<Question>
A palindrome is a number that reads the same forwards and backwards. The sum of a particular set of three consecutive positive integers is a three-digit palindrome. If the sum is less than 220, what is the greatest possible value for the largest of the three integers in the set?
</Question>

<think>
<Plan_1>
Okay, let's try to solve this problem step by step. The question is about finding the greatest possible value for the largest of three consecutive positive integers, given that their sum is a three-digit palindrome less than 220.

First, let me make sure I understand the problem correctly. We have three consecutive numbers. Let's call them n, n+1, and n+2. Their sum would be n + (n+1) + (n+2) which simplifies to 3n + 3.
So the sum is 3n + 3. This sum is a three-digit palindrome, and it's less than 220. We need to find the maximum possible value of the largest number in the set, which is n+2.

A palindrome is a number that reads the same forwards and backwards. For a three-digit number, that means the first digit is the same as the last digit. So, examples of three-digit palindromes less than 220 would be 101, 111, 121, 131, ...., up to 212. Wait, 212 is the last three-digit palindrome before 220. But the sum has to be less than 220, so the maximum possible palindrome would be 212.

So, the possible palindromes we can have are from 101 up to 212.
Let me list them out to be thorough. Starting from 101, then 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212.

<prompt_1.1> List all three-digit palindromes less than 220. </prompt_1.1>

Since the sum is 3n + 3, which equals the palindrome, we can write 3n + 3 = palindrome. Therefore, n = (palindrome - 3)/3. Since n must be a positive integer, (palindrome - 3) must be divisible by 3. So, we need to check which of these palindromes minus 3 is divisible by 3, and then find the corresponding n, and from there the largest number n+2.

Our goal is to maximize the largest number, which is n+2. So, we should look for the largest possible palindrome less than 220, which when subtracted by 3 gives a multiple of 3, and then compute n and check if it's a positive integer.
The largest palindrome less than 220 is 212. Let's check if 212 - 3 = 209 is divisible by 3. 209 divided by 3 is approximately 69.666...., which is not an integer. So 212 is not a valid sum.

Next palindrome is 202. Let's check 202 - 3 = 199.
199 divided by 3 is approximately 66.333...., not an integer. So 202 is out.

Next is 191. 191 - 3 = 188.

<prompt_1.2> Check if 191, 181, and 171 meet the divisibility condition and result in positive integers for n. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
Wait, let me think. For three-digit numbers, the first digit can be 1 or 2. But since the sum is less than 220, the first digit can be 1 or 2. If the first digit is 2, then the last digit must also be 2, so the number would be 2X2, where X is the middle digit. The maximum such number less than 220 would be 212 because 222 is already over 220. So yes, the palindromes between 101 and 212.
</execution_1.1>

<execution_1.2>
188 divided by 3 is about 62.666...., not an integer. So nope.

Next palindrome: 181. 181 - 3 = 178. 178/3 ≈ 59.333...., not an integer.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.2:
171 - 3 = 168. 168 divided by 3 is 56. That's an integer! So n = 56. Therefore, the three consecutive numbers would be 56, 57, 58. Their sum is 56 + 57 + 58 = 171.
Wait, but 171 is a palindrome, and 58 is the largest number. But maybe there's a larger palindrome that also works? Let's check the palindromes between 171 and 212 to see if there are any others where (palindrome - 3) is divisible by 3.

<prompt_2.1> Check palindromes between 171 and 212 to find other possible values for the sum. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Wait, after 171, the next palindrome is 181, which we checked earlier, but 181 - 3 = 178, which isn't divisible by 3. Then 191: 191 - 3 = 188, not divisible by 3. Then 202: 202 - 3 = 199, not divisible by 3. Then 212: 212 - 3 = 209, not divisible by 3. So the only palindrome in that range where (palindrome - 3) is divisible by 3 is 171.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, but that seems too low. Let me double-check. Maybe I missed some palindromes? Let's list all three-digit palindromes less than 220 again:

<prompt_3.1> Re-evaluate the list of palindromes and verify calculations for divisibility by 3. </prompt_3.1>

But wait, let me confirm. Let's check if there are other palindromes between 171 and 212 where (palindrome - 3) is divisible by 3. Let's take 151.

<prompt_3.2> Re-evaluate palindromes between 171 and 212 for divisibility by 3. </prompt_3.2>

Hmm. So 171 is the only palindrome in that range that fits the criteria. Therefore, the answer would be 58.

<prompt_3.3> Verify the solution by checking calculations for the sum and divisibility. </prompt_3.3>
</Plan_3>

<Execution_3>
<execution_3.1>
101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212.

Wait, that's correct. So between 171 and 212, the next palindrome is 202, then 212. As checked earlier, neither 202 nor 212 work.
</execution_3.1>

<execution_3.2>
151 - 3 = 148. 148/3 ≈ 49.333...., not integer. 141 - 3 = 138. 138/3 = 46. So n = 46. Then the numbers are 46, 47, 48. Sum is 138, which is a palindrome. But that's smaller. Then 131 -3=128, not divisible by 3. 121-3=118, nope. 111-3=108, 108/3=36. So n=36. Numbers: 36,37,38. Sum 111. Also smaller. 101-3=98, 98/3≈32.666, not integer.
</execution_3.2>

