therefore, I think my answer is correct. The first term greater than 125 is 129, which is term10.

But to be absolutely thorough, let me check if maybe the question is from a source where the answer is different. However, since I don't have external resources, I need to rely on my own calculations.

Alternatively, maybe I made a mistake in the term numbers? Let me check:

Term1: 0

Term2: 1

Term3: 1

Term4: 3

Term5: 6

Term6: 9

Term7: 27

Term8: 31

Term9: 124

Term10: 129

Yes, that's 10 terms. So term10 is the 10th term. Therefore, unless there's a confusion between zero-based or one-based indexing, but the problem says "the first term is 0", so term1 is 0, term2 is 1, etc. So term10 is the 10th term. Therefore, 129 is the answer.

Therefore, I think the answer is 129, so the value of the first term greater than 125 is \boxed{129}.

**Final Answer**
\boxed{129}

To determine the first term in the given sequence that is greater than 125, we must first analyze the rule that governs the sequence generation.

---

### **Sequence Rules and Pattern**

The sequence starts with:

- **Term 1:** 0

Subsequent terms are generated **alternately by adding and multiplying** each successive integer, starting from **1**. The pattern of operations and numbers is consistent as follows:

- **Term 2:** Add 1 → Term1 + 1 = 0 + 1 = **1**
- **Term 3:** Multiply by 1 → Term2 × 1 = 1 × 1 = **1**
- **Term 4:** Add 2 → Term3 + 2 = 1 + 2 = **3**
- **Term 5:** Multiply by 2 → Term4 × 2 = 3 × 2 = **6**
- **Term 6:** Add 3 → Term5 + 3 = 6 + 3 = **9**
- **Term 7:** Multiply by 3 → Term6 × 3 = 9 × 3 = **27**
- **Term 8:** Add 4 → Term7 + 4 = 27 + 4 = **31**
- **Term 9:** Multiply by 4 → Term8 × 4 = 31 × 4 = **124**
- **Term 10:** Add 5 → Term9 + 5 = 124 + 5 = **129**
- **Term 11:** Multiply by 5 → Term10 × 5 = 129 × 5 = **645**
- **Term 12:** Add 6 → Term11 + 6 = 645 + 6 = **651**
- **Term 13:** Multiply by 6 → Term12 × 6 = 651 × 6 = **3906**

---

### **Identifying the First Term Greater than 125**

From the above calculations:

- **Term 9:** 124 (**less than 125**)
- **Term 10:** 129 (**greater than 125**)

So, the **first term in the sequence that is greater than 125** is at **Term 10**, and its value is:

$$
\boxed{129}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this sequence problem. So, the sequence starts with 0, and then each subsequent term is alternately adding and multiplying by each successive integer starting from 1. Let me try to write down the terms step by step to understand the pattern.

First term: 0.

Then the second term is produced by adding 1 to the first term. So, 0 + 1 = 1. That makes sense. The third term is produced by multiplying the second term by 1. Wait, so after adding, we multiply? Let me check. The problem says "alternately adding and multiplying by each successive integer beginning with 1." So, starting from the first term, the operations alternate between adding and multiplying? Or is it that each term is generated by either adding or multiplying, alternating between the two operations, and each time using the next integer?

Let me parse the problem again. "Subsequent terms are produced by alternately adding and multiplying by each successive integer beginning with 1." For instance, the second term is produced by adding 1 to the first term; the third term is produced by multiplying the second term by 1; the fourth term is produced by adding 2 to the third term; and so on.

So, operation alternates between add and multiply, starting with adding. The numbers to add or multiply by are successive integers starting with 1. So:

Term 1: 0

Term 2: Term1 + 1 = 0 + 1 = 1

Term3: Term2 * 1 = 1 * 1 = 1

Term4: Term3 + 2 = 1 + 2 = 3

Term5: Term4 * 2 = 3 * 2 = 6

Term6: Term5 + 3 = 6 + 3 = 9

Term7: Term6 * 3 = 9 * 3 = 27

Term8: Term7 + 4 = 27 + 4 = 31

Term9: Term8 * 4 = 31 * 4 = 124

Term10: Term9 + 5 = 124 + 5 = 129

Term11: Term10 * 5 = 129 * 5 = 645

Wait, so according to this, the terms go 0, 1, 1, 3, 6, 9, 27, 31, 124, 129, 645,...

So, the question is asking for the first term that is greater than 125. Looking at the terms I just calculated, term 10 is 129, which is greater than 125. But let me check if I did all the steps correctly.

Let me write them down again step by step to make sure I haven't made an error.

Term 1: 0

Term 2: Term1 + 1 = 0 + 1 = 1

Term 3: Term2 * 1 = 1 * 1 = 1

Term 4: Term3 + 2 = 1 + 2 = 3

Term 5: Term4 * 2 = 3 * 2 = 6

Term 6: Term5 + 3 = 6 + 3 = 9

Term 7: Term6 * 3 = 9 * 3 = 27

Term 8: Term7 + 4 = 27 + 4 = 31

Term 9: Term8 * 4 = 31 * 4 = 124

Term 10: Term9 + 5 = 124 + 5 = 129

Term 11: Term10 * 5 = 129 * 5 = 645

Term 12: Term11 + 6 = 645 + 6 = 651

Term 13: Term12 * 6 = 651 * 6 = 3906

So, according to this, term 10 is 129, which is the first term greater than 125. However, let me check whether the operations are indeed alternating starting with add, then multiply, then add, then multiply, etc. So the operations for term 2 is add, term3 multiply, term4 add, term5 multiply, term6 add, term7 multiply, term8 add, term9 multiply, term10 add, term11 multiply, term12 add, term13 multiply, etc. So yes, the operations alternate between add and multiply, starting with add for term2. So for term n, where n ≥ 2, if n is even, is it add or multiply? Let me check.

Term2: add 1 (operation is add, since starting with add)

Term3: multiply by 1

Term4: add 2

Term5: multiply by 2

Term6: add 3

Term7: multiply by 3

Term8: add 4

Term9: multiply by 4

Term10: add 5

Term11: multiply by 5

So, for term number k, the operation depends on whether k is even or odd? Wait, term2 is add, term3 is multiply, term4 is add, term5 is multiply... So for term numbers starting from 2:

If term number is even (term2, term4, term6, ...), then the operation is add?

Wait term2 is add, term4 is add, term6 is add, term8 is add, term10 is add, etc. So for term numbers that are even (starting from term2), it's adding. Then term numbers that are odd (term3, term5, term7, etc.) are multiplying. But note that the number to add or multiply by is the integer starting from 1, increasing by 1 each time. So for term2, we add 1; term3, multiply by 1; term4, add 2; term5, multiply by 2; term6, add 3; term7, multiply by 3; term8, add 4; term9, multiply by 4; term10, add 5; term11, multiply by 5; term12, add 6; etc. So the number to add or multiply by is (n-1)/2 when n is odd? Wait, maybe not. Let's think.

Looking at term2: add 1, term3: multiply by 1, term4: add 2, term5: multiply by 2, term6: add 3, term7: multiply by 3, term8: add 4, term9: multiply by 4, term10: add 5, term11: multiply by 5, term12: add 6, term13: multiply by 6,...

So for term number k ≥ 2, if k is even, then we add (k/2) to the previous term. If k is odd, then we multiply the previous term by ( (k-1)/2 ). Wait let me check:

For term2: k=2, even, so add 1. k/2 = 1. Correct.

term3: k=3, odd, multiply by (3-1)/2 = 1. Correct.

term4: k=4, even, add 4/2=2. Correct.

term5: k=5, odd, multiply by (5-1)/2=2. Correct.

term6: k=6, even, add 6/2=3. Correct. Yes, so generalizing:

For term k (starting from k=2):

If k is even, then term_k = term_{k-1} + (k/2)

If k is odd, then term_k = term_{k-1} * ((k-1)/2)

But let's check that with term3:

k=3, odd, term3 = term2 * ((3-1)/2) = term2 * 1 = 1*1=1. Correct.

term5: k=5, odd, term5 = term4 * ((5-1)/2)= term4 * 2 = 3*2=6. Correct.

term7: term6 * 3 = 9*3=27. Correct. So that formula seems correct.

Therefore, the sequence can be generated by the following rules:

Term1 = 0

For k ≥ 2:

If k is even, term_k = term_{k-1} + (k/2)

If k is odd, term_k = term_{k-1} * ((k-1)/2)

Alternatively, another way to think is that for each operation, starting with adding 1, then multiplying by 1, then adding 2, multiplying by 2, adding 3, multiplying by 3, etc. So alternately adding and multiplying, with the number increasing by 1 each time. So the operation alternates between add and multiply, and the number to add/multiply alternates between 1, 1, 2, 2, 3, 3, 4, 4,...

Therefore, the sequence is generated as follows:

Start with 0.

Then for each step, alternate between adding and multiplying, with the number to add/multiply being 1, then 1, then 2, then 2, then 3, then 3, etc. So for step 1 (to get term2), add 1; step2 (to get term3), multiply by 1; step3 (to get term4), add 2; step4 (to get term5), multiply by 2; step5 (to get term6), add 3; step6 (to get term7), multiply by 3; and so on. So each pair of steps (add and multiply) uses the same number, increasing by 1 each pair. So that's another way to think about it.

So given that, let me make sure that my previous calculations were correct.

Term1: 0

Term2: Term1 + 1 = 0 + 1 = 1

Term3: Term2 * 1 = 1 * 1 = 1

Term4: Term3 + 2 = 1 + 2 = 3

Term5: Term4 * 2 = 3 * 2 = 6

Term6: Term5 + 3 = 6 + 3 = 9

Term7: Term6 * 3 = 9 * 3 = 27

Term8: Term7 + 4 = 27 + 4 = 31

Term9: Term8 * 4 = 31 * 4 = 124

Term10: Term9 + 5 = 124 + 5 = 129

Term11: Term10 * 5 = 129 * 5 = 645

Term12: Term11 + 6 = 645 + 6 = 651

Term13: Term12 * 6 = 651 * 6 = 3906

Term14: Term13 + 7 = 3906 + 7 = 3913

Term15: Term14 * 7 = 3913 * 7 = 27391

So, the terms are: 0, 1, 1, 3, 6, 9, 27, 31, 124, 129, 645, 651, 3906, 3913, 27391,...

So looking for the first term greater than 125. Term10 is 129, which is greater than 125. Term9 is 124, which is less than 125. Therefore, term10 is the first term greater than 125, so the answer is 129?

Wait, but let me check again. Maybe I miscalculated some terms. Let me verify term by term again:

Term1: 0

Term2: 0 + 1 = 1

Term3: 1 * 1 = 1

Term4: 1 + 2 = 3

Term5: 3 * 2 = 6

Term6: 6 + 3 = 9

Term7: 9 * 3 = 27

Term8: 27 + 4 = 31

Term9: 31 * 4 = 124

Term10: 124 + 5 = 129

Yes, that's correct. Term10 is 129. Term9 is 124. So 129 is the first term exceeding 125. Therefore, the answer is 129. However, let me check if the problem says "the first term that is greater than 125", so 129 is the first such term. So maybe the answer is 129.

Wait, but maybe I need to check the problem statement again. The problem says "the first term that is greater than 125". So 129 is greater than 125, and the previous term was 124, which is less than 125. Therefore, term10 is 129. But let me check the problem statement once again to make sure that I didn't misinterpret the operation.

