Explanation:
Okay, so I need to figure out how many points with integer coordinates are strictly inside this rectangle. The rectangle has vertices at (5,4), (-5,4), (-5,-4), and (5,-4). Let me visualize this first. The rectangle is centered at the origin, stretching from -5 to 5 on the x-axis and from -4 to 4 on the y-axis. So, the sides are vertical and horizontal lines.

Since the points have to be strictly inside the rectangle, I can't include the boundary points. That means the x-coordinates have to be greater than -5 and less than 5, and the y-coordinates have to be greater than -4 and less than 4. Wait, let me confirm that.

If the rectangle's vertices are at (5,4), (-5,4), (-5,-4), (5,-4), then the sides are the lines x = 5, x = -5, y = 4, and y = -4. So the region inside the rectangle would be all points (x, y) such that -5 < x < 5 and -4 < y < 4. Since the problem says "strictly in the interior," that means the coordinates can't be on the edges. So, for the x-values, they must be integers between -4 and 4, but not including -5 and 5. Similarly, the y-values must be integers between -3 and 3, not including -4 and 4.

Wait, let me think again. If the original rectangle goes from x = -5 to x = 5, then the interior points would have x strictly greater than -5 and strictly less than 5. Since x must be an integer, the possible integer values for x would be from -4 up to 4. Because -5 is excluded, so the next integer is -4, and similarly, 5 is excluded, so the next integer is 4.

Similarly for y: the rectangle goes from y = -4 to y = 4. So the interior points must have y strictly greater than -4 and strictly less than 4. Therefore, integer y-coordinates would be -3, -2, -1, 0, 1, 2, 3. So from -3 up to 3.

Therefore, to compute the number of integer points strictly inside the rectangle, I need to compute the number of possible x-values multiplied by the number of possible y-values.

So first, let me find how many integer x-values are there between -4 and 4 inclusive. Let me count: starting from -4, -3, -2, -1, 0, 1, 2, 3, 4. That's 9 values. Wait, let me check: from -4 to 4 inclusive, the number of integers is 4 - (-4) + 1 = 9? Wait, formula for number of integers from a to b inclusive is b - a + 1. So here, from -4 to 4: 4 - (-4) + 1 = 4 + 4 + 1 = 9. Yes. So 9 x-values.

Similarly for y-values: from -3 to 3 inclusive. Let me compute that. 3 - (-3) + 1 = 3 + 3 + 1 = 7. So 7 y-values.

Therefore, total number of interior points is 9 * 7 = 63.

Wait, but let me make sure I didn't make a mistake here. Because sometimes when counting, especially with negative numbers, it's easy to slip up. Let me count the x-values again. The integers strictly between -5 and 5. So integers greater than -5 and less than 5. So starting from -4, -3, -2, -1, 0, 1, 2, 3, 4. So that's 9 numbers. Similarly for y: strictly between -4 and 4. So integers greater than -4 and less than 4. So -3, -2, -1, 0, 1, 2, 3. That's 7 numbers. So 9*7 is indeed 63. Hmm.

Alternatively, another way to think about it: the rectangle has length and width. The length along the x-axis is from -5 to 5, which is 10 units. But since we are excluding the boundaries, the interior length is 10 - 2 = 8 units? Wait, but how does that translate to integer points? Wait, maybe not. Because if you have a line from -5 to 5, the number of integer points on the boundary is 11 (from -5 to 5 inclusive). But if you want strictly inside, you exclude those on the boundary. However, in terms of the number of integer points inside, it's not directly the length minus 2. Because the length in terms of units is 10, but the number of integer points is different.

But perhaps my initial approach is correct. Let me think again. For the x-coordinate, the interior points must satisfy -5 < x < 5. Since x is integer, x can be -4, -3, -2, -1, 0, 1, 2, 3, 4. So that's 9 points. Similarly for y: -4 < y < 4, so y can be -3, -2, -1, 0, 1, 2, 3. That's 7 points. Multiplying them gives 9*7 = 63. So that seems correct.

But just to be absolutely sure, maybe I can think of a smaller example. For example, suppose the rectangle is from (0,0) to (2,2). Then the interior points would be (1,1), so only 1 point. Using the same method: for x, strictly between 0 and 2, so x=1, which is 1 value. Similarly y=1, so 1*1=1. Correct.

Another example: rectangle from (-1, -1) to (1, 1). Then interior points would be strictly inside, so x: -0. So x must be between -1 and 1, so x can be 0. Similarly y can be 0. So only (0,0). So 1*1=1. Correct. If the rectangle were from (-2, -2) to (2, 2), then interior points would be x from -1, 0, 1 (3 values), y similarly, so 3*3=9. Which is correct. For example, points (-1,-1), (-1,0), (-1,1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1). That's 9 points. So that seems to check out.

So applying the same logic to the original problem, with x from -4 to 4 inclusive (9 values) and y from -3 to 3 inclusive (7 values). So 9*7=63. Therefore, I think that is the answer.

But let me cross-verify with another method. Maybe using the formula for the number of integer points in a rectangle. The total number of integer points in the rectangle including boundaries is (number of x's from -5 to 5 inclusive) times (number of y's from -4 to 4 inclusive). That would be (5 - (-5) + 1) * (4 - (-4) + 1) = (11)*(9) = 99. But this includes the boundary points. Then, the number of interior points would be total points minus the boundary points. Wait, but how do we calculate the boundary points?

Alternatively, maybe it's easier to calculate interior points directly as we did before. But just to check, let me try this alternative method.

Total number of points in the rectangle (including boundary): for x from -5 to 5 inclusive, that's 11 points. For y from -4 to 4 inclusive, that's 9 points. So total points: 11*9=99. Now, how many interior points? Well, interior points are those not on the boundary. The boundary includes all points on the edges. But how do we compute that?

Alternatively, using inclusion-exclusion, but that might be complicated. But maybe another way. The interior points are those strictly inside, which is the same as (x from -4 to 4) and (y from -3 to 3), which gives 9*7=63. Therefore, if total points is 99, then the boundary points would be 99 - 63 = 36. Let me check if that makes sense.

Alternatively, compute the boundary points. The rectangle has four sides. The total boundary points can be calculated as the sum of the four sides, but subtracting the corners which are counted twice.

So, for the top side: y=4, x from -5 to 5. That's 11 points. Similarly, bottom side: y=-4, x from -5 to 5. Also 11 points. Left side: x=-5, y from -4 to 4. That's 9 points. Right side: x=5, y from -4 to 4. 9 points. So total boundary points would be 11 + 11 + 9 + 9 = 40. But then we have to subtract the corners which are counted twice. Each corner is counted in two sides. There are four corners. So subtract 4. So total boundary points: 40 - 4 = 36. Which matches the previous number. So total interior points = total points - boundary points = 99 - 36 = 63. Which matches our original calculation. Therefore, 63 is correct.

So, that gives me more confidence that 63 is the right answer.

But just to make sure, let me think of another way. For example, how many integer points are strictly inside? For the x-coordinate, the interior is from -5 < x < 5. Since x is integer, that is x = -4, -3, -2, -1, 0, 1, 2, 3, 4. So that is 9 values. Similarly for y: -4 < y < 4, so y = -3, -2, -1, 0, 1, 2, 3. Which is 7 values. So 9 * 7 = 63. Yep.

Another sanity check: if the rectangle was from (-a, -b) to (a, b), then the number of interior points would be (2a - 1) * (2b - 1). Wait, let's check with the previous examples.