<execution_3.3>
But wait, maybe there's a mistake here. Let me check another way. Let's see, 3n + 3 is the sum. So 3(n + 1) is the sum. Therefore, the sum must be divisible by 3. So the palindrome must be divisible by 3. Wait, but we subtracted 3 to get 3n. Wait, actually, if 3n + 3 is divisible by 3, then the palindrome is divisible by 3. So perhaps another approach is to find palindromes less than 220 that are divisible by 3, and then check if (palindrome - 3)/3 is an integer. Wait, no. If 3n + 3 is the palindrome, then palindrome ≡ 0 mod 3. Because 3n is divisible by 3, and adding 3 (which is also divisible by 3) gives another multiple of 3. Therefore, the palindrome must be divisible by 3. Therefore, instead of checking (palindrome - 3) divisible by 3, it's actually the palindrome itself must be divisible by 3. Because 3n + 3 = 3(n + 1). So the sum is a multiple of 3. Therefore, the palindrome must be divisible by 3. Therefore, we need to find three-digit palindromes less than 220 that are divisible by 3. Then, the largest such palindrome would give the largest n. Let's see. That approach might be more straightforward.

So, first, list all three-digit palindromes less than 220, then check which ones are divisible by 3, then find the maximum among them. Let's try this approach.

Three-digit palindromes less than 220: The first digit is 1 or 2. If the first digit is 1, the number is 1X1, where X is any digit from 0 to 9. So 101, 111, 121, 131, 141, 151, 161, 171, 181, 191. If the first digit is 2, then the number is 2X2, but since the sum is less than 220, 2X2 must be less than 220. So 202, 212. Because 222 is 222, which is over 220. So palindromes are 202 and 212.

So the list of palindromes is: 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212.

Now, check which of these are divisible by 3. A number is divisible by 3 if the sum of its digits is divisible by 3. Let's compute the digit sums:
</execution_3.3>
</Execution_3>

<Plan_4>
Based on execution_3.3:
- 101: 1 + 0 + 1 = 2 → Not divisible by 3
- 111: 1 + 1 + 1 = 3 → Divisible by 3
- 121: 1 + 2 + 1 = 4 → Not
- 131: 1 + 3 + 1 = 5 → Not
- 141: 1 + 4 + 1 = 6 → Divisible by 3
- 151: 1 + 5 + 1 = 7 → Not
- 161: 1 + 6 + 1 = 8 → Not
- 171: 1 + 7 + 1 = 9 → Divisible by 3
- 181: 1 + 8 + 1 = 10 → Not
- 191: 1 + 9 + 1 = 11 → Not
- 202: 2 + 0 + 2 = 4 → Not
- 212: 2 + 1 + 2 = 5 → Not

So the palindromes divisible by 3 are: 111, 141, 171.

<prompt_4.1> Identify the largest palindrome and calculate the largest number in the set. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Among these, the largest is 171. So the sum is 171. Then, the largest number in the set would be (171 - 3)/3 + 1. Wait, let's see. Wait, the sum is 3n + 3. So 3n + 3 = 171. Therefore, 3n = 171 - 3 = 168. So n = 168 / 3 = 56. Therefore, the three numbers are 56, 57, 58. The largest is 58. So that's the answer. But wait, are there any other palindromes between 171 and 220 that are divisible by 3? From the list above, the next palindrome after 171 is 181, which isn't divisible by 3, then 191 (not), 202 (not), 212 (not). So 171 is the largest such palindrome. Therefore, the answer is 58.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But let me just check once more. Maybe there's a mistake in the digit sum for 212 or 202? Let's verify:

<prompt_5.1> Verify the calculations for divisibility and confirm the largest possible value. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
212: 2 + 1 + 2 = 5. 5 isn't divisible by 3. 202: 2 + 0 + 2 = 4. Also not. So no. So yes, 171 is the largest palindrome less than 220 that's divisible by 3. Therefore, the largest number is 58.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
But hold on, wait a second. Let me check if the problem says "three consecutive positive integers". So n must be a positive integer. In our case, n = 56, which is positive, so that's okay. But just to make sure, let's check if there's a palindrome between 171 and 220 that's divisible by 3. For example, 180 is not a palindrome, 183, 186, etc. , but they wouldn't be palindromes. The next palindrome after 171 is 181, which we already checked. So no, there's no other palindrome in that range. Therefore, 171 is indeed the largest possible sum, leading to the largest number being 58.

But let me just verify once more. Let's take n = 56. Then the numbers are 56, 57, 58. Sum is 56 + 57 + 58 = 171. 171 is a palindrome (reads the same forwards and backwards), and 171 is less than 220. So that satisfies all conditions. Therefore, the greatest possible value for the largest number is 58.

Wait, but just to be thorough, is there a case where maybe the sum is a palindrome but not divisible by 3? Wait, no. Because the sum is 3n + 3, which is 3(n + 1). So it's inherently divisible by 3. Therefore, the sum must be a three-digit palindrome divisible by 3. Therefore, our approach was correct.

<prompt_6.1> Finalize and confirm the solution. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
So the answer is 58. I think that's solid. I don't see any mistakes here. The key steps were recognizing that the sum must be a multiple of 3 (since it's 3n + 3), listing all three-digit palindromes less than 220, checking which ones are divisible by 3, and selecting the largest such palindrome. Then solving for n gives the sequence, and the largest number is n + 2. All steps check out.
</execution_6.1>
</Execution_6>

<Final_answer>
58. Thus, the greatest possible value for the largest of the three integers in the set is $\boxed{58}$.
</Final_answer>
</think>