Original problem statement:

"In the sequence 0, 1, 1, 3, 6, 9, 27, ..., the first term is 0. Subsequent terms are produced by alternately adding and multiplying by each successive integer beginning with 1. For instance, the second term is produced by adding 1 to the first term; the third term is produced by multiplying the second term by 1; the fourth term is produced by adding 2 to the third term; and so on. What is the value of the first term that is greater than 125?"

Yes, that seems to match with my calculations. Let me check if there is a term between term9 and term10 that could be greater than 125? But term9 is 124, term10 is 129. So no. So term10 is 129. Therefore, the answer is 129?

But wait, let me check if I have the correct operations for term numbers. Let me think again. Maybe I miscounted the term numbers? Let me list them again with term numbers:

Term1: 0

Term2: 0 + 1 = 1

Term3: 1 * 1 = 1

Term4: 1 + 2 = 3

Term5: 3 * 2 = 6

Term6: 6 + 3 = 9

Term7: 9 * 3 = 27

Term8: 27 + 4 = 31

Term9: 31 * 4 = 124

Term10: 124 + 5 = 129

Yes, so term numbers 1 through 10 are as above. So term10 is 129. Therefore, the first term greater than 125 is term10, which is 129.

But wait, let me check term 11: 129 * 5 = 645. So term11 is 645. Term10 is 129, which is the first term exceeding 125. Therefore, the answer is 129.

But wait, maybe there's a mistake here. Let me check if the operations alternate starting with adding, then multiplying, then adding, etc. So starting from term1, the operations to get term2 is add, term3 is multiply, term4 add, term5 multiply, term6 add, term7 multiply, term8 add, term9 multiply, term10 add, term11 multiply, etc. Therefore, for term numbers:

Term2: add 1

Term3: multiply by 1

Term4: add 2

Term5: multiply by 2

Term6: add 3

Term7: multiply by 3

Term8: add 4

Term9: multiply by 4

Term10: add 5

Term11: multiply by 5

Term12: add 6

So, each time, the operation alternates, and the number to add/multiply is increasing by 1 every two terms. So, first add 1, multiply by 1, add 2, multiply by 2, add 3, multiply by 3, etc. Therefore, the numbers to add/multiply are 1,1,2,2,3,3,4,4,5,5,6,6,...

So that seems to be consistent with the operations.

So, given that, term10 is adding 5 to term9, which was 124, so 124 + 5 = 129. Term11 is multiplying term10 by 5, so 129 * 5 = 645. Therefore, term10 is 129, which is the first term exceeding 125. Therefore, the answer should be 129.

But wait, let me check if maybe there is a term before term10 that is greater than 125? Let's check term7: 27, term8: 31, term9: 124, term10: 129. So term9 is 124, which is less than 125, term10 is 129, which is greater. So yes, 129 is the first term greater than 125.

But just to make sure, let me re-express the sequence:

Term1: 0

Term2: 0 + 1 = 1

Term3: 1 × 1 = 1

Term4: 1 + 2 = 3

Term5: 3 × 2 = 6

Term6: 6 + 3 = 9

Term7: 9 × 3 = 27

Term8: 27 + 4 = 31

Term9: 31 × 4 = 124

Term10: 124 + 5 = 129

Term11: 129 × 5 = 645

Term12: 645 + 6 = 651

So, yes, term10 is 129. Therefore, the answer is 129. Wait, but maybe the problem counts the first term greater than 125 as term10, but maybe I need to check if there is a term between term9 and term10? But no, there are no terms in between. Each term is generated one after another. So term9 is 124, term10 is 129, so term10 is the first term after 124 that is greater than 125.

Therefore, I think the answer is 129. However, let me check once again, maybe my initial calculations were wrong. Let me verify term8 and term9 again.

Term7: 27

Term8: Term7 + 4 = 27 + 4 = 31. Correct.

Term9: Term8 × 4 = 31 × 4 = 124. Correct.

Term10: Term9 + 5 = 124 + 5 = 129. Correct.

Yes, that's right. So term10 is 129, which is greater than 125. Therefore, the answer is 129. However, I feel like maybe the problem expects a different answer, so let me check if I misread the question.

Wait, the problem says "the first term that is greater than 125." So 129 is greater than 125, and there is no term between term9 and term10. So, yes, 129 is the answer.

But let me check if there is a different interpretation of the problem. For example, maybe the operations are alternately adding and multiplying, but using the successive integers starting from 1, meaning that for term2, you add 1; term3, you multiply by 2; term4, you add 3; term5, you multiply by 4; etc. But that would be different.

Wait, the problem says: "alternately adding and multiplying by each successive integer beginning with 1."

The example says:

The second term is produced by adding 1 to the first term; the third term is produced by multiplying the second term by 1; the fourth term is produced by adding 2 to the third term; and so on.

Therefore, the numbers to add/multiply are 1, 1, 2, 2, 3, 3,... So for term2: add 1; term3: multiply by 1; term4: add 2; term5: multiply by 2; term6: add 3; term7: multiply by 3; etc. So the numbers are 1, 1, 2, 2, 3, 3, 4, 4,... So my original interpretation was correct.

Therefore, term10 is 129. So the answer is 129.

But let me check another way. Maybe I can model this sequence with a formula or find a pattern.

Looking at the sequence:

Term1: 0

Term2: 1

Term3: 1

Term4: 3

Term5: 6

Term6: 9

Term7: 27

Term8: 31

Term9: 124

Term10: 129

Term11: 645

Term12: 651

Term13: 3906

Term14: 3913

Term15: 27391

So looking at this, after term7, which is 27, term8 is adding 4, term9 is multiplying by 4, term10 is adding 5, term11 multiplying by 5, etc.

So, if we consider that every pair of operations (add and multiply) with the same number, like for number 1: add 1 then multiply by 1; number 2: add 2 then multiply by 2; etc. However, starting from term2, which is adding 1, then term3 is multiplying by 1. Then term4 is adding 2, term5 is multiplying by 2. So for each number n ≥1, we have an add operation with n and then a multiply operation with n. Except for n=1, since the first add is with n=1, and the first multiply is with n=1. Then for n=2, add 2 and multiply by 2, etc.

So, if we think of it as for n ≥1, we do:

Add n to the previous term, then multiply by n. Wait, but in the sequence, term2 is adding 1, term3 is multiplying by 1. Then term4 is adding 2, term5 is multiplying by 2. So for each n, the operations are add n and multiply by n, but for n=1, that's term2 and term3, for n=2, term4 and term5, etc. So the sequence is built by pairs of operations. So if I group them:

Group 1 (n=1): term2 (add 1), term3 (multiply by 1)

Group 2 (n=2): term4 (add 2), term5 (multiply by 2)

Group 3 (n=3): term6 (add 3), term7 (multiply by 3)

Group 4 (n=4): term8 (add 4), term9 (multiply by 4)

Group 5 (n=5): term10 (add 5), term11 (multiply by 5)

Group 6 (n=6): term12 (add 6), term13 (multiply by 6)

and so on.

So each group corresponds to a value of n, starting at 1, and for each group, we perform an add and a multiply operation. So starting from term1 = 0, then:

Group 1 (n=1):

term2 = term1 + 1 = 0 + 1 = 1

term3 = term2 * 1 = 1 * 1 = 1

Group 2 (n=2):

term4 = term3 + 2 = 1 + 2 = 3

term5 = term4 * 2 = 3 * 2 = 6

Group 3 (n=3):

term6 = term5 + 3 = 6 + 3 = 9

term7 = term6 * 3 = 9 * 3 = 27

Group 4 (n=4):

term8 = term7 + 4 = 27 + 4 = 31

term9 = term8 * 4 = 31 * 4 = 124

Group 5 (n=5):

term10 = term9 + 5 = 124 + 5 = 129

term11 = term10 * 5 = 129 * 5 = 645

Group 6 (n=6):

term12 = term11 + 6 = 645 + 6 = 651

term13 = term12 * 6 = 651 * 6 = 3906

...

So, in this grouping, each group n (starting at n=1) performs an add and a multiply. So, for group n, the terms are:

term_{2n} = term_{2n-1} + n

term_{2n+1} = term_{2n} * n

Wait, let me check with n=1:

term2 = term1 + 1 = 0 + 1 = 1 (which is term2)

term3 = term2 * 1 = 1 * 1 = 1 (term3)

Similarly, for n=2:

term4 = term3 + 2 = 1 + 2 = 3 (term4)

term5 = term4 * 2 = 3 * 2 = 6 (term5)

Yes. So for each n ≥ 1, term_{2n} is term_{2n - 1} + n, and term_{2n + 1} is term_{2n} * n. Therefore, this seems to be a way to model it.

Therefore, if we think in groups, each group n consists of two terms:

term_{2n} = term_{2n - 1} + n

term_{2n + 1} = term_{2n} * n

But how can we express term_{2n} and term_{2n +1} in terms of previous terms? It might not be straightforward, but maybe we can find a pattern or formula.

Alternatively, since we need only to find the first term greater than 125, which is term10 (129), and term10 is the result of adding 5 to term9 (which is 124). Term9 is from multiplying term8 by 4. Term8 is adding 4 to term7 (which is 27). Term7 is multiplying term6 by 3, term6 is adding 3 to term5 (6), term5 is multiplying term4 by 2, term4 is adding 2 to term3 (1), term3 is multiplying term2 by 1, term2 is adding 1 to term1 (0). So the sequence is built up step by step. So as calculated earlier, term10 is 129. So unless there's a miscalculation here, which I don't see, 129 should be the correct answer.

Alternatively, maybe the problem is expecting a different answer if there's a different interpretation. Let me check again.

Another possible misinterpretation: Maybe the operations are alternately adding and multiplying, but the numbers to add/multiply are 1, 2, 3, 4,... So for term2: add 1, term3: multiply by 2, term4: add 3, term5: multiply by 4, etc. But according to the problem statement, the example says:

"the second term is produced by adding 1 to the first term; the third term is produced by multiplying the second term by 1; the fourth term is produced by adding 2 to the third term; and so on."

Therefore, the numbers are 1, 1, 2, 2, 3, 3,... So the example explicitly says that term3 is multiplied by 1, which matches with the numbers increasing every two operations, not every operation.

Therefore, the correct interpretation is that for each pair of operations (add and multiply), the number increases by 1. Therefore, my previous calculations hold.

Therefore, I think my answer is correct. The first term greater than 125 is 129, which is term10.

But to be absolutely thorough, let me check if maybe the question is from a source where the answer is different. However, since I don't have external resources, I need to rely on my own calculations.

Alternatively, maybe I made a mistake in the term numbers? Let me check:

Term1: 0

Term2: 1

Term3: 1

Term4: 3

Term5: 6

Term6: 9

Term7: 27

Term8: 31

Term9: 124

Term10: 129

Yes, that's 10 terms. So term10 is the 10th term. Therefore, unless there's a confusion between zero-based or one-based indexing, but the problem says "the first term is 0", so term1 is 0, term2 is 1, etc. So term10 is the 10th term. Therefore, 129 is the answer.

Therefore, I think the answer is 129, so the value of the first term greater than 125 is \boxed{129}.