In the example where the rectangle was from (-1, -1) to (1, 1), the interior points would be (0,0), which is 1 point. According to the formula: (2*1 -1)*(2*1 -1) = 1*1=1. Correct. For the rectangle from (-2, -2) to (2,2), the interior points would be (x from -1, 0, 1) and y similarly. So 3*3=9. Formula: (2*2 -1)*(2*2 -1)= 3*3=9. Correct. So in general, for a rectangle from (-a, -b) to (a, b), the number of interior points is (2a -1)*(2b -1). Wait, but in our problem, the rectangle is from (-5, -4) to (5, 4). Therefore, a=5, b=4. Then the formula gives (2*5 -1)*(2*4 -1) = (10 -1)*(8 -1)= 9*7=63. Which matches our answer. So that formula is also confirming that 63 is correct. Therefore, this seems solid.

Therefore, I think the answer is 63.

**Final Answer**
The number of integer coordinate points strictly inside the rectangular region is \boxed{63}.

To determine how many integer-coordinate points lie **strictly in the interior** of the rectangular region defined by the vertices $(-5,4)$, $(-5,-4)$, $(5,-4)$, and $(5,4)$, we begin by analyzing the boundaries of this rectangle.

---

### Step 1: Understand the Bounding Region

This is a rectangle centered at the origin:

- The **x-coordinates** of the boundary range from $x = -5$ to $x = 5$,
- The **y-coordinates** of the boundary range from $y = -4$ to $y = 4$.

We are interested in **integer-coordinate points strictly inside** the rectangle, i.e., points whose coordinates satisfy:

$$
-5 < x < 5 \quad \text{and} \quad -4 < y < 4
$$

---

### Step 2: Determine the Possible Integer Values

- For **x-values**:
  The integers strictly between $-5$ and $5$ are:
  $$
  -4, -3, -2, -1, 0, 1, 2, 3, 4 \quad \Rightarrow \quad 9 \text{ values}
  $$

- For **y-values**:
  The integers strictly between $-4$ and $4$ are:
  $$
  -3, -2, -1, 0, 1, 2, 3 \quad \Rightarrow \quad 7 \text{ values}
  $$

---

### Step 3: Compute the Total Number of Interior Points

Since every valid $x$-coordinate can pair with every valid $y$-coordinate (and all such pairs are strictly inside the rectangle), the total number of interior integer-coordinate points is:

$$
9 \times 7 = 63
$$

---

### Final Answer

$$
\boxed{63}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many points with integer coordinates are strictly inside this rectangle. The rectangle has vertices at (5,4), (-5,4), (-5,-4), and (5,-4). Let me visualize this first. The rectangle is centered at the origin, stretching from -5 to 5 on the x-axis and from -4 to 4 on the y-axis. So, the sides are vertical and horizontal lines.

Since the points have to be strictly inside the rectangle, I can't include the boundary points. That means the x-coordinates have to be greater than -5 and less than 5, and the y-coordinates have to be greater than -4 and less than 4. Wait, let me confirm that.

If the rectangle's vertices are at (5,4), (-5,4), (-5,-4), (5,-4), then the sides are the lines x = 5, x = -5, y = 4, and y = -4. So the region inside the rectangle would be all points (x, y) such that -5 < x < 5 and -4 < y < 4. Since the problem says "strictly in the interior," that means the coordinates can't be on the edges. So, for the x-values, they must be integers between -4 and 4, but not including -5 and 5. Similarly, the y-values must be integers between -3 and 3, not including -4 and 4.

Wait, let me think again. If the original rectangle goes from x = -5 to x = 5, then the interior points would have x strictly greater than -5 and strictly less than 5. Since x must be an integer, the possible integer values for x would be from -4 up to 4. Because -5 is excluded, so the next integer is -4, and similarly, 5 is excluded, so the next integer is 4.

Similarly for y: the rectangle goes from y = -4 to y = 4. So the interior points must have y strictly greater than -4 and strictly less than 4. Therefore, integer y-coordinates would be -3, -2, -1, 0, 1, 2, 3. So from -3 up to 3.

Therefore, to compute the number of integer points strictly inside the rectangle, I need to compute the number of possible x-values multiplied by the number of possible y-values.

So first, let me find how many integer x-values are there between -4 and 4 inclusive. Let me count: starting from -4, -3, -2, -1, 0, 1, 2, 3, 4. That's 9 values. Wait, let me check: from -4 to 4 inclusive, the number of integers is 4 - (-4) + 1 = 9? Wait, formula for number of integers from a to b inclusive is b - a + 1. So here, from -4 to 4: 4 - (-4) + 1 = 4 + 4 + 1 = 9. Yes. So 9 x-values.

Similarly for y-values: from -3 to 3 inclusive. Let me compute that. 3 - (-3) + 1 = 3 + 3 + 1 = 7. So 7 y-values.

Therefore, total number of interior points is 9 * 7 = 63.

Wait, but let me make sure I didn't make a mistake here. Because sometimes when counting, especially with negative numbers, it's easy to slip up. Let me count the x-values again. The integers strictly between -5 and 5. So integers greater than -5 and less than 5. So starting from -4, -3, -2, -1, 0, 1, 2, 3, 4. So that's 9 numbers. Similarly for y: strictly between -4 and 4. So integers greater than -4 and less than 4. So -3, -2, -1, 0, 1, 2, 3. That's 7 numbers. So 9*7 is indeed 63. Hmm.

Alternatively, another way to think about it: the rectangle has length and width. The length along the x-axis is from -5 to 5, which is 10 units. But since we are excluding the boundaries, the interior length is 10 - 2 = 8 units? Wait, but how does that translate to integer points? Wait, maybe not. Because if you have a line from -5 to 5, the number of integer points on the boundary is 11 (from -5 to 5 inclusive). But if you want strictly inside, you exclude those on the boundary. However, in terms of the number of integer points inside, it's not directly the length minus 2. Because the length in terms of units is 10, but the number of integer points is different.

But perhaps my initial approach is correct. Let me think again. For the x-coordinate, the interior points must satisfy -5 < x < 5. Since x is integer, x can be -4, -3, -2, -1, 0, 1, 2, 3, 4. So that's 9 points. Similarly for y: -4 < y < 4, so y can be -3, -2, -1, 0, 1, 2, 3. That's 7 points. Multiplying them gives 9*7 = 63. So that seems correct.

But just to be absolutely sure, maybe I can think of a smaller example. For example, suppose the rectangle is from (0,0) to (2,2). Then the interior points would be (1,1), so only 1 point. Using the same method: for x, strictly between 0 and 2, so x=1, which is 1 value. Similarly y=1, so 1*1=1. Correct.

Another example: rectangle from (-1, -1) to (1, 1). Then interior points would be strictly inside, so x: -0. So x must be between -1 and 1, so x can be 0. Similarly y can be 0. So only (0,0). So 1*1=1. Correct. If the rectangle were from (-2, -2) to (2, 2), then interior points would be x from -1, 0, 1 (3 values), y similarly, so 3*3=9. Which is correct. For example, points (-1,-1), (-1,0), (-1,1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1). That's 9 points. So that seems to check out.

So applying the same logic to the original problem, with x from -4 to 4 inclusive (9 values) and y from -3 to 3 inclusive (7 values). So 9*7=63. Therefore, I think that is the answer.

But let me cross-verify with another method. Maybe using the formula for the number of integer points in a rectangle. The total number of integer points in the rectangle including boundaries is (number of x's from -5 to 5 inclusive) times (number of y's from -4 to 4 inclusive). That would be (5 - (-5) + 1) * (4 - (-4) + 1) = (11)*(9) = 99. But this includes the boundary points. Then, the number of interior points would be total points minus the boundary points. Wait, but how do we calculate the boundary points?

Alternatively, maybe it's easier to calculate interior points directly as we did before. But just to check, let me try this alternative method.

Total number of points in the rectangle (including boundary): for x from -5 to 5 inclusive, that's 11 points. For y from -4 to 4 inclusive, that's 9 points. So total points: 11*9=99. Now, how many interior points? Well, interior points are those not on the boundary. The boundary includes all points on the edges. But how do we compute that?

Alternatively, using inclusion-exclusion, but that might be complicated. But maybe another way. The interior points are those strictly inside, which is the same as (x from -4 to 4) and (y from -3 to 3), which gives 9*7=63. Therefore, if total points is 99, then the boundary points would be 99 - 63 = 36. Let me check if that makes sense.

Alternatively, compute the boundary points. The rectangle has four sides. The total boundary points can be calculated as the sum of the four sides, but subtracting the corners which are counted twice.

So, for the top side: y=4, x from -5 to 5. That's 11 points. Similarly, bottom side: y=-4, x from -5 to 5. Also 11 points. Left side: x=-5, y from -4 to 4. That's 9 points. Right side: x=5, y from -4 to 4. 9 points. So total boundary points would be 11 + 11 + 9 + 9 = 40. But then we have to subtract the corners which are counted twice. Each corner is counted in two sides. There are four corners. So subtract 4. So total boundary points: 40 - 4 = 36. Which matches the previous number. So total interior points = total points - boundary points = 99 - 36 = 63. Which matches our original calculation. Therefore, 63 is correct.

So, that gives me more confidence that 63 is the right answer.

But just to make sure, let me think of another way. For example, how many integer points are strictly inside? For the x-coordinate, the interior is from -5 < x < 5. Since x is integer, that is x = -4, -3, -2, -1, 0, 1, 2, 3, 4. So that is 9 values. Similarly for y: -4 < y < 4, so y = -3, -2, -1, 0, 1, 2, 3. Which is 7 values. So 9 * 7 = 63. Yep.

Another sanity check: if the rectangle was from (-a, -b) to (a, b), then the number of interior points would be (2a - 1) * (2b - 1). Wait, let's check with the previous examples.

In the example where the rectangle was from (-1, -1) to (1, 1), the interior points would be (0,0), which is 1 point. According to the formula: (2*1 -1)*(2*1 -1) = 1*1=1. Correct. For the rectangle from (-2, -2) to (2,2), the interior points would be (x from -1, 0, 1) and y similarly. So 3*3=9. Formula: (2*2 -1)*(2*2 -1)= 3*3=9. Correct. So in general, for a rectangle from (-a, -b) to (a, b), the number of interior points is (2a -1)*(2b -1). Wait, but in our problem, the rectangle is from (-5, -4) to (5, 4). Therefore, a=5, b=4. Then the formula gives (2*5 -1)*(2*4 -1) = (10 -1)*(8 -1)= 9*7=63. Which matches our answer. So that formula is also confirming that 63 is correct. Therefore, this seems solid.

Therefore, I think the answer is 63.

**Final Answer**
The number of integer coordinate points strictly inside the rectangular region is \boxed{63}.

To determine how many integer-coordinate points lie **strictly in the interior** of the rectangular region defined by the vertices $(-5,4)$, $(-5,-4)$, $(5,-4)$, and $(5,4)$, we begin by analyzing the boundaries of this rectangle.

---

### Step 1: Understand the Bounding Region

This is a rectangle centered at the origin:

- The **x-coordinates** of the boundary range from $x = -5$ to $x = 5$,
- The **y-coordinates** of the boundary range from $y = -4$ to $y = 4$.

We are interested in **integer-coordinate points strictly inside** the rectangle, i.e., points whose coordinates satisfy:

$$
-5 < x < 5 \quad \text{and} \quad -4 < y < 4
$$

---

### Step 2: Determine the Possible Integer Values

- For **x-values**:
  The integers strictly between $-5$ and $5$ are:
  $$
  -4, -3, -2, -1, 0, 1, 2, 3, 4 \quad \Rightarrow \quad 9 \text{ values}
  $$

- For **y-values**:
  The integers strictly between $-4$ and $4$ are:
  $$
  -3, -2, -1, 0, 1, 2, 3 \quad \Rightarrow \quad 7 \text{ values}
  $$

---

### Step 3: Compute the Total Number of Interior Points

Since every valid $x$-coordinate can pair with every valid $y$-coordinate (and all such pairs are strictly inside the rectangle), the total number of interior integer-coordinate points is:

$$
9 \times 7 = 63
$$

---

### Final Answer

$$
\boxed{63}
$$

Solution:
The rectangular region is 10 units by 8 units, resulting in an 8 by 6 rectangular region in the interior, which forms a 9-by-7 array of lattice points. That's $\boxed{63}$ points with integer coordinates, as shown in the figure. [asy]
import olympiad; size(150); defaultpen(linewidth(0.8));
add(grid(10,8));
draw((1,1)--(9,1)--(9,7)--(1,7)--cycle,linewidth(1.2));
[/asy] Note: We are counting points, not squares. It is a common mistake to count the interior squares, for 48, instead of interior lattice points, which gives the correct answer of 63.


Explanation:
Okay, so I need to find the value of x_{531} + x_{753} + x_{975} given this recurrence relation. Let me start by understanding the problem step by step.

First, the initial terms are given: x₁ = 211, x₂ = 375, x₃ = 420, x₄ = 523. Then, for n ≥ 5, each term is defined by the recurrence relation xₙ = xₙ₋₁ - xₙ₋₂ + xₙ₋₃ - xₙ₋₄. So, each term depends on the previous four terms.

My goal is to find x_{531} + x_{753} + x_{975}. Since calculating each term up to 975 directly would be impractical, there must be some pattern or periodicity in the sequence that I can exploit.

Let me recall that linear recurrence relations often have periodic solutions or eventually become periodic if the characteristic equation has roots that are roots of unity or something like that. Maybe I can compute the terms until I find a repeating cycle, then use that cycle to find the values at positions 531, 753, and 975.

First, let me try to compute some terms of the sequence to see if there's a pattern.

Given:
x₁ = 211
x₂ = 375
x₃ = 420
x₄ = 523

Now, compute x₅:

x₅ = x₄ - x₃ + x₂ - x₁ = 523 - 420 + 375 - 211

Let me calculate that step by step:

523 - 420 = 103

103 + 375 = 478

478 - 211 = 267

So x₅ = 267

Now x₆ = x₅ - x₄ + x₃ - x₂

Compute:

267 - 523 + 420 - 375

Compute step by step:

267 - 523 = -256

-256 + 420 = 164

164 - 375 = -211

So x₆ = -211

x₇ = x₆ - x₅ + x₄ - x₃

-211 - 267 + 523 - 420

Compute:

-211 - 267 = -478

-478 + 523 = 45

45 - 420 = -375

x₇ = -375

x₈ = x₇ - x₆ + x₅ - x₄

-375 - (-211) + 267 - 523

Which is -375 + 211 + 267 - 523

Compute step by step:

-375 + 211 = -164

-164 + 267 = 103

103 - 523 = -420

x₈ = -420

x₉ = x₈ - x₇ + x₆ - x₅

-420 - (-375) + (-211) - 267

Which is -420 + 375 - 211 - 267

Compute:

-420 + 375 = -45

-45 - 211 = -256

-256 - 267 = -523

x₉ = -523

x₁₀ = x₉ - x₈ + x₇ - x₆

-523 - (-420) + (-375) - (-211)

Which is -523 + 420 - 375 + 211

Calculating step by step:

-523 + 420 = -103

-103 - 375 = -478

-478 + 211 = -267

x₁₀ = -267

x₁₁ = x₁₀ - x₉ + x₈ - x₇

-267 - (-523) + (-420) - (-375)

Which is -267 + 523 - 420 + 375

Compute:

-267 + 523 = 256

256 - 420 = -164

-164 + 375 = 211

x₁₁ = 211

Wait a minute! x₁₁ is 211, which is the same as x₁. Let me check if this is a cycle.