**Final Answer**
\boxed{129}

To determine the first term in the given sequence that is greater than 125, we must first analyze the rule that governs the sequence generation.

---

### **Sequence Rules and Pattern**

The sequence starts with:

- **Term 1:** 0

Subsequent terms are generated **alternately by adding and multiplying** each successive integer, starting from **1**. The pattern of operations and numbers is consistent as follows:

- **Term 2:** Add 1 → Term1 + 1 = 0 + 1 = **1**
- **Term 3:** Multiply by 1 → Term2 × 1 = 1 × 1 = **1**
- **Term 4:** Add 2 → Term3 + 2 = 1 + 2 = **3**
- **Term 5:** Multiply by 2 → Term4 × 2 = 3 × 2 = **6**
- **Term 6:** Add 3 → Term5 + 3 = 6 + 3 = **9**
- **Term 7:** Multiply by 3 → Term6 × 3 = 9 × 3 = **27**
- **Term 8:** Add 4 → Term7 + 4 = 27 + 4 = **31**
- **Term 9:** Multiply by 4 → Term8 × 4 = 31 × 4 = **124**
- **Term 10:** Add 5 → Term9 + 5 = 124 + 5 = **129**
- **Term 11:** Multiply by 5 → Term10 × 5 = 129 × 5 = **645**
- **Term 12:** Add 6 → Term11 + 6 = 645 + 6 = **651**
- **Term 13:** Multiply by 6 → Term12 × 6 = 651 × 6 = **3906**

---

### **Identifying the First Term Greater than 125**

From the above calculations:

- **Term 9:** 124 (**less than 125**)
- **Term 10:** 129 (**greater than 125**)

So, the **first term in the sequence that is greater than 125** is at **Term 10**, and its value is:

$$
\boxed{129}
$$

Solution: Continuing this sequence from the 27, we add four to make 31, then multiply 31 by four to make 124, then add five to 124 to make 129. Thus, $\boxed{129}$ is the first term that is greater than 125.


Explanation:
Okay, let's try to tackle this problem step by step. So, Steve is thinking of a polynomial with all positive integer roots, and the polynomial is given in the form P(x) = 2x³ - 2a x² + (a² - 81)x - c, where a and c are positive integers. Jon first says there's more than one such polynomial, which makes Steve give a specific value of a, but even then, there are still two possible values of c. We need to find the sum of those two possible values of c.

First, let me recall that if a polynomial has roots r, s, t (all positive integers), then it can be written as P(x) = k(x - r)(x - s)(x - t), where k is the leading coefficient. In this case, the polynomial is given as 2x³ - 2a x² + (a² - 81)x - c. So comparing this with the standard form, we can see that k = 2. Therefore, the polynomial can be written as 2(x - r)(x - s)(x - t).

Expanding 2(x - r)(x - s)(x - t) should give us the polynomial in the given form. Let me do that expansion.

First, expand (x - r)(x - s)(x - t):

First multiply two factors: (x - r)(x - s) = x² - (r + s)x + rs. Then multiply by (x - t):

(x² - (r + s)x + rs)(x - t) = x³ - (r + s + t)x² + (rs + rt + st)x - rst.

So, multiplying by 2:

2x³ - 2(r + s + t)x² + 2(rs + rt + st)x - 2rst.

Comparing this with the given polynomial: 2x³ - 2a x² + (a² - 81)x - c.

So, equating coefficients:

1. Coefficient of x²: -2(r + s + t) = -2a ⇒ r + s + t = a.

2. Coefficient of x: 2(rs + rt + st) = a² - 81 ⇒ rs + rt + st = (a² - 81)/2.

3. Constant term: -2rst = -c ⇒ c = 2rst.

So, we have three equations:

1. r + s + t = a.

2. rs + rt + st = (a² - 81)/2.

3. c = 2rst.

Also, r, s, t are positive integers, and a and c are positive integers.

So, since r, s, t are positive integers, then their sum a must be at least 3 (since the smallest sum is 1 + 1 + 1 = 3). Similarly, c = 2rst is also a positive integer.

Now, the problem says that initially, Jon says there's more than one such polynomial. So, there are multiple triples (r, s, t) of positive integers that satisfy these equations for some a and c. Then, after Steve gives a specific value of a, Jon still can't determine c uniquely, meaning there are two different triples (r, s, t) with the same a but different c's.

Our goal is to find those two c's and add them.

So, first, let's think about how to relate these equations. Since we have r + s + t = a and rs + rt + st = (a² - 81)/2. Let me think about how to relate these. Perhaps using Vieta's formula.

Alternatively, maybe we can think of rs + rt + st as related to (r + s + t)^2 - (r² + s² + t²). But not sure if that helps. Alternatively, think of the equations:

From equation 1 and 2, we can express rs + rt + st in terms of a. Let me denote S = r + s + t = a, and Q = rs + rt + st = (a² - 81)/2. Then, also, the product rst = c / 2.

But how do we get more information? Maybe think of the roots r, s, t. Since they are positive integers, perhaps we can find possible values of a such that (a² - 81) is even, since Q must be integer. Therefore, (a² - 81) must be even. Therefore, a² must be odd + 81. Wait, 81 is odd, so a² must be odd + odd = even? Wait:

Wait, if (a² - 81) is even, then a² ≡ 81 mod 2. Since 81 is odd, so a² is odd. Therefore, a must be odd. Therefore, a is an odd positive integer. So that tells us that a is odd.

Therefore, possible values for a are odd integers greater than or equal to 3 (since minimum sum of three positive integers is 3). Also, since Q is positive, (a² - 81)/2 must be positive. Therefore, a² - 81 > 0 ⇒ a² > 81 ⇒ a > 9. Since a is a positive integer, so a ≥ 10. However, a must be odd and greater than 9. Therefore, possible values for a are 11, 13, 15,... But Jon says there are multiple such polynomials, meaning multiple triples (r, s, t) with same a? Wait, no. Wait, initially, Jon says there is more than one such polynomial. So, initially, there are multiple (a, c) pairs? Or multiple triples (r, s, t) with different a and c? Wait, the problem says:

Steve says to Jon, "I am thinking of a polynomial... for some positive integers a and c." Then, Jon says "There is more than one such polynomial." So, initially, there are multiple polynomials (i.e., multiple (a, c) pairs) that satisfy the given form with roots all positive integers. Then Steve gives a specific value of a, but even then, there are two possible c's. So, that suggests that for the given a, there are two different triples (r, s, t) leading to different c's. Therefore, we need to find such an a where there are two different triples (r, s, t) with the same a but different c's. Then, after Steve tells Jon the value of a, Jon can't determine c uniquely, so there are two possible c's. Our job is to find those two c's and sum them.

Therefore, first, we need to find all possible a's such that there are multiple triples (r, s, t) with r + s + t = a and rs + rt + st = (a² - 81)/2, and then find the a that is unique in the sense that once a is given, there are two different c's. Then, sum those c's.

Alternatively, maybe we need to first find possible a's with multiple triples, then find which a has exactly two triples leading to different c's. So, perhaps the steps are:

1. Find all possible a's (odd integers ≥ 11) where there are multiple triples (r, s, t) of positive integers with r + s + t = a and rs + rt + st = (a² - 81)/2.

2. Among those a's, find the ones where there are exactly two different triples (r, s, t), leading to different c's. Then, sum those c's.

But how do we find such a's?

Alternatively, perhaps we can consider that if we have three numbers r, s, t, with sum a and sum of products Q = (a² - 81)/2. Let me think of Vieta's formula. If we think of r, s, t as roots of the cubic equation x³ - a x² + Q x - (c/2) = 0. However, since the original polynomial is 2x³ - 2a x² + (a² - 81)x - c, which is 2 times the monic polynomial. So the monic polynomial is x³ - a x² + [(a² - 81)/2]x - (c/2). Therefore, the roots satisfy Vieta's formula. However, since the problem states that all roots are positive integers, so we can consider the cubic equation with roots r, s, t.

But how do we find such triples? Maybe consider that for given a, Q = (a² - 81)/2. So, given a, we can compute Q and then find triples (r, s, t) with sum a and sum of products Q. Alternatively, maybe we can find relations between r, s, t and a.

Alternatively, think about the cubic equation. Let me think. Let me denote that for given a, the sum of roots is a, the sum of products is Q, and the product is c/2. Also, there's a relation between these. Let me think if we can find another equation or constraint.

Alternatively, think of the roots as integers. Let me consider that r, s, t are positive integers, so maybe we can find possible triples.

Alternatively, note that for the cubic with roots r, s, t, the discriminant must be non-negative, but since all roots are real, it's automatically satisfied. Not sure if that helps.

Alternatively, think of the relation between the sum of squares. Since (r + s + t)^2 = r² + s² + t² + 2(rs + rt + st). Therefore:

a² = r² + s² + t² + 2Q ⇒ r² + s² + t² = a² - 2Q.

But Q is (a² - 81)/2, so:

r² + s² + t² = a² - 2*(a² - 81)/2 = a² - (a² - 81) = 81.

Wow! That's interesting. Therefore, the sum of the squares of the roots is 81. So, r² + s² + t² = 81. So, we have:

Given that r, s, t are positive integers such that:

1. r + s + t = a

2. r² + s² + t² = 81

So, that's a crucial relation. So, this reduces the problem to finding all triples of positive integers (r, s, t) with r² + s² + t² = 81. Then, for each such triple, we can compute a = r + s + t, and c = 2rst. Then, since Jon initially says there's more than one such polynomial, meaning there are multiple triples (r, s, t) with different a's? Or same a's? Wait, the original problem says that Steve says "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form... for some positive integers a and c." Then Jon says "There is more than one such polynomial." So, there are multiple polynomials (i.e., multiple (a, c) pairs) that satisfy the given form with roots being positive integers. Therefore, there are multiple triples (r, s, t) with different a's and c's. However, when Steve gives a specific value of a, then there are two different c's. Therefore, that particular a must be such that there are two different triples (r, s, t) that have the same sum a but different products rst (hence different c's). Therefore, our first step is to find all triples (r, s, t) of positive integers with r² + s² + t² = 81, compute their sums a = r + s + t, and then see which a's have multiple triples. Then, for those a's, check if they have two different c's (i.e., different products rst). Then, sum those c's.

Therefore, the problem reduces to:

Find all positive integer triples (r, s, t) with r² + s² + t² = 81, compute their sums a and products c = 2rst, then find which a's have at least two triples leading to different c's, and then sum those c's.

Therefore, let's first find all the positive integer triples (r, s, t) such that r² + s² + t² = 81.

First, since r, s, t are positive integers, each at least 1. Let me note that 81 is 9², so possible squares less than 81 are 1, 4, 9, 16, 25, 36, 49, 64.

So, let me find all triples of squares adding up to 81.

Note that since r, s, t are positive integers, let's consider all possible ordered triples where order doesn't matter (since roots are indistinct in Vieta's formula; but since polynomials are determined by the multiset of roots, we can consider unordered triples. However, since we need to consider different ordered triples if they result in different products, but in reality, different permutations of the same numbers would give the same product rst, so we can consider unordered triples.

So, first, list all unordered triples of positive integers (r, s, t), r ≤ s ≤ t, such that r² + s² + t² = 81.

Let me proceed step by step.