Let me write down all the terms so far:

n: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,...

x: 211, 375, 420, 523, 267, -211, -375, -420, -523, -267, 211,...

So x₁₁ is 211, which is same as x₁. Let me check if the next terms repeat the original sequence.

Let me compute x₁₂:

x₁₂ = x₁₁ - x₁₀ + x₉ - x₈

211 - (-267) + (-523) - (-420)

Which is 211 + 267 - 523 + 420

Calculate:

211 + 267 = 478

478 - 523 = -45

-45 + 420 = 375

x₁₂ = 375, which is same as x₂.

Similarly, x₁₃ would be x₁₂ - x₁₁ + x₁₀ - x₉

375 - 211 + (-267) - (-523)

Which is 375 - 211 - 267 + 523

Compute:

375 - 211 = 164

164 - 267 = -103

-103 + 523 = 420

Which is x₃ = 420. So x₁₃ = x₃.

So it seems like starting from x₁, the sequence repeats every 10 terms? Wait, because x₁₁ is x₁, x₁₂ is x₂, x₁₃ is x₃, etc. So the period is 10? Let me confirm with x₁₄:

x₁₄ = x₁₃ - x₁₂ + x₁₁ - x₁₀

420 - 375 + 211 - (-267)

Which is 420 - 375 + 211 + 267

Compute:

420 - 375 = 45

45 + 211 = 256

256 + 267 = 523, which is x₄. So x₁₄ = x₄.

Therefore, the sequence is periodic with period 10. Let me check one more term to be sure. x₁₅:

x₁₅ = x₁₄ - x₁₃ + x₁₂ - x₁₁ = 523 - 420 + 375 - 211

Which is the same as x₅, which was 267. So yes, the pattern repeats every 10 terms.

So the sequence has period 10. That is, for any n, x_{n} = x_{n+10}.

Therefore, if I can find the value of n modulo 10, I can find the corresponding term.

Therefore, to compute x_{531}, x_{753}, x_{975}, I need to find 531 mod 10, 753 mod 10, 975 mod 10. Then look up the corresponding term in the sequence.

But let me confirm the period again. Since x₁ = x₁₁, x₂ = x₁₂, etc., so the period is 10. Let me check if the period is 10 or maybe a smaller period. For example, maybe it's 5? Let me check.

Looking at the terms:

x₁ = 211, x₂ = 375, x₃ = 420, x₄ = 523, x₅ = 267, x₆ = -211, x₇ = -375, x₈ = -420, x₉ = -523, x₁₀ = -267, x₁₁ = 211...

If I check if the sequence repeats every 10 terms, then yes. But maybe there is a smaller period? Let me check if the sequence repeats before 10. For example, check if x₁ = x₁ + 5? Let's see:

x₁ = 211, x₆ = -211, which is -x₁. Not the same. Similarly, x₂ = 375, x₇ = -375. So negative of x₂. Similarly for others. So the period isn't 5, but 10. Because if you go another 5 terms, you get back to original terms. For example, x₁₁ = x₁, x₁₂ = x₂, etc. So the period is 10.

Therefore, the sequence is periodic with period 10. So to find x_n, we can compute n mod 10. However, since the period is 10, we need to check if n mod 10 is 0, then it would correspond to x_{10}, otherwise it is the remainder.

But let me check. For example, n = 10: 10 mod 10 is 0, and x_{10} = -267. If n = 20, then x_{20} should be same as x_{10}, which is -267. Let me verify by computing x₁₅, x₁₆,...

But maybe I can just go with the period 10.

Thus, for any n, x_n = x_{n mod 10}, but with the caveat that if n mod 10 is 0, then it refers to x_{10} and not x_0 (since we don't have x_0). So, the mapping is:

n mod 10 = 1 → x₁

n mod 10 = 2 → x₂

...

n mod 10 = 0 → x_{10}

Therefore, for n = 531, let's compute 531 divided by 10. Let me compute 531 / 10: 10*53=530, so 531 = 10*53 +1 → remainder 1. Therefore, 531 mod 10 = 1. Therefore, x_{531} = x₁ = 211.

Similarly, 753 divided by 10: 10*75=750, so 753 = 10*75 + 3 → remainder 3. Therefore, 753 mod 10 = 3. Therefore, x_{753} = x₃ = 420.

Similarly, 975 divided by 10: 10*97 = 970, so 975 = 10*97 +5 → remainder 5. Therefore, 975 mod 10 =5. Therefore, x_{975} =x₅=267.

Therefore, x_{531} + x_{753} + x_{975} = 211 + 420 + 267.

Let me compute that:

211 + 420 = 631

631 + 267 = 898

Wait, so the answer would be 898?

But let me double-check my calculations. Because sometimes when dealing with modular arithmetic, it's easy to make an off-by-one error, especially with the modulus.

Let me verify the mod calculations:

For 531:

Divide 531 by 10: 10*53 = 530. 531 - 530 =1. So remainder 1. So yes, 531 mod10=1. Therefore, x_{531} =x₁=211. Correct.

For 753:

Divide 753 by 10: 10*75=750, 753-750=3. So remainder 3. Therefore, x_{753}=x₃=420. Correct.

For 975: 10*97=970, 975-970=5. Remainder 5. Therefore, x_{975}=x₅=267. Correct.

Therefore the sum is 211 + 420 + 267. Let me check that again:

211 + 420:

200 + 400 = 600, 11 + 20 = 31, so total 631. Then 631 + 267:

600 + 200 = 800, 31 + 67 = 98, so total 898. Yes, that's correct.

Wait, but before I get too confident, let me check if my assumption of period 10 is correct. Because if the sequence is periodic with period 10, then starting from term 1, every 10 terms it repeats.

But let me check another term beyond x₁₁. Let me compute x₁₂ (we already did), which was 375, which is x₂. x₁₃ = 420 =x₃. x₁₄=523=x₄. x₁₅=267=x₅. x₁₆=-211=x₆. x₁₇=-375=x₇. x₁₈=-420=x₈. x₁₉=-523=x₉. x₂₀=-267=x₁₀. x₂₁=211=x₁. Yep, so that's consistent. Therefore, the period is indeed 10. Therefore, the calculation is correct.

Therefore, the answer is 898. Wait, but let me check again the original terms:

x₁ = 211, x₂=375, x₃=420, x₄=523, x₅=267. So yes, x_{531} is x₁=211, x_{753} is x₃=420, x_{975} is x₅=267. Sum is 211+420+267=898.

But hold on, let me check if there is any possibility that the modulus is 10 but the mapping is different? For example, sometimes when people talk about modulus, they might consider 0 to be equivalent to 10. But since our terms start at n=1, so if the remainder is 0, then it's equivalent to term 10. So, for example, n=10: 10 mod10=0, so we take term 10. n=20: same as n=10. So that's correct.

Alternatively, if I think in terms of indices, the sequence has period 10, so x_{n} = x_{n+10} for all n ≥1. Therefore, to find x_n, compute n mod 10, and if it's 0, use term 10, else use the term corresponding to the remainder. Therefore, yes, the calculation is correct.

Therefore, I think the answer is 898. Wait, but just to be thorough, let me verify once again with another term. Let me compute x_{11} which should be x₁. Earlier, we saw x₁₁ =211. Let me compute x_{11} again to confirm.

x₁₁ = x₁₀ - x₉ + x₈ - x₇.

x₁₀ was -267, x₉ was -523, x₈ was -420, x₇ was -375. Therefore:

-267 - (-523) + (-420) - (-375) = -267 + 523 - 420 + 375.

Calculate:

-267 + 523 is 256. Then 256 - 420 is -164. Then -164 + 375 is 211. Yep, correct. So x₁₁ is 211, same as x₁. Therefore, our calculation is correct.

Therefore, confident that the period is 10. Therefore, the answer is 898.

Wait, but hold on. Let me check if maybe there's a miscalculation in the initial terms. Let me recheck the calculations for x₅ through x₁₀:

x₁ = 211

x₂ = 375

x₃ = 420

x₄ = 523

x₅ = x₄ - x₃ + x₂ - x₁ = 523 - 420 + 375 - 211 = (523 - 420) + (375 - 211) = 103 + 164 = 267. Correct.

x₆ = x₅ - x₄ + x₃ - x₂ = 267 - 523 + 420 - 375. Let me compute again:

267 - 523 = -256. Then, -256 + 420 = 164. Then, 164 - 375 = -211. Correct.

x₇ = x₆ - x₅ + x₄ - x₃ = -211 - 267 + 523 - 420.

-211 - 267 = -478. Then, -478 + 523 = 45. Then, 45 - 420 = -375. Correct.

x₈ = x₇ - x₆ + x₅ - x₄ = -375 - (-211) + 267 - 523 = -375 + 211 + 267 - 523.

-375 + 211 is -164. Then, -164 + 267 is 103. Then, 103 - 523 is -420. Correct.

x₉ = x₈ - x₇ + x₆ - x₅ = -420 - (-375) + (-211) - 267 = -420 + 375 -211 -267.

-420 + 375 is -45. Then, -45 -211 is -256. Then, -256 -267 is -523. Correct.

x₁₀ = x₉ - x₈ + x₇ - x₆ = -523 - (-420) + (-375) - (-211) = -523 + 420 - 375 + 211.

-523 + 420 is -103. -103 - 375 is -478. -478 + 211 is -267. Correct.

x₁₁ = x₁₀ - x₉ + x₈ - x₇ = -267 - (-523) + (-420) - (-375) = -267 + 523 - 420 + 375.

Which is same as before, which gives 211. Correct.

So all calculations are correct, so the period is indeed 10. Hence, the answer is 898.

Wait, but just to check, let me try with n=10. If n=10, then mod 10 is 0, so x_{10} is -267, which is correct. So for n=10, it's term 10. So for n=531, 531 mod 10=1, so term 1, which is 211. For n=753, mod 10=3, term 3 is 420. n=975, mod 10=5, term 5 is 267.

Adding them up: 211 + 420 is 631, 631 + 267 is 898.

Therefore, I think the answer is 898. However, to make sure, let me check with another example. For example, let's compute x_{10 + 1} = x_{11} = x₁ = 211. Similarly, if I take n=10 + 2 = 12, x_{12}=x₂=375, which matches our earlier calculation. So yes, the periodicity is correct.

Therefore, I think the answer is 898.

**Final Answer**
\boxed{898}

To solve the problem, we are given a recurrence relation:

$$
x_1 = 211,\quad x_2 = 375,\quad x_3 = 420,\quad x_4 = 523,
$$
$$
x_n = x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4} \quad \text{for } n \geq 5.
$$

We are asked to compute:

$$
x_{531} + x_{753} + x_{975}.
$$

---

### Step 1: Identify Periodicity of the Sequence

Since computing $ x_n $ for $ n = 531, 753, 975 $ directly would be impractical, we examine if the sequence satisfies a periodic pattern. Start by computing a few more terms in the sequence:

- $ x_5 = x_4 - x_3 + x_2 - x_1 = 523 - 420 + 375 - 211 = 267 $
- $ x_6 = x_5 - x_4 + x_3 - x_2 = 267 - 523 + 420 - 375 = -211 $
- $ x_7 = x_6 - x_5 + x_4 - x_3 = -211 - 267 + 523 - 420 = -375 $
- $ x_8 = x_7 - x_6 + x_5 - x_4 = -375 + 211 + 267 - 523 = -420 $
- $ x_9 = x_8 - x_7 + x_6 - x_5 = -420 + 375 - 211 - 267 = -523 $
- $ x_{10} = x_9 - x_8 + x_7 - x_6 = -523 + 420 - 375 + 211 = -267 $
- $ x_{11} = x_{10} - x_9 + x_8 - x_7 = -267 + 523 - 420 + 375 = 211 $

Now observe:
- $ x_{11} = x_1 = 211 $
- $ x_{12} = x_2 = 375 $

This suggests **periodicity with a period of 10**: the sequence repeats every 10 terms.

---

### Step 2: Use Periodicity to Compute Relevant Terms

Given the periodicity of 10, we use the property:

$$
x_n = x_{n \mod 10}, \quad \text{where if } n \mod 10 = 0, \text{then } x_{n} = x_{10}
$$

Compute the modulo 10 for each required index:

- $ 531 \mod 10 = 1 \Rightarrow x_{531} = x_1 = 211 $
- $ 753 \mod 10 = 3 \Rightarrow x_{753} = x_3 = 420 $
- $ 975 \mod 10 = 5 \Rightarrow x_{975} = x_5 = 267 $

---

### Step 3: Add the Selected Terms

Now compute the sum:

$$
x_{531} + x_{753} + x_{975} = 211 + 420 + 267 = \boxed{898}
$$

---

### Final Answer

$$
\boxed{898}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of x_{531} + x_{753} + x_{975} given this recurrence relation. Let me start by understanding the problem step by step.

First, the initial terms are given: x₁ = 211, x₂ = 375, x₃ = 420, x₄ = 523. Then, for n ≥ 5, each term is defined by the recurrence relation xₙ = xₙ₋₁ - xₙ₋₂ + xₙ₋₃ - xₙ₋₄. So, each term depends on the previous four terms.

My goal is to find x_{531} + x_{753} + x_{975}. Since calculating each term up to 975 directly would be impractical, there must be some pattern or periodicity in the sequence that I can exploit.

Let me recall that linear recurrence relations often have periodic solutions or eventually become periodic if the characteristic equation has roots that are roots of unity or something like that. Maybe I can compute the terms until I find a repeating cycle, then use that cycle to find the values at positions 531, 753, and 975.

First, let me try to compute some terms of the sequence to see if there's a pattern.

Given:
x₁ = 211
x₂ = 375
x₃ = 420
x₄ = 523

Now, compute x₅:

x₅ = x₄ - x₃ + x₂ - x₁ = 523 - 420 + 375 - 211

Let me calculate that step by step:

523 - 420 = 103

103 + 375 = 478

478 - 211 = 267

So x₅ = 267

Now x₆ = x₅ - x₄ + x₃ - x₂

Compute:

267 - 523 + 420 - 375

Compute step by step:

267 - 523 = -256

-256 + 420 = 164

164 - 375 = -211

So x₆ = -211

x₇ = x₆ - x₅ + x₄ - x₃

-211 - 267 + 523 - 420

Compute:

-211 - 267 = -478

-478 + 523 = 45

45 - 420 = -375

x₇ = -375

x₈ = x₇ - x₆ + x₅ - x₄

-375 - (-211) + 267 - 523

Which is -375 + 211 + 267 - 523

Compute step by step:

-375 + 211 = -164

-164 + 267 = 103

103 - 523 = -420

x₈ = -420

x₉ = x₈ - x₇ + x₆ - x₅

-420 - (-375) + (-211) - 267

Which is -420 + 375 - 211 - 267

Compute:

-420 + 375 = -45

-45 - 211 = -256

-256 - 267 = -523

x₉ = -523

x₁₀ = x₉ - x₈ + x₇ - x₆

-523 - (-420) + (-375) - (-211)

Which is -523 + 420 - 375 + 211

Calculating step by step:

-523 + 420 = -103

-103 - 375 = -478

-478 + 211 = -267

x₁₀ = -267

x₁₁ = x₁₀ - x₉ + x₈ - x₇

-267 - (-523) + (-420) - (-375)

Which is -267 + 523 - 420 + 375

Compute:

-267 + 523 = 256

256 - 420 = -164

-164 + 375 = 211

x₁₁ = 211

Wait a minute! x₁₁ is 211, which is the same as x₁. Let me check if this is a cycle.