Start with t as the largest value. Since t² ≤ 81 -1 -1 = 79, so t ≤ 8 since 8²=64, 9²=81. But if t ≥ 9, then t² ≥81, so t² + ... would be more than 81. So t can be from 1 to 8. However, since t is the largest, and r ≤ s ≤ t, let's check for each t from 1 upwards.

But let's start with t from 8 downwards, since larger t would give us fewer possibilities.

First, t=8: t²=64. Then, r² + s² = 81 -64=17. Now, find positive integers r ≤ s ≤8 with r² + s²=17.

Possible squares up to 17: 1, 4, 9, 16.

Check possible pairs:

s can be from ceil(sqrt(17/2)) up to floor(sqrt(17)).

sqrt(17) ≈4.12. So s can be 4 or less?

Wait, for r ≤ s, so r ≤ s, and r² + s²=17.

Let me check possible (r, s):

Check r=1: s²=17-1=16 ⇒ s=4. So (1,4,8). Check 1² +4² +8²=1+16+64=81. Yes. Valid.

r=2: 2² + s²=17 ⇒ 4 + s²=17 ⇒ s²=13 ⇒ not integer.

r=3: 9 + s²=17 ⇒ s²=8 ⇒ nope.

So only (1,4,8). So one triple here.

t=7: t²=49. Then r² + s²=81-49=32.

Find r ≤ s ≤7 with r² + s²=32.

Possible squares up to 32: 1,4,9,16,25.

Check possible pairs:

s can be from ceil(sqrt(32/2)) to floor(sqrt(32)).

sqrt(32) ≈5.65, so s can be up to 5.

Check s=5: s²=25. Then r²=32-25=7 ⇒ not square.

s=4: s²=16 ⇒ r²=32-16=16 ⇒ r=4. So (4,4,7). Check 4² +4² +7²=16+16+49=81. Yes.

s=3: s²=9 ⇒ r²=32-9=23 ⇒ nope.

s=2: r²=32-4=28 ⇒ nope.

s=1: r²=31 ⇒ nope.

So only (4,4,7).

t=6: t²=36. Then r² + s²=81-36=45.

Find r ≤ s ≤6 with r² + s²=45.

Squares up to 45: 1,4,9,16,25,36.

Check:

s can be up to 6.

Check s=6: s²=36 ⇒ r²=45-36=9 ⇒ r=3. So (3,6,6). Check 9 + 36 + 36 = 81. Yes.

s=5: s²=25 ⇒ r²=45-25=20 ⇒ nope.

s=4: 16 ⇒ r²=45 -16=29 ⇒ nope.

s=3: 9 ⇒ r²=45 -9=36 ⇒ r=6. But since r ≤ s, if s=3, r would have to be ≤3, but r²=36 ⇒ r=6 which is more than s=3. Not allowed. So invalid.

s=2: 4 ⇒ r²=41, nope. s=1: r²=44, nope. So only (3,6,6).

t=5: t²=25. Then r² + s²=81 -25=56.

Find r ≤ s ≤5 with r² + s²=56.

s can be up to 5.

s=5: 25 ⇒ r²=56-25=31 ⇒ nope.

s=4: 16 ⇒ r²=56 -16 =40 ⇒ nope.

s=3: 9 ⇒ r²=56 -9=47 ⇒ nope.

s=2: 4 ⇒ r²=52 ⇒ nope.

s=1: 1 ⇒ r²=55 ⇒ nope. No solutions here.

t=4: t²=16. Then r² + s²=81 -16=65.

Looking for r ≤ s ≤4 with r² + s²=65. But maximum s is 4, so s²=16. So r² + 16=65 ⇒ r²=49 ⇒ r=7. But r=7 is greater than s=4, which contradicts r ≤ s. So no solution.

Similarly for t=3, t=2, t=1: even smaller numbers, so r² + s² would need to be even larger, which is impossible. Therefore, t cannot be less than 4 for this to be possible.

Wait, wait, when t=5, we had r² + s²=56, but with r ≤ s ≤5. However, maybe if we consider different orderings? No, because we are considering r ≤ s ≤ t, so s cannot exceed t=5, so we checked that.

So the possible triples are:

(1,4,8), (4,4,7), (3,6,6).

Wait, are there any more? Let me check for t=7, s= something else?

Wait, for t=7, we had only (4,4,7). For t=8, (1,4,8). For t=6, (3,6,6). What about other triples?

Wait, let me check t=5 again. Wait, 5² + 5² + 5² = 75, which is more than 81. Wait, but maybe other combinations?

Wait, for example, what about (2, 5, 6)? Let me check 2² +5² +6² = 4 +25 +36=65. Not 81. (2, 6, 7): 4 + 36 + 49=89. Too big. (3, 4, 8): 9 + 16 + 64=89. (3, 5, 7): 9 +25 +49=83. (3, 5, 6): 9+25+36=70. (4, 5, 6): 16 +25 +36=77. (2, 3, 8): 4 + 9 + 64=77. (2, 4, 7): 4 + 16 + 49=69. (1,5, 7): 1 +25 +49=75. (1, 3, 8): 1 +9 +64=74. (1, 2, 8): 1 +4 +64=69. Hmm. How about (5, 5, 5): 75. (5, 5, 6): 25 +25 + 36=86. (5, 4, 6): same as (4,5,6). Wait, maybe I missed some?

Wait, let me think again. Are there other triples besides the three I found?

Wait, let's check for t=7:

We had (4,4,7). What about (2, 6, 7)? 4 + 36 + 49 = 89. No. (3, 5, 7): 9 +25 +49=83. (1, 6, 7): 1 + 36 + 49=86. Not 81. How about t=7, s= something else?

Wait, t=7, so r² + s²=32. We found (4,4,7). Are there other pairs?

For example, if r= 2 and s= sqrt(32 -4)=sqrt(28) which isn't integer. Or r= 1, s= sqrt(31). Nope. So only (4,4,7).

For t=8, only (1,4,8). For t=6, only (3,6,6). What about other possibilities where t is not the maximum?

Wait, but we considered all t's from 1 to 8. Wait, but maybe there are other triples with different orderings? For example, (4,1,8) is same as (1,4,8). So as we considered unordered triples, we have only three triples:

(1,4,8), (4,4,7), (3,6,6).

Wait, but let me check for other triples. For example, suppose two numbers are same. Let's check (5, 5, 5): sum of squares 75. Not 81. What about (5, 5, 6): sum of squares 25 +25 + 36 = 86. Not. (5, 5, 7): 25 +25 + 49 = 99. Not. (5, 5, 4): same as (4,5,5), which we have already considered? Wait, in our previous search, we had for t=5, but we didn't find that. Wait, let me check again.

Wait, for t=5, the sum of squares of r and s would be 56. If we consider r and s as 5 and something?

Wait, if t=5, then r² + s² = 81 -25=56. Suppose one of r or s is 5. Then the other would be sqrt(56 - 25) = sqrt(31), which is not integer. So no. If r=5 and s=5, then 5² +5² = 50, which is less than 56, so r² + s²=56 would require one more. Wait, if r=5 and s= sqrt(56 - 25)=sqrt(31). Not integer. So no. Therefore, no such triples here.

How about (2, 3, sqrt(81 -4 -9)) = sqrt(68). Nope. (2, 4, sqrt(81 -4 -16)) = sqrt(61). No. (2,5, sqrt(81 -4 -25)) = sqrt(52). No. (2,6, sqrt(81 -4 -36)) = sqrt(41). No. (2,7, sqrt(81 -4 -49))=sqrt(28). No. Similarly for (3, ...). Let me check (3, 4, sqrt(81 -9 -16))=sqrt(56). Nope. (3, 5, sqrt(81 -9 -25))=sqrt(47). (3, 6, sqrt(81 -9 -36))=sqrt(36)=6. So that gives (3,6,6). Which is already in our list. So that's covered.

Similarly, (4, 5, sqrt(81 - 16 -25))=sqrt(40). No. (4, 6, sqrt(81 -16 -36))=sqrt(29). No. (4,7, sqrt(81 -16 -49))=sqrt(16)=4. So r=4, s=7, t=4? But we need r ≤ s ≤ t. So if t=7, then the triple would be (4,4,7). Which is already considered.

Similarly, (5, 5, sqrt(81 -25 -25))=sqrt(31). No. So seems like we have only three triples:

(1,4,8), (3,6,6), (4,4,7).

Wait, but wait, another thought. What about (2, 2, sqrt(81 -4 -4)) = sqrt(73). Not integer. What about (2, 5, 6)? 4 +25 +36=65. (2, 5, 7): 4 +25 +49=78. (2, 5, 8): 4 +25 +64=93. (2, 6, 7): 4 +36 +49=89. (2, 6, 8): 4 +36 +64=104. (2, 7, 8): 4 +49 +64=117. Nope. How about (3, 3, sqrt(81 -9 -9))=sqrt(63). No. (3, 4, sqrt(...)): already checked. (3, 5, sqrt(...)): checked.

So seems only three triples. Wait, but wait, let me think again. What about (5, 5, 5)? No, sum of squares 75. (5, 6, 6): 25 + 36 + 36 = 97. (5, 4, 6): 25 + 16 + 36 = 77. Still not 81. (5, 5, 7): 25 +25 +49=99. (5, 7, 7): 25 +49 +49=123. So nope.

Wait, so three triples. Therefore, these three triples. So then, for each triple, compute a = sum of roots, and c = 2 * product.

Let me compute them:

First triple: (1, 4, 8). Then:

a = 1 + 4 + 8 = 13.

c = 2 * 1 * 4 * 8 = 2 * 32 = 64.

Second triple: (4, 4, 7). Sum: 4 + 4 + 7 = 15. Product: 4*4*7=112. c=2*112=224.

Third triple: (3, 6, 6). Sum: 3 + 6 + 6 = 15. Product: 3*6*6=108. c=2*108=216.

Wait, so here, there are two triples with a=15: (4,4,7) and (3,6,6). Therefore, for a=15, there are two different c's: 224 and 216. Hence, when Steve tells Jon the value of a, which is 15, Jon can't determine c, since there are two possibilities. Then the problem says "Find the sum of the two possible values of c." So, 224 + 216 = 440? Wait, but wait, the problem says "the value of a" and then Jon says there are two possible c's. So, the answer would be 216 + 224 = 440? But wait, let me check.

Wait, but why do we have two triples for a=15? Because the first triple (1,4,8) gives a=13, and the other two triples give a=15. So initially, there are three polynomials: one with a=13, c=64; one with a=15, c=216; another with a=15, c=224. Therefore, Jon initially says there is more than one such polynomial, which is true because there are two polynomials with a=15 and one with a=13. Then, when Steve gives a specific value of a, which is 15 (since otherwise, if he gave a=13, Jon would know c=64 uniquely). Therefore, Jon is told a=15, and then there are two possible c's: 216 and 224. Hence, their sum is 216 + 224 = 440. Is that the answer?

Wait, but let me check if there are other triples. Maybe I missed some?

Wait, let's check again. For example, could there be another triple with different numbers?