Let me write down all the terms so far:

n: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,...

x: 211, 375, 420, 523, 267, -211, -375, -420, -523, -267, 211,...

So x₁₁ is 211, which is same as x₁. Let me check if the next terms repeat the original sequence.

Let me compute x₁₂:

x₁₂ = x₁₁ - x₁₀ + x₉ - x₈

211 - (-267) + (-523) - (-420)

Which is 211 + 267 - 523 + 420

Calculate:

211 + 267 = 478

478 - 523 = -45

-45 + 420 = 375

x₁₂ = 375, which is same as x₂.

Similarly, x₁₃ would be x₁₂ - x₁₁ + x₁₀ - x₉

375 - 211 + (-267) - (-523)

Which is 375 - 211 - 267 + 523

Compute:

375 - 211 = 164

164 - 267 = -103

-103 + 523 = 420

Which is x₃ = 420. So x₁₃ = x₃.

So it seems like starting from x₁, the sequence repeats every 10 terms? Wait, because x₁₁ is x₁, x₁₂ is x₂, x₁₃ is x₃, etc. So the period is 10? Let me confirm with x₁₄:

x₁₄ = x₁₃ - x₁₂ + x₁₁ - x₁₀

420 - 375 + 211 - (-267)

Which is 420 - 375 + 211 + 267

Compute:

420 - 375 = 45

45 + 211 = 256

256 + 267 = 523, which is x₄. So x₁₄ = x₄.

Therefore, the sequence is periodic with period 10. Let me check one more term to be sure. x₁₅:

x₁₅ = x₁₄ - x₁₃ + x₁₂ - x₁₁ = 523 - 420 + 375 - 211

Which is the same as x₅, which was 267. So yes, the pattern repeats every 10 terms.

So the sequence has period 10. That is, for any n, x_{n} = x_{n+10}.

Therefore, if I can find the value of n modulo 10, I can find the corresponding term.

Therefore, to compute x_{531}, x_{753}, x_{975}, I need to find 531 mod 10, 753 mod 10, 975 mod 10. Then look up the corresponding term in the sequence.

But let me confirm the period again. Since x₁ = x₁₁, x₂ = x₁₂, etc., so the period is 10. Let me check if the period is 10 or maybe a smaller period. For example, maybe it's 5? Let me check.

Looking at the terms:

x₁ = 211, x₂ = 375, x₃ = 420, x₄ = 523, x₅ = 267, x₆ = -211, x₇ = -375, x₈ = -420, x₉ = -523, x₁₀ = -267, x₁₁ = 211...

If I check if the sequence repeats every 10 terms, then yes. But maybe there is a smaller period? Let me check if the sequence repeats before 10. For example, check if x₁ = x₁ + 5? Let's see:

x₁ = 211, x₆ = -211, which is -x₁. Not the same. Similarly, x₂ = 375, x₇ = -375. So negative of x₂. Similarly for others. So the period isn't 5, but 10. Because if you go another 5 terms, you get back to original terms. For example, x₁₁ = x₁, x₁₂ = x₂, etc. So the period is 10.

Therefore, the sequence is periodic with period 10. So to find x_n, we can compute n mod 10. However, since the period is 10, we need to check if n mod 10 is 0, then it would correspond to x_{10}, otherwise it is the remainder.

But let me check. For example, n = 10: 10 mod 10 is 0, and x_{10} = -267. If n = 20, then x_{20} should be same as x_{10}, which is -267. Let me verify by computing x₁₅, x₁₆,...

But maybe I can just go with the period 10.

Thus, for any n, x_n = x_{n mod 10}, but with the caveat that if n mod 10 is 0, then it refers to x_{10} and not x_0 (since we don't have x_0). So, the mapping is:

n mod 10 = 1 → x₁

n mod 10 = 2 → x₂

...

n mod 10 = 0 → x_{10}

Therefore, for n = 531, let's compute 531 divided by 10. Let me compute 531 / 10: 10*53=530, so 531 = 10*53 +1 → remainder 1. Therefore, 531 mod 10 = 1. Therefore, x_{531} = x₁ = 211.

Similarly, 753 divided by 10: 10*75=750, so 753 = 10*75 + 3 → remainder 3. Therefore, 753 mod 10 = 3. Therefore, x_{753} = x₃ = 420.

Similarly, 975 divided by 10: 10*97 = 970, so 975 = 10*97 +5 → remainder 5. Therefore, 975 mod 10 =5. Therefore, x_{975} =x₅=267.

Therefore, x_{531} + x_{753} + x_{975} = 211 + 420 + 267.

Let me compute that:

211 + 420 = 631

631 + 267 = 898

Wait, so the answer would be 898?

But let me double-check my calculations. Because sometimes when dealing with modular arithmetic, it's easy to make an off-by-one error, especially with the modulus.

Let me verify the mod calculations:

For 531:

Divide 531 by 10: 10*53 = 530. 531 - 530 =1. So remainder 1. So yes, 531 mod10=1. Therefore, x_{531} =x₁=211. Correct.

For 753:

Divide 753 by 10: 10*75=750, 753-750=3. So remainder 3. Therefore, x_{753}=x₃=420. Correct.

For 975: 10*97=970, 975-970=5. Remainder 5. Therefore, x_{975}=x₅=267. Correct.

Therefore the sum is 211 + 420 + 267. Let me check that again:

211 + 420:

200 + 400 = 600, 11 + 20 = 31, so total 631. Then 631 + 267:

600 + 200 = 800, 31 + 67 = 98, so total 898. Yes, that's correct.

Wait, but before I get too confident, let me check if my assumption of period 10 is correct. Because if the sequence is periodic with period 10, then starting from term 1, every 10 terms it repeats.

But let me check another term beyond x₁₁. Let me compute x₁₂ (we already did), which was 375, which is x₂. x₁₃ = 420 =x₃. x₁₄=523=x₄. x₁₅=267=x₅. x₁₆=-211=x₆. x₁₇=-375=x₇. x₁₈=-420=x₈. x₁₉=-523=x₉. x₂₀=-267=x₁₀. x₂₁=211=x₁. Yep, so that's consistent. Therefore, the period is indeed 10. Therefore, the calculation is correct.

Therefore, the answer is 898. Wait, but let me check again the original terms:

x₁ = 211, x₂=375, x₃=420, x₄=523, x₅=267. So yes, x_{531} is x₁=211, x_{753} is x₃=420, x_{975} is x₅=267. Sum is 211+420+267=898.

But hold on, let me check if there is any possibility that the modulus is 10 but the mapping is different? For example, sometimes when people talk about modulus, they might consider 0 to be equivalent to 10. But since our terms start at n=1, so if the remainder is 0, then it's equivalent to term 10. So, for example, n=10: 10 mod10=0, so we take term 10. n=20: same as n=10. So that's correct.

Alternatively, if I think in terms of indices, the sequence has period 10, so x_{n} = x_{n+10} for all n ≥1. Therefore, to find x_n, compute n mod 10, and if it's 0, use term 10, else use the term corresponding to the remainder. Therefore, yes, the calculation is correct.

Therefore, I think the answer is 898. Wait, but just to be thorough, let me verify once again with another term. Let me compute x_{11} which should be x₁. Earlier, we saw x₁₁ =211. Let me compute x_{11} again to confirm.

x₁₁ = x₁₀ - x₉ + x₈ - x₇.

x₁₀ was -267, x₉ was -523, x₈ was -420, x₇ was -375. Therefore:

-267 - (-523) + (-420) - (-375) = -267 + 523 - 420 + 375.

Calculate:

-267 + 523 is 256. Then 256 - 420 is -164. Then -164 + 375 is 211. Yep, correct. So x₁₁ is 211, same as x₁. Therefore, our calculation is correct.

Therefore, confident that the period is 10. Therefore, the answer is 898.

Wait, but hold on. Let me check if maybe there's a miscalculation in the initial terms. Let me recheck the calculations for x₅ through x₁₀:

x₁ = 211

x₂ = 375

x₃ = 420

x₄ = 523

x₅ = x₄ - x₃ + x₂ - x₁ = 523 - 420 + 375 - 211 = (523 - 420) + (375 - 211) = 103 + 164 = 267. Correct.

x₆ = x₅ - x₄ + x₃ - x₂ = 267 - 523 + 420 - 375. Let me compute again:

267 - 523 = -256. Then, -256 + 420 = 164. Then, 164 - 375 = -211. Correct.

x₇ = x₆ - x₅ + x₄ - x₃ = -211 - 267 + 523 - 420.

-211 - 267 = -478. Then, -478 + 523 = 45. Then, 45 - 420 = -375. Correct.

x₈ = x₇ - x₆ + x₅ - x₄ = -375 - (-211) + 267 - 523 = -375 + 211 + 267 - 523.

-375 + 211 is -164. Then, -164 + 267 is 103. Then, 103 - 523 is -420. Correct.

x₉ = x₈ - x₇ + x₆ - x₅ = -420 - (-375) + (-211) - 267 = -420 + 375 -211 -267.

-420 + 375 is -45. Then, -45 -211 is -256. Then, -256 -267 is -523. Correct.

x₁₀ = x₉ - x₈ + x₇ - x₆ = -523 - (-420) + (-375) - (-211) = -523 + 420 - 375 + 211.

-523 + 420 is -103. -103 - 375 is -478. -478 + 211 is -267. Correct.

x₁₁ = x₁₀ - x₉ + x₈ - x₇ = -267 - (-523) + (-420) - (-375) = -267 + 523 - 420 + 375.

Which is same as before, which gives 211. Correct.

So all calculations are correct, so the period is indeed 10. Hence, the answer is 898.

Wait, but just to check, let me try with n=10. If n=10, then mod 10 is 0, so x_{10} is -267, which is correct. So for n=10, it's term 10. So for n=531, 531 mod 10=1, so term 1, which is 211. For n=753, mod 10=3, term 3 is 420. n=975, mod 10=5, term 5 is 267.

Adding them up: 211 + 420 is 631, 631 + 267 is 898.

Therefore, I think the answer is 898. However, to make sure, let me check with another example. For example, let's compute x_{10 + 1} = x_{11} = x₁ = 211. Similarly, if I take n=10 + 2 = 12, x_{12}=x₂=375, which matches our earlier calculation. So yes, the periodicity is correct.

Therefore, I think the answer is 898.

**Final Answer**
\boxed{898}

To solve the problem, we are given a recurrence relation:

$$
x_1 = 211,\quad x_2 = 375,\quad x_3 = 420,\quad x_4 = 523,
$$
$$
x_n = x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4} \quad \text{for } n \geq 5.
$$

We are asked to compute:

$$
x_{531} + x_{753} + x_{975}.
$$

---

### Step 1: Identify Periodicity of the Sequence

Since computing $ x_n $ for $ n = 531, 753, 975 $ directly would be impractical, we examine if the sequence satisfies a periodic pattern. Start by computing a few more terms in the sequence:

- $ x_5 = x_4 - x_3 + x_2 - x_1 = 523 - 420 + 375 - 211 = 267 $
- $ x_6 = x_5 - x_4 + x_3 - x_2 = 267 - 523 + 420 - 375 = -211 $
- $ x_7 = x_6 - x_5 + x_4 - x_3 = -211 - 267 + 523 - 420 = -375 $
- $ x_8 = x_7 - x_6 + x_5 - x_4 = -375 + 211 + 267 - 523 = -420 $
- $ x_9 = x_8 - x_7 + x_6 - x_5 = -420 + 375 - 211 - 267 = -523 $
- $ x_{10} = x_9 - x_8 + x_7 - x_6 = -523 + 420 - 375 + 211 = -267 $
- $ x_{11} = x_{10} - x_9 + x_8 - x_7 = -267 + 523 - 420 + 375 = 211 $

Now observe:
- $ x_{11} = x_1 = 211 $
- $ x_{12} = x_2 = 375 $

This suggests **periodicity with a period of 10**: the sequence repeats every 10 terms.

---

### Step 2: Use Periodicity to Compute Relevant Terms

Given the periodicity of 10, we use the property:

$$
x_n = x_{n \mod 10}, \quad \text{where if } n \mod 10 = 0, \text{then } x_{n} = x_{10}
$$

Compute the modulo 10 for each required index:

- $ 531 \mod 10 = 1 \Rightarrow x_{531} = x_1 = 211 $
- $ 753 \mod 10 = 3 \Rightarrow x_{753} = x_3 = 420 $
- $ 975 \mod 10 = 5 \Rightarrow x_{975} = x_5 = 267 $

---

### Step 3: Add the Selected Terms

Now compute the sum:

$$
x_{531} + x_{753} + x_{975} = 211 + 420 + 267 = \boxed{898}
$$

---

### Final Answer

$$
\boxed{898}
$$

Solution: Shifting the recurrence over by one and adding, we have: \[\begin{aligned} x_n &= x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4} \\ x_{n-1} &= x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5} \\ \implies x_n + x_{n-1} &= x_{n-1} - x_{n-5} \end{aligned}\]so $x_n = -x_{n-5}$ for all $n.$ In particular, $x_n = -x_{n-5} = -(-x_{n-10}) = x_{n-10},$ so the sequence repeats with period $10.$ Thus, \[\begin{aligned} x_{531} + x_{753} + x_{975} &= x_1 + x_3 + x_5 \\ &= x_1 + x_3 + (x_4-x_3+x_2-x_1) \\ &= x_2 + x_4 \\ &= 375 + 523 = \boxed{898}. \end{aligned}\]


Explanation:
Okay, so I need to find three consecutive positive integers where their sum is a three-digit palindrome, and the sum is less than 220. Then, I have to find the greatest possible value for the largest of those three integers. Alright, let me start breaking this down.

First, let me recall what a palindrome is. A palindrome is a number that reads the same forwards and backwards. For example, 121 or 1331. Since the problem mentions a three-digit palindrome, that means the sum has to be between 100 and 999, but specifically less than 220. So the possible sums are three-digit palindromes from 101 up to 219. Wait, but palindromes have the first and last digits the same. So for three-digit numbers, the first digit is the same as the third. Let me list all three-digit palindromes less than 220.