Wait, think about 81 as sum of squares. Let me check all possible combinations again. Maybe (2, 6, 7)? 4 + 36 + 49 = 89. No. (2, 3, 8): 4 + 9 + 64 =77. (2, 5, 7): 4 + 25 +49=78. (3, 4, 8): 9 + 16 +64=89. (3, 5, 7): 9 + 25 +49=83. (4, 5, 6): 16 +25 +36=77. (5, 5, 7): 25 +25 +49=99. (5, 6, 6): 25 + 36 +36=97. (1, 3, 8): 1 +9 +64=74. (1, 5, 7): 1 +25 +49=75. (1, 6, 7): 1 + 36 +49=86. (1, 2, 8): 1 +4 +64=69. So no, seems like only three triples. So that seems correct.

Wait, but then why does the problem say that Jon says there is more than one polynomial? Well, because there are multiple polynomials (for example, the one with a=13, c=64; and two with a=15, c=216 and 224). Therefore, when Steve gives the value of a, which must be 15 (as otherwise, if a=13, Jon could uniquely determine c=64, but since he says there are two possible c's, then the a must be 15). Therefore, the answer is 216 + 224 = 440.

Wait, but let me check if there are any other triples. For example, let me check (sqrt(9), sqrt(9), sqrt(63))? No, since they need to be integers. Alternatively, maybe (something else). Wait, but we already considered all possibilities. Wait, 81 can also be expressed as 9² + 0² + 0², but 0 isn't positive. So no. Therefore, only the three triples we found.

Therefore, the two c's are 216 and 224. Their sum is 216 + 224 = 440.

But let me check if the problem says that Jon is told that there are two possible values of c after Steve gives a. Therefore, the answer is 440. However, I need to check my calculations again.

First, check the triples:

Triple 1: 1, 4, 8. Sum a = 1 + 4 + 8 = 13. c = 2*1*4*8 = 64. Correct.

Triple 2: 4, 4, 7. Sum a=4 + 4 + 7 = 15. c = 2*4*4*7 = 2*112 = 224. Correct.

Triple 3: 3, 6, 6. Sum a=3 + 6 + 6 = 15. c = 2*3*6*6 = 2*108 = 216. Correct.

Yes, so two triples for a=15. Therefore, the sum is 216 + 224 = 440.

But wait, I have a doubt here. Because the problem says that Steve says, "Here is the value of a." Then Jon says, "There are still two possible values of c." So, that implies that when Steve gives a, there are two c's. Therefore, the value of a must be 15. Therefore, the answer is 216 + 224 = 440.

But let me check once again if there are other triples. Maybe I missed one? Let me think.

Wait, what about (5, 5, sqrt(81 - 25 -25)) = sqrt(31). Not integer. (5, 4, sqrt(81 - 25 - 16)) = sqrt(40). Not integer. What about (something with 9? Wait, 9² is 81. If one of the roots is 9, then r² + s² + t² = 81. So if one root is 9, then the other two roots must be zero, but they have to be positive integers. Therefore, no.

Another idea: could there be triples with numbers larger than 8? For example, if there is a root larger than 8, say 9, then as above, the sum would be too big. But since the sum of squares is 81, the maximum possible root is 9 (as 9²=81, and the others would have to be zero). So no. So no other triples.

Therefore, only three triples. Therefore, the answer should be 440. However, I feel that maybe there is a mistake here because 440 seems a bit large. Let me check the problem again.

Wait, the problem says the polynomial is 2x³ - 2a x² + (a² - 81)x - c. For the triple (1,4,8), a is 13. Then, the coefficient of x is (a² -81) = (169 - 81) = 88. Let's check if expanding 2(x - 1)(x - 4)(x - 8) gives the correct coefficients.

First, compute (x - 1)(x - 4)(x - 8):

First multiply (x - 1)(x - 4) = x² -5x + 4.

Multiply by (x - 8):

(x² -5x + 4)(x - 8) = x³ -8x² -5x² +40x +4x -32 = x³ -13x² +44x -32.

Multiply by 2: 2x³ -26x² +88x -64. Which matches the given polynomial with a=13 and c=64: since P(x) = 2x³ -2a x² + (a² -81)x -c. So with a=13, -2a is -26, and a² -81 is 169 -81=88, and c=64. Yes, matches. So that's correct.

For the triple (4,4,7), a=15, so the polynomial is 2x³ -2*15x² + (225 -81)x -c = 2x³ -30x² +144x -c. Let's compute 2(x -4)(x -4)(x -7). First, (x -4)^2 = x² -8x +16. Multiply by (x -7):

(x² -8x +16)(x -7) = x³ -7x² -8x² +56x +16x -112 = x³ -15x² +72x -112. Multiply by 2: 2x³ -30x² +144x -224. Therefore, c=224. Correct.

For the triple (3,6,6): a=15, so same polynomial 2x³ -30x² +144x -c. Compute 2(x -3)(x -6)(x -6). First compute (x -6)^2 = x² -12x +36. Multiply by (x -3):

(x² -12x +36)(x -3) = x³ -3x² -12x² +36x +36x -108 = x³ -15x² +72x -108. Multiply by 2: 2x³ -30x² +144x -216. Therefore, c=216. Correct.

Therefore, these are correct. Therefore, the two c's are 216 and 224. Their sum is 216 + 224 = 440.

But wait, the problem says "the sum of the two possible values of c". So, 440. But let me check if that's the answer. However, I recall that sometimes problems like this might have answers that are smaller numbers. Maybe I made a mistake in considering the triples?

Wait, let me check if there are any more triples. For example, maybe (2, 2, sqrt(81 -4 -4)) = sqrt(73). No. (2, 3, sqrt(81 -4 -9))=sqrt(68). No. (2, 5, sqrt(81 -4 -25))=sqrt(52). No. (3, 3, sqrt(81 -9 -9))=sqrt(63). No. (3, 4, sqrt(81 -9 -16))=sqrt(56). No. (3,5, sqrt(81 -9 -25))=sqrt(47). No. (4,5, sqrt(81 -16 -25))=sqrt(40). No. (4, 6, sqrt(81 -16 -36))=sqrt(29). No. (5, 6, sqrt(81 -25 -36))=sqrt(20). No. So no other triples. So seems correct.

Alternatively, perhaps I missed some permutations? But no, because we considered unordered triples. So yes.

Alternatively, maybe there is another a where multiple triples exist? For example, is there another a with multiple triples? But according to our previous analysis, only a=15 has two triples. The other a=13 has only one triple. Therefore, when Steve gave a specific value of a (which must be 15, since otherwise there would be only one c), Jon says there are two c's. Therefore, the answer is 216 + 224 = 440. Therefore, I think this is correct.

But let me check once again. Maybe the problem says "the sum of the two possible values of c", and maybe the answer is different. Let me check if 440 is correct. Alternatively, maybe there are more triples?

Wait, here's a thought: maybe the triple (2, 2, 7). 2² + 2² +7²=4 +4 +49=57≠81. No. (2, 3, 7): 4 +9 +49=62. (2, 4, 7): 4 +16 +49=69. (2, 5, 7): 4 +25 +49=78. (2, 6, 7): 4 +36 +49=89. No. (3, 3, 6): 9 +9 +36=54. (3, 4, 6): 9 +16 +36=61. (3, 5, 6): 9 +25 +36=70. (4, 4, 5): 16 +16 +25=57. No. So no other triples. Hence, only three triples. Therefore, answer is 440. Maybe the answer is 440. However, since the problem is from an Olympiad, maybe the answer is different? Wait, but I can't see any mistake in my logic. Therefore, I think the answer is 440.

But let me check once again the calculation for the sum of squares. For the triple (4,4,7): 16 +16 +49=81. Correct. For (3,6,6): 9 + 36 + 36=81. Correct. For (1,4,8): 1 + 16 + 64=81. Correct. All correct. Therefore, yes. Then the answer should be 216 + 224 = 440.

But let me check the problem statement again. It says that Steve says "I am thinking of a polynomial whose roots are all positive integers." So there are multiple such polynomials. Then Jon says there's more than one. Then Steve gives a value of a, and Jon still has two c's. Therefore, the answer is 440. So, I think that is the correct answer.

**Final Answer**
\boxed{440}

To solve this problem, we analyze the polynomial:

$$
P(x) = 2x^3 - 2a x^2 + (a^2 - 81)x - c
$$

Given that all the roots of this cubic polynomial are **positive integers**, we can express the polynomial in factored form as:

$$
P(x) = 2(x - r)(x - s)(x - t)
$$

where $ r, s, t $ are positive integers. Expanding this expression:

$$
P(x) = 2 \left[ x^3 - (r+s+t)x^2 + (rs + rt + st)x - rst \right]
$$

Now, equating to the given polynomial:

- Coefficient of $ x^2 $: $ -2(r + s + t) = -2a \Rightarrow r + s + t = a $
- Coefficient of $ x $: $ 2(rs + rt + st) = a^2 - 81 \Rightarrow rs + rt + st = \frac{a^2 - 81}{2} $
- Constant term: $ -2rst = -c \Rightarrow c = 2rst $

We conclude that:
- The sum of the roots: $ a = r + s + t $
- The sum of product pairs: $ rs + rt + st = \frac{a^2 - 81}{2} $
- The product of roots: $ rst = \frac{c}{2} $

---

### Step 1: Use the identity for the sum of squares

We take the identity:

$$
(r + s + t)^2 = r^2 + s^2 + t^2 + 2(rs + rt + st)
$$

Multiply both sides by 1 and rearrange:

$$
r^2 + s^2 + t^2 = a^2 - 2(rs + rt + st)
$$

Substitute $ rs + rt + st = \frac{a^2 - 81}{2} $:

$$
r^2 + s^2 + t^2 = a^2 - 2\left( \frac{a^2 - 81}{2} \right) = a^2 - (a^2 - 81) = 81
$$

So the **sum of squares of the roots is fixed**:

$$
r^2 + s^2 + t^2 = 81
$$

---

### Step 2: Find all positive integer solutions to $ r^2 + s^2 + t^2 = 81 $

We systematically find all sets of positive integers $ (r, s, t) $, with $ r \leq s \leq t $, that satisfy the sum of squares condition.

We test all combinations:

- $ (1, 4, 8) $: $ 1^2 + 4^2 + 8^2 = 1 + 16 + 64 = 81 $
- $ (3, 6, 6) $: $ 9 + 36 + 36 = 81 $
- $ (4, 4, 7) $: $ 16 + 16 + 49 = 81 $

These are the **only** valid positive integer triples.

---

### Step 3: Compute corresponding values of $ a $ and $ c $

Now for each triple:

#### Triple (1, 4, 8):

- $ a = 1 + 4 + 8 = 13 $
- $ c = 2 \times 1 \times 4 \times 8 = 64 $

#### Triple (3, 6, 6):

- $ a = 3 + 6 + 6 = 15 $
- $ c = 2 \times 3 \times 6 \times 6 = 216 $

#### Triple (4, 4, 7):

- $ a = 4 + 4 + 7 = 15 $
- $ c = 2 \times 4 \times 4 \times 7 = 224 $

---

### Step 4: Identify when there are two distinct values of $ c $

Note that:

- There is **only one** value of $ c $ for $ a = 13 $: **64**
- There are **two** values of $ c $ for $ a = 15 $: **216** and **224**

So, if Steve gives the value $ a = 15 $, Jon cannot uniquely determine $ c $ — there are **two possible values**.

Hence, the **sum of the two possible values of $ c $** is:

$$
216 + 224 = \boxed{440}
$$

---

### ✅ Final Answer:

$$
\boxed{440}
$$

[Assistant]
Explanation:
Okay, let's try to tackle this problem step by step. So, Steve is thinking of a polynomial with all positive integer roots, and the polynomial is given in the form P(x) = 2x³ - 2a x² + (a² - 81)x - c, where a and c are positive integers. Jon first says there's more than one such polynomial, which makes Steve give a specific value of a, but even then, there are still two possible values of c. We need to find the sum of those two possible values of c.

First, let me recall that if a polynomial has roots r, s, t (all positive integers), then it can be written as P(x) = k(x - r)(x - s)(x - t), where k is the leading coefficient. In this case, the polynomial is given as 2x³ - 2a x² + (a² - 81)x - c. So comparing this with the standard form, we can see that k = 2. Therefore, the polynomial can be written as 2(x - r)(x - s)(x - t).

Expanding 2(x - r)(x - s)(x - t) should give us the polynomial in the given form. Let me do that expansion.

First, expand (x - r)(x - s)(x - t):

First multiply two factors: (x - r)(x - s) = x² - (r + s)x + rs. Then multiply by (x - t):

(x² - (r + s)x + rs)(x - t) = x³ - (r + s + t)x² + (rs + rt + st)x - rst.

So, multiplying by 2:

2x³ - 2(r + s + t)x² + 2(rs + rt + st)x - 2rst.

Comparing this with the given polynomial: 2x³ - 2a x² + (a² - 81)x - c.

So, equating coefficients:

1. Coefficient of x²: -2(r + s + t) = -2a ⇒ r + s + t = a.

2. Coefficient of x: 2(rs + rt + st) = a² - 81 ⇒ rs + rt + st = (a² - 81)/2.

3. Constant term: -2rst = -c ⇒ c = 2rst.

So, we have three equations:

1. r + s + t = a.

2. rs + rt + st = (a² - 81)/2.

3. c = 2rst.

Also, r, s, t are positive integers, and a and c are positive integers.

So, since r, s, t are positive integers, then their sum a must be at least 3 (since the smallest sum is 1 + 1 + 1 = 3). Similarly, c = 2rst is also a positive integer.

Now, the problem says that initially, Jon says there's more than one such polynomial. So, there are multiple triples (r, s, t) of positive integers that satisfy these equations for some a and c. Then, after Steve gives a specific value of a, Jon still can't determine c uniquely, meaning there are two different triples (r, s, t) with the same a but different c's.

Our goal is to find those two c's and add them.

So, first, let's think about how to relate these equations. Since we have r + s + t = a and rs + rt + st = (a² - 81)/2. Let me think about how to relate these. Perhaps using Vieta's formula.

Alternatively, maybe we can think of rs + rt + st as related to (r + s + t)^2 - (r² + s² + t²). But not sure if that helps. Alternatively, think of the equations:

From equation 1 and 2, we can express rs + rt + st in terms of a. Let me denote S = r + s + t = a, and Q = rs + rt + st = (a² - 81)/2. Then, also, the product rst = c / 2.

But how do we get more information? Maybe think of the roots r, s, t. Since they are positive integers, perhaps we can find possible values of a such that (a² - 81) is even, since Q must be integer. Therefore, (a² - 81) must be even. Therefore, a² must be odd + 81. Wait, 81 is odd, so a² must be odd + odd = even? Wait:

Wait, if (a² - 81) is even, then a² ≡ 81 mod 2. Since 81 is odd, so a² is odd. Therefore, a must be odd. Therefore, a is an odd positive integer. So that tells us that a is odd.

Therefore, possible values for a are odd integers greater than or equal to 3 (since minimum sum of three positive integers is 3). Also, since Q is positive, (a² - 81)/2 must be positive. Therefore, a² - 81 > 0 ⇒ a² > 81 ⇒ a > 9. Since a is a positive integer, so a ≥ 10. However, a must be odd and greater than 9. Therefore, possible values for a are 11, 13, 15,... But Jon says there are multiple such polynomials, meaning multiple triples (r, s, t) with same a? Wait, no. Wait, initially, Jon says there is more than one such polynomial. So, initially, there are multiple (a, c) pairs? Or multiple triples (r, s, t) with different a and c? Wait, the problem says:

Steve says to Jon, "I am thinking of a polynomial... for some positive integers a and c." Then, Jon says "There is more than one such polynomial." So, initially, there are multiple polynomials (i.e., multiple (a, c) pairs) that satisfy the given form with roots all positive integers. Then Steve gives a specific value of a, but even then, there are two possible c's. So, that suggests that for the given a, there are two different triples (r, s, t) leading to different c's. Therefore, we need to find such an a where there are two different triples (r, s, t) with the same a but different c's. Then, after Steve tells Jon the value of a, Jon can't determine c uniquely, so there are two possible c's. Our job is to find those two c's and sum them.

Therefore, first, we need to find all possible a's such that there are multiple triples (r, s, t) with r + s + t = a and rs + rt + st = (a² - 81)/2, and then find the a that is unique in the sense that once a is given, there are two different c's. Then, sum those c's.

Alternatively, maybe we need to first find possible a's with multiple triples, then find which a has exactly two triples leading to different c's. So, perhaps the steps are:

1. Find all possible a's (odd integers ≥ 11) where there are multiple triples (r, s, t) of positive integers with r + s + t = a and rs + rt + st = (a² - 81)/2.

2. Among those a's, find the ones where there are exactly two different triples (r, s, t), leading to different c's. Then, sum those c's.

But how do we find such a's?

Alternatively, perhaps we can consider that if we have three numbers r, s, t, with sum a and sum of products Q = (a² - 81)/2. Let me think of Vieta's formula. If we think of r, s, t as roots of the cubic equation x³ - a x² + Q x - (c/2) = 0. However, since the original polynomial is 2x³ - 2a x² + (a² - 81)x - c, which is 2 times the monic polynomial. So the monic polynomial is x³ - a x² + [(a² - 81)/2]x - (c/2). Therefore, the roots satisfy Vieta's formula. However, since the problem states that all roots are positive integers, so we can consider the cubic equation with roots r, s, t.

But how do we find such triples? Maybe consider that for given a, Q = (a² - 81)/2. So, given a, we can compute Q and then find triples (r, s, t) with sum a and sum of products Q. Alternatively, maybe we can find relations between r, s, t and a.

Alternatively, think about the cubic equation. Let me think. Let me denote that for given a, the sum of roots is a, the sum of products is Q, and the product is c/2. Also, there's a relation between these. Let me think if we can find another equation or constraint.

Alternatively, think of the roots as integers. Let me consider that r, s, t are positive integers, so maybe we can find possible triples.

Alternatively, note that for the cubic with roots r, s, t, the discriminant must be non-negative, but since all roots are real, it's automatically satisfied. Not sure if that helps.

Alternatively, think of the relation between the sum of squares. Since (r + s + t)^2 = r² + s² + t² + 2(rs + rt + st). Therefore:

a² = r² + s² + t² + 2Q ⇒ r² + s² + t² = a² - 2Q.

But Q is (a² - 81)/2, so:

r² + s² + t² = a² - 2*(a² - 81)/2 = a² - (a² - 81) = 81.

Wow! That's interesting. Therefore, the sum of the squares of the roots is 81. So, r² + s² + t² = 81. So, we have:

Given that r, s, t are positive integers such that:

1. r + s + t = a

2. r² + s² + t² = 81

So, that's a crucial relation. So, this reduces the problem to finding all triples of positive integers (r, s, t) with r² + s² + t² = 81. Then, for each such triple, we can compute a = r + s + t, and c = 2rst. Then, since Jon initially says there's more than one such polynomial, meaning there are multiple triples (r, s, t) with different a's? Or same a's? Wait, the original problem says that Steve says "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form... for some positive integers a and c." Then Jon says "There is more than one such polynomial." So, there are multiple polynomials (i.e., multiple (a, c) pairs) that satisfy the given form with roots being positive integers. Therefore, there are multiple triples (r, s, t) with different a's and c's. However, when Steve gives a specific value of a, then there are two different c's. Therefore, that particular a must be such that there are two different triples (r, s, t) that have the same sum a but different products rst (hence different c's). Therefore, our first step is to find all triples (r, s, t) of positive integers with r² + s² + t² = 81, compute their sums a = r + s + t, and then see which a's have multiple triples. Then, for those a's, check if they have two different c's (i.e., different products rst). Then, sum those c's.

Therefore, the problem reduces to:

Find all positive integer triples (r, s, t) with r² + s² + t² = 81, compute their sums a and products c = 2rst, then find which a's have at least two triples leading to different c's, and then sum those c's.

Therefore, let's first find all the positive integer triples (r, s, t) such that r² + s² + t² = 81.

First, since r, s, t are positive integers, each at least 1. Let me note that 81 is 9², so possible squares less than 81 are 1, 4, 9, 16, 25, 36, 49, 64.

So, let me find all triples of squares adding up to 81.

Note that since r, s, t are positive integers, let's consider all possible ordered triples where order doesn't matter (since roots are indistinct in Vieta's formula; but since polynomials are determined by the multiset of roots, we can consider unordered triples. However, since we need to consider different ordered triples if they result in different products, but in reality, different permutations of the same numbers would give the same product rst, so we can consider unordered triples.

So, first, list all unordered triples of positive integers (r, s, t), r ≤ s ≤ t, such that r² + s² + t² = 81.

Let me proceed step by step.

Start with t as the largest value. Since t² ≤ 81 -1 -1 = 79, so t ≤ 8 since 8²=64, 9²=81. But if t ≥ 9, then t² ≥81, so t² + ... would be more than 81. So t can be from 1 to 8. However, since t is the largest, and r ≤ s ≤ t, let's check for each t from 1 upwards.

But let's start with t from 8 downwards, since larger t would give us fewer possibilities.

First, t=8: t²=64. Then, r² + s² = 81 -64=17. Now, find positive integers r ≤ s ≤8 with r² + s²=17.

Possible squares up to 17: 1, 4, 9, 16.

Check possible pairs:

s can be from ceil(sqrt(17/2)) up to floor(sqrt(17)).

sqrt(17) ≈4.12. So s can be 4 or less?

Wait, for r ≤ s, so r ≤ s, and r² + s²=17.

Let me check possible (r, s):

Check r=1: s²=17-1=16 ⇒ s=4. So (1,4,8). Check 1² +4² +8²=1+16+64=81. Yes. Valid.

r=2: 2² + s²=17 ⇒ 4 + s²=17 ⇒ s²=13 ⇒ not integer.

r=3: 9 + s²=17 ⇒ s²=8 ⇒ nope.

So only (1,4,8). So one triple here.

t=7: t²=49. Then r² + s²=81-49=32.

Find r ≤ s ≤7 with r² + s²=32.

Possible squares up to 32: 1,4,9,16,25.

Check possible pairs:

s can be from ceil(sqrt(32/2)) to floor(sqrt(32)).

sqrt(32) ≈5.65, so s can be up to 5.

Check s=5: s²=25. Then r²=32-25=7 ⇒ not square.

s=4: s²=16 ⇒ r²=32-16=16 ⇒ r=4. So (4,4,7). Check 4² +4² +7²=16+16+49=81. Yes.

s=3: s²=9 ⇒ r²=32-9=23 ⇒ nope.

s=2: r²=32-4=28 ⇒ nope.

s=1: r²=31 ⇒ nope.

So only (4,4,7).

t=6: t²=36. Then r² + s²=81-36=45.

Find r ≤ s ≤6 with r² + s²=45.

Squares up to 45: 1,4,9,16,25,36.

Check:

s can be up to 6.

Check s=6: s²=36 ⇒ r²=45-36=9 ⇒ r=3. So (3,6,6). Check 9 + 36 + 36 = 81. Yes.

s=5: s²=25 ⇒ r²=45-25=20 ⇒ nope.

s=4: 16 ⇒ r²=45 -16=29 ⇒ nope.

s=3: 9 ⇒ r²=45 -9=36 ⇒ r=6. But since r ≤ s, if s=3, r would have to be ≤3, but r²=36 ⇒ r=6 which is more than s=3. Not allowed. So invalid.

s=2: 4 ⇒ r²=41, nope. s=1: r²=44, nope. So only (3,6,6).

t=5: t²=25. Then r² + s²=81 -25=56.

Find r ≤ s ≤5 with r² + s²=56.

s can be up to 5.

s=5: 25 ⇒ r²=56-25=31 ⇒ nope.

s=4: 16 ⇒ r²=56 -16 =40 ⇒ nope.

s=3: 9 ⇒ r²=56 -9=47 ⇒ nope.

s=2: 4 ⇒ r²=52 ⇒ nope.

s=1: 1 ⇒ r²=55 ⇒ nope. No solutions here.

t=4: t²=16. Then r² + s²=81 -16=65.

Looking for r ≤ s ≤4 with r² + s²=65. But maximum s is 4, so s²=16. So r² + 16=65 ⇒ r²=49 ⇒ r=7. But r=7 is greater than s=4, which contradicts r ≤ s. So no solution.

Similarly for t=3, t=2, t=1: even smaller numbers, so r² + s² would need to be even larger, which is impossible. Therefore, t cannot be less than 4 for this to be possible.

Wait, wait, when t=5, we had r² + s²=56, but with r ≤ s ≤5. However, maybe if we consider different orderings? No, because we are considering r ≤ s ≤ t, so s cannot exceed t=5, so we checked that.

So the possible triples are:

(1,4,8), (4,4,7), (3,6,6).

Wait, are there any more? Let me check for t=7, s= something else?

Wait, for t=7, we had only (4,4,7). For t=8, (1,4,8). For t=6, (3,6,6). What about other triples?

Wait, let me check t=5 again. Wait, 5² + 5² + 5² = 75, which is more than 81. Wait, but maybe other combinations?

Wait, for example, what about (2, 5, 6)? Let me check 2² +5² +6² = 4 +25 +36=65. Not 81. (2, 6, 7): 4 + 36 + 49=89. Too big. (3, 4, 8): 9 + 16 + 64=89. (3, 5, 7): 9 +25 +49=83. (3, 5, 6): 9+25+36=70. (4, 5, 6): 16 +25 +36=77. (2, 3, 8): 4 + 9 + 64=77. (2, 4, 7): 4 + 16 + 49=69. (1,5, 7): 1 +25 +49=75. (1, 3, 8): 1 +9 +64=74. (1, 2, 8): 1 +4 +64=69. Hmm. How about (5, 5, 5): 75. (5, 5, 6): 25 +25 + 36=86. (5, 4, 6): same as (4,5,6). Wait, maybe I missed some?

Wait, let me think again. Are there other triples besides the three I found?

Wait, let's check for t=7:

We had (4,4,7). What about (2, 6, 7)? 4 + 36 + 49 = 89. No. (3, 5, 7): 9 +25 +49=83. (1, 6, 7): 1 + 36 + 49=86. Not 81. How about t=7, s= something else?

Wait, t=7, so r² + s²=32. We found (4,4,7). Are there other pairs?

For example, if r= 2 and s= sqrt(32 -4)=sqrt(28) which isn't integer. Or r= 1, s= sqrt(31). Nope. So only (4,4,7).

For t=8, only (1,4,8). For t=6, only (3,6,6). What about other possibilities where t is not the maximum?

Wait, but we considered all t's from 1 to 8. Wait, but maybe there are other triples with different orderings? For example, (4,1,8) is same as (1,4,8). So as we considered unordered triples, we have only three triples:

(1,4,8), (4,4,7), (3,6,6).

Wait, but let me check for other triples. For example, suppose two numbers are same. Let's check (5, 5, 5): sum of squares 75. Not 81. What about (5, 5, 6): sum of squares 25 +25 + 36 = 86. Not. (5, 5, 7): 25 +25 + 49 = 99. Not. (5, 5, 4): same as (4,5,5), which we have already considered? Wait, in our previous search, we had for t=5, but we didn't find that. Wait, let me check again.

Wait, for t=5, the sum of squares of r and s would be 56. If we consider r and s as 5 and something?

Wait, if t=5, then r² + s² = 81 -25=56. Suppose one of r or s is 5. Then the other would be sqrt(56 - 25) = sqrt(31), which is not integer. So no. If r=5 and s=5, then 5² +5² = 50, which is less than 56, so r² + s²=56 would require one more. Wait, if r=5 and s= sqrt(56 - 25)=sqrt(31). Not integer. So no. Therefore, no such triples here.

How about (2, 3, sqrt(81 -4 -9)) = sqrt(68). Nope. (2, 4, sqrt(81 -4 -16)) = sqrt(61). No. (2,5, sqrt(81 -4 -25)) = sqrt(52). No. (2,6, sqrt(81 -4 -36)) = sqrt(41). No. (2,7, sqrt(81 -4 -49))=sqrt(28). No. Similarly for (3, ...). Let me check (3, 4, sqrt(81 -9 -16))=sqrt(56). Nope. (3, 5, sqrt(81 -9 -25))=sqrt(47). (3, 6, sqrt(81 -9 -36))=sqrt(36)=6. So that gives (3,6,6). Which is already in our list. So that's covered.

Similarly, (4, 5, sqrt(81 - 16 -25))=sqrt(40). No. (4, 6, sqrt(81 -16 -36))=sqrt(29). No. (4,7, sqrt(81 -16 -49))=sqrt(16)=4. So r=4, s=7, t=4? But we need r ≤ s ≤ t. So if t=7, then the triple would be (4,4,7). Which is already considered.

Similarly, (5, 5, sqrt(81 -25 -25))=sqrt(31). No. So seems like we have only three triples:

(1,4,8), (3,6,6), (4,4,7).

Wait, but wait, another thought. What about (2, 2, sqrt(81 -4 -4)) = sqrt(73). Not integer. What about (2, 5, 6)? 4 +25 +36=65. (2, 5, 7): 4 +25 +49=78. (2, 5, 8): 4 +25 +64=93. (2, 6, 7): 4 +36 +49=89. (2, 6, 8): 4 +36 +64=104. (2, 7, 8): 4 +49 +64=117. Nope. How about (3, 3, sqrt(81 -9 -9))=sqrt(63). No. (3, 4, sqrt(...)): already checked. (3, 5, sqrt(...)): checked.

So seems only three triples. Wait, but wait, let me think again. What about (5, 5, 5)? No, sum of squares 75. (5, 6, 6): 25 + 36 + 36 = 97. (5, 4, 6): 25 + 16 + 36 = 77. Still not 81. (5, 5, 7): 25 +25 +49=99. (5, 7, 7): 25 +49 +49=123. So nope.

Wait, so three triples. Therefore, these three triples. So then, for each triple, compute a = sum of roots, and c = 2 * product.

Let me compute them:

First triple: (1, 4, 8). Then:

a = 1 + 4 + 8 = 13.

c = 2 * 1 * 4 * 8 = 2 * 32 = 64.

Second triple: (4, 4, 7). Sum: 4 + 4 + 7 = 15. Product: 4*4*7=112. c=2*112=224.

Third triple: (3, 6, 6). Sum: 3 + 6 + 6 = 15. Product: 3*6*6=108. c=2*108=216.

Wait, so here, there are two triples with a=15: (4,4,7) and (3,6,6). Therefore, for a=15, there are two different c's: 224 and 216. Hence, when Steve tells Jon the value of a, which is 15, Jon can't determine c, since there are two possibilities. Then the problem says "Find the sum of the two possible values of c." So, 224 + 216 = 440? Wait, but wait, the problem says "the value of a" and then Jon says there are two possible c's. So, the answer would be 216 + 224 = 440? But wait, let me check.

Wait, but why do we have two triples for a=15? Because the first triple (1,4,8) gives a=13, and the other two triples give a=15. So initially, there are three polynomials: one with a=13, c=64; one with a=15, c=216; another with a=15, c=224. Therefore, Jon initially says there is more than one such polynomial, which is true because there are two polynomials with a=15 and one with a=13. Then, when Steve gives a specific value of a, which is 15 (since otherwise, if he gave a=13, Jon would know c=64 uniquely). Therefore, Jon is told a=15, and then there are two possible c's: 216 and 224. Hence, their sum is 216 + 224 = 440. Is that the answer?

Wait, but let me check if there are other triples. Maybe I missed some?

Wait, let's check again. For example, could there be another triple with different numbers?

Wait, think about 81 as sum of squares. Let me check all possible combinations again. Maybe (2, 6, 7)? 4 + 36 + 49 = 89. No. (2, 3, 8): 4 + 9 + 64 =77. (2, 5, 7): 4 + 25 +49=78. (3, 4, 8): 9 + 16 +64=89. (3, 5, 7): 9 + 25 +49=83. (4, 5, 6): 16 +25 +36=77. (5, 5, 7): 25 +25 +49=99. (5, 6, 6): 25 + 36 +36=97. (1, 3, 8): 1 +9 +64=74. (1, 5, 7): 1 +25 +49=75. (1, 6, 7): 1 + 36 +49=86. (1, 2, 8): 1 +4 +64=69. So no, seems like only three triples. So that seems correct.

Wait, but then why does the problem say that Jon says there is more than one polynomial? Well, because there are multiple polynomials (for example, the one with a=13, c=64; and two with a=15, c=216 and 224). Therefore, when Steve gives the value of a, which must be 15 (as otherwise, if a=13, Jon could uniquely determine c=64, but since he says there are two possible c's, then the a must be 15). Therefore, the answer is 216 + 224 = 440.

Wait, but let me check if there are any other triples. For example, let me check (sqrt(9), sqrt(9), sqrt(63))? No, since they need to be integers. Alternatively, maybe (something else). Wait, but we already considered all possibilities. Wait, 81 can also be expressed as 9² + 0² + 0², but 0 isn't positive. So no. Therefore, only the three triples we found.

Therefore, the two c's are 216 and 224. Their sum is 216 + 224 = 440.

But let me check if the problem says that Jon is told that there are two possible values of c after Steve gives a. Therefore, the answer is 440. However, I need to check my calculations again.

First, check the triples:

Triple 1: 1, 4, 8. Sum a = 1 + 4 + 8 = 13. c = 2*1*4*8 = 64. Correct.

Triple 2: 4, 4, 7. Sum a=4 + 4 + 7 = 15. c = 2*4*4*7 = 2*112 = 224. Correct.

Triple 3: 3, 6, 6. Sum a=3 + 6 + 6 = 15. c = 2*3*6*6 = 2*108 = 216. Correct.

Yes, so two triples for a=15. Therefore, the sum is 216 + 224 = 440.

But wait, I have a doubt here. Because the problem says that Steve says, "Here is the value of a." Then Jon says, "There are still two possible values of c." So, that implies that when Steve gives a, there are two c's. Therefore, the value of a must be 15. Therefore, the answer is 216 + 224 = 440.

But let me check once again if there are other triples. Maybe I missed one? Let me think.

Wait, what about (5, 5, sqrt(81 - 25 -25)) = sqrt(31). Not integer. (5, 4, sqrt(81 - 25 - 16)) = sqrt(40). Not integer. What about (something with 9? Wait, 9² is 81. If one of the roots is 9, then r² + s² + t² = 81. So if one root is 9, then the other two roots must be zero, but they have to be positive integers. Therefore, no.

Another idea: could there be triples with numbers larger than 8? For example, if there is a root larger than 8, say 9, then as above, the sum would be too big. But since the sum of squares is 81, the maximum possible root is 9 (as 9²=81, and the others would have to be zero). So no. So no other triples.

Therefore, only three triples. Therefore, the answer should be 440. However, I feel that maybe there is a mistake here because 440 seems a bit large. Let me check the problem again.

Wait, the problem says the polynomial is 2x³ - 2a x² + (a² - 81)x - c. For the triple (1,4,8), a is 13. Then, the coefficient of x is (a² -81) = (169 - 81) = 88. Let's check if expanding 2(x - 1)(x - 4)(x - 8) gives the correct coefficients.

First, compute (x - 1)(x - 4)(x - 8):

First multiply (x - 1)(x - 4) = x² -5x + 4.

Multiply by (x - 8):

(x² -5x + 4)(x - 8) = x³ -8x² -5x² +40x +4x -32 = x³ -13x² +44x -32.

Multiply by 2: 2x³ -26x² +88x -64. Which matches the given polynomial with a=13 and c=64: since P(x) = 2x³ -2a x² + (a² -81)x -c. So with a=13, -2a is -26, and a² -81 is 169 -81=88, and c=64. Yes, matches. So that's correct.

For the triple (4,4,7), a=15, so the polynomial is 2x³ -2*15x² + (225 -81)x -c = 2x³ -30x² +144x -c. Let's compute 2(x -4)(x -4)(x -7). First, (x -4)^2 = x² -8x +16. Multiply by (x -7):

(x² -8x +16)(x -7) = x³ -7x² -8x² +56x +16x -112 = x³ -15x² +72x -112. Multiply by 2: 2x³ -30x² +144x -224. Therefore, c=224. Correct.

For the triple (3,6,6): a=15, so same polynomial 2x³ -30x² +144x -c. Compute 2(x -3)(x -6)(x -6). First compute (x -6)^2 = x² -12x +36. Multiply by (x -3):

(x² -12x +36)(x -3) = x³ -3x² -12x² +36x +36x -108 = x³ -15x² +72x -108. Multiply by 2: 2x³ -30x² +144x -216. Therefore, c=216. Correct.

Therefore, these are correct. Therefore, the two c's are 216 and 224. Their sum is 216 + 224 = 440.

But wait, the problem says "the sum of the two possible values of c". So, 440. But let me check if that's the answer. However, I recall that sometimes problems like this might have answers that are smaller numbers. Maybe I made a mistake in considering the triples?

Wait, let me check if there are any more triples. For example, maybe (2, 2, sqrt(81 -4 -4)) = sqrt(73). No. (2, 3, sqrt(81 -4 -9))=sqrt(68). No. (2, 5, sqrt(81 -4 -25))=sqrt(52). No. (3, 3, sqrt(81 -9 -9))=sqrt(63). No. (3, 4, sqrt(81 -9 -16))=sqrt(56). No. (3,5, sqrt(81 -9 -25))=sqrt(47). No. (4,5, sqrt(81 -16 -25))=sqrt(40). No. (4, 6, sqrt(81 -16 -36))=sqrt(29). No. (5, 6, sqrt(81 -25 -36))=sqrt(20). No. So no other triples. So seems correct.

Alternatively, perhaps I missed some permutations? But no, because we considered unordered triples. So yes.

Alternatively, maybe there is another a where multiple triples exist? For example, is there another a with multiple triples? But according to our previous analysis, only a=15 has two triples. The other a=13 has only one triple. Therefore, when Steve gave a specific value of a (which must be 15, since otherwise there would be only one c), Jon says there are two c's. Therefore, the answer is 216 + 224 = 440. Therefore, I think this is correct.

But let me check once again. Maybe the problem says "the sum of the two possible values of c", and maybe the answer is different. Let me check if 440 is correct. Alternatively, maybe there are more triples?

Wait, here's a thought: maybe the triple (2, 2, 7). 2² + 2² +7²=4 +4 +49=57≠81. No. (2, 3, 7): 4 +9 +49=62. (2, 4, 7): 4 +16 +49=69. (2, 5, 7): 4 +25 +49=78. (2, 6, 7): 4 +36 +49=89. No. (3, 3, 6): 9 +9 +36=54. (3, 4, 6): 9 +16 +36=61. (3, 5, 6): 9 +25 +36=70. (4, 4, 5): 16 +16 +25=57. No. So no other triples. Hence, only three triples. Therefore, answer is 440. Maybe the answer is 440. However, since the problem is from an Olympiad, maybe the answer is different? Wait, but I can't see any mistake in my logic. Therefore, I think the answer is 440.

But let me check once again the calculation for the sum of squares. For the triple (4,4,7): 16 +16 +49=81. Correct. For (3,6,6): 9 + 36 + 36=81. Correct. For (1,4,8): 1 + 16 + 64=81. Correct. All correct. Therefore, yes. Then the answer should be 216 + 224 = 440.

But let me check the problem statement again. It says that Steve says "I am thinking of a polynomial whose roots are all positive integers." So there are multiple such polynomials. Then Jon says there's more than one. Then Steve gives a value of a, and Jon still has two c's. Therefore, the answer is 440. So, I think that is the correct answer.

**Final Answer**
\boxed{440}

To solve this problem, we analyze the polynomial:

$$
P(x) = 2x^3 - 2a x^2 + (a^2 - 81)x - c
$$

Given that all the roots of this cubic polynomial are **positive integers**, we can express the polynomial in factored form as:

$$
P(x) = 2(x - r)(x - s)(x - t)
$$

where $ r, s, t $ are positive integers. Expanding this expression:

$$
P(x) = 2 \left[ x^3 - (r+s+t)x^2 + (rs + rt + st)x - rst \right]
$$

Now, equating to the given polynomial:

- Coefficient of $ x^2 $: $ -2(r + s + t) = -2a \Rightarrow r + s + t = a $
- Coefficient of $ x $: $ 2(rs + rt + st) = a^2 - 81 \Rightarrow rs + rt + st = \frac{a^2 - 81}{2} $
- Constant term: $ -2rst = -c \Rightarrow c = 2rst $

We conclude that:
- The sum of the roots: $ a = r + s + t $
- The sum of product pairs: $ rs + rt + st = \frac{a^2 - 81}{2} $
- The product of roots: $ rst = \frac{c}{2} $

---

### Step 1: Use the identity for the sum of squares

We take the identity:

$$
(r + s + t)^2 = r^2 + s^2 + t^2 + 2(rs + rt + st)
$$

Multiply both sides by 1 and rearrange:

$$
r^2 + s^2 + t^2 = a^2 - 2(rs + rt + st)
$$

Substitute $ rs + rt + st = \frac{a^2 - 81}{2} $:

$$
r^2 + s^2 + t^2 = a^2 - 2\left( \frac{a^2 - 81}{2} \right) = a^2 - (a^2 - 81) = 81
$$

So the **sum of squares of the roots is fixed**:

$$
r^2 + s^2 + t^2 = 81
$$

---

### Step 2: Find all positive integer solutions to $ r^2 + s^2 + t^2 = 81 $

We systematically find all sets of positive integers $ (r, s, t) $, with $ r \leq s \leq t $, that satisfy the sum of squares condition.

We test all combinations:

- $ (1, 4, 8) $: $ 1^2 + 4^2 + 8^2 = 1 + 16 + 64 = 81 $
- $ (3, 6, 6) $: $ 9 + 36 + 36 = 81 $
- $ (4, 4, 7) $: $ 16 + 16 + 49 = 81 $

These are the **only** valid positive integer triples.

---

### Step 3: Compute corresponding values of $ a $ and $ c $

Now for each triple:

#### Triple (1, 4, 8):

- $ a = 1 + 4 + 8 = 13 $
- $ c = 2 \times 1 \times 4 \times 8 = 64 $

#### Triple (3, 6, 6):

- $ a = 3 + 6 + 6 = 15 $
- $ c = 2 \times 3 \times 6 \times 6 = 216 $

#### Triple (4, 4, 7):

- $ a = 4 + 4 + 7 = 15 $
- $ c = 2 \times 4 \times 4 \times 7 = 224 $

---

### Step 4: Identify when there are two distinct values of $ c $

Note that:

- There is **only one** value of $ c $ for $ a = 13 $: **64**
- There are **two** values of $ c $ for $ a = 15 $: **216** and **224**

So, if Steve gives the value $ a = 15 $, Jon cannot uniquely determine $ c $ — there are **two possible values**.

Hence, the **sum of the two possible values of $ c $** is:

$$
216 + 224 = \boxed{440}
$$

---

### ✅ Final Answer:

$$
\boxed{440}
$$

Solution:
Let $r, s, t$ be the three positive integer roots of $P(x).$ Then by Vieta's formulas, \[\begin{aligned} r+s+t &= a, \\ rs+st+rt &= \frac{a^2-81}{2}, \\ rst &= \frac{c}{2}. \end{aligned}\]Substituting the first equation into the second to eliminate $a,$ we have \[rs+st+rt = \frac{(r+s+t)^2 - 81}{2} = \frac{(r^2+s^2+t^2) + 2(rs+st+rt) - 81}{2}.\]This simplifies to \[r^2 + s^2 + t^2 = 81.\]Therefore, each of $r, s, t$ lies in the set $\{1, 2, \ldots, 9\}.$ Assuming without loss of generality that $r \le s \le t,$ we have $81=r^2+s^2+t^2 \le 3t^2,$ so $t^2 \ge 27,$ and $t \ge 6.$ We take cases:

If $t = 6,$ then $r^2+s^2 = 81 - 6^2 = 45;$ the only solution where $r \le s \le 6$ is $(r, s) = (3, 6).$
If $t = 7,$ then $r^2+s^2 = 81-7^2 = 32;$ the only solution where $r \le s \le 7$ is $(r, s) = (4, 4).$
If $t = 8,$ then $r^2+s^2 = 81-8^2 = 17;$ the only solution where $r \le s \le 8$ is $(r, s) = (1, 4).$

Therefore, the possible sets of roots of such a polynomial are $(3, 6, 6), (4, 4, 7),$ and $(1, 4, 8).$ Calculating $a = r+s+t$ and $c=2rst$ for each set, we have $(a, c) = (15, 216), (15, 224), (13, 64).$

Since, given the value of $a,$ there are still two possible values of $c,$ it must be that $a = 15,$ since two of the pairs $(a, c)$ have $a = 15,$ but only one has $a = 13.$ Then the sum of the two possible values of $c$ is \[216 + 224 = \boxed{440}.\]