Three-digit palindromes less than 220 would start with 1 or 2. Let's see:

- Starting with 1: The first digit is 1, the third digit is also 1. The middle digit can be anything from 0-9. So these are 101, 111, 121, 131, 141, 151, 161, 171, 181, 191. Then starting with 2, the first digit is 2, so the third digit must be 2 as well. But the number has to be less than 220. So the first digit is 2, the second digit must be less than 2, because 220 is excluded. Wait, the number is less than 220. So if the palindrome starts with 2, then the number is between 200 and 219. Therefore, the second digit must be 0 or 1. Let me check:

- If first digit is 2, third digit is 2. The middle digit can be 0 or 1? Because 202 is okay, but 212 is 212 which is less than 220. Wait, 212 is less than 220? Yes, 212 is 212, and 220 is excluded. So palindromes starting with 2 can be 202, 212. Then 222 would be over 220, so not included. So the three-digit palindromes less than 220 are:

101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212.

Okay, so that's my list of possible sums. Now, the problem is about three consecutive positive integers. Let me denote them as n-1, n, n+1. Wait, but actually, they are three consecutive integers. Let me think. Let me denote the three numbers as x, x+1, x+2. Then their sum is x + (x+1) + (x+2) = 3x + 3 = 3(x + 1). Therefore, the sum of three consecutive integers is always a multiple of 3. Therefore, the sum S = 3(x + 1). Therefore, S must be divisible by 3.

Therefore, among the palindromes I listed above, which ones are divisible by 3? Let me check each of them:

First, 101: 1+0+1=2, not divisible by 3. 101 divided by 3 is 33.666... Not integer. So not divisible by 3.

111: 1+1+1=3, which is divisible by 3. So 111 is divisible by 3.

121: 1+2+1=4, not divisible by 3.

131: 1+3+1=5, nope.

141: 1+4+1=6, yes, divisible by 3.

151: 1+5+1=7, nope.

161: 1+6+1=8, nope.

171: 1+7+1=9, yes.

181: 1+8+1=10, nope.

191: 1+9+1=11, nope.

202: 2+0+2=4, nope.

212: 2+1+2=5, nope.

So the palindromes that are divisible by 3 are 111, 141, 171. Wait, let me check if there are others? Wait, 111, 141, 171. Then 202? 2+0+2=4, not. 212? 2+1+2=5. No. So only those three? Wait, maybe I made a mistake. Let me check again:

Original palindromes less than 220:

101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212.

Sum digits:

101: 1+0+1=2

111: 1+1+1=3 ✔

121: 1+2+1=4

131: 1+3+1=5

141: 1+4+1=6 ✔

151: 1+5+1=7

161: 1+6+1=8

171: 1+7+1=9 ✔

181: 1+8+1=10

191: 1+9+1=11

202: 2+0+2=4

212: 2+1+2=5

Yes. So only 111, 141, 171. So those are the possible sums. Therefore, the sum can be 111, 141, or 171. Wait, but wait, what about numbers like 121? Wait, 121 is not divisible by 3. Let me check 121 divided by 3. 121 / 3 is 40.333... So no, not an integer. So only those three.

Wait, but hold on, is that all? Let me check if there are other palindromes that are divisible by 3. For example, let me check 212. 212 / 3 is 70.666. No. 202 / 3 is 67.333. No. So yeah, only 111, 141, 171. Wait, but maybe I missed some palindromes? Let me check again.

Wait, three-digit palindromes less than 220: starting with 1, the middle digit varies from 0 to 9, so 101, 111, 121,..., up to 191. Then starting with 2, the middle digit can be 0 or 1, so 202 and 212. So that's all. So only 111, 141, 171 are divisible by 3. Therefore, the possible sums are 111, 141, 171. Wait, but why are there only three? Because only those three palindromes have digit sums divisible by 3. So that seems correct.

But let me check if there's something wrong here. Let me check 111: sum is 111. Then 141, 171. So, the three possible sums. Therefore, for each of these sums, we can find the three consecutive numbers. Let me do that.

First, if the sum is S = 3(x + 1). Therefore, x + 1 = S / 3. Therefore, x = (S / 3) - 1. Then the three numbers are x, x + 1, x + 2. Therefore, the largest number is x + 2 = (S / 3) - 1 + 2 = (S / 3) + 1.

Therefore, for each sum S, the largest number is (S / 3) + 1. Therefore, if I want the greatest possible value for the largest of the three integers, I should take the largest possible sum S, which is 171. Therefore, (171 / 3) + 1 = 57 + 1 = 58. Therefore, 58? Wait, but that seems low. Wait, let me check.

Wait, if the sum is 171, then the three numbers would be?

Since sum is 3(x + 1) = 171, so x + 1 = 57, so x = 56. Therefore, numbers are 56, 57, 58. Their sum is 56 + 57 + 58 = 171. Correct. So the largest number is 58. But maybe there's a larger possible sum? Wait, but we thought only 171 is the largest possible sum. Wait, is that right? Let me check if there are any other palindromes that I missed which are divisible by 3. Wait, maybe I missed some palindromes? Let me think.

Wait, three-digit palindromes less than 220. Let me think again. For example, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. So that's all. And only 111, 141, 171 are divisible by 3. So 171 is the largest. Therefore, according to this, the largest possible value of the largest integer is 58. But wait, that seems low. The problem says "the sum is less than 220", but maybe we need to check other possibilities?

Wait, perhaps there is a mistake here. Let me think again. Wait, maybe there are other palindromes which are divisible by 3? Let me check for example 212. 212 divided by 3 is 70.666... Not integer. 202 / 3 is 67.333... Also not. 191 / 3? 191 divided by 3 is 63.666... Not integer. 181 / 3? 181 / 3 = 60.333... Not. 171 / 3 = 57, which is okay. 161 / 3? 53.666... 151 / 3? 50.333... 141 / 3 = 47. 131 / 3? 43.666... 121 / 3? 40.333... 111 / 3 = 37. 101 / 3? 33.666... So yeah, only 111, 141, 171. So that seems correct.

But 58 seems low. Let me think again. Wait, maybe there is a mistake in how I set up the problem. The problem says "three consecutive positive integers". So if the sum is S, then S = 3x + 3, where x is the first integer. Therefore, the largest integer is x + 2. So the largest integer is (S / 3 - 1) + 2 = S / 3 + 1. So for S = 171, largest integer is 171 / 3 + 1 = 57 + 1 = 58. So that seems right. But 58? Let me think, is there a way to get a larger number?

Wait, maybe there's another approach. Let me check if there are other palindromic sums that I might have missed. Wait, suppose that I didn't consider that the sum could be something else? Wait, but the problem says the sum is a three-digit palindrome less than 220, so only those numbers. But maybe the problem allows the sum to be a three-digit palindrome, but not necessarily that the sum is 3(x + 1). Wait, no, the sum of three consecutive integers is always 3 times the middle number. So if we have three consecutive integers, the middle one is (x + 1), so sum is 3(x + 1). Therefore, the sum must be divisible by 3. Therefore, only those three sums. Therefore, the maximum sum is 171, leading to the largest number being 58. But maybe there's a mistake here? Let me check with the sum of 171.

Let me check if there is another set of numbers where the sum is a palindrome less than 220, but maybe not necessarily following that structure? Wait, no, the sum of three consecutive numbers is always three times the middle number. Therefore, it must be divisible by 3. Therefore, only those three sums. Therefore, the answer would be 58. But maybe I need to check if there's a higher possible sum. Wait, what if there's a palindrome that is not divisible by 3, but the sum is equal to that palindrome? Wait, but if the sum is not divisible by 3, then there can't be three consecutive integers adding up to that sum. Because sum of three consecutive numbers is always divisible by 3. So that's a key point. Therefore, only palindromes divisible by 3 can be sums. Therefore, only 111, 141, 171. Hence, maximum sum is 171. Hence, largest number is 58.

But wait, maybe there is another way. Wait, maybe the problem is not about three consecutive integers starting from x, x+1, x+2? Wait, but consecutive integers must be like that. For example, if the numbers are n-1, n, n+1, their sum is 3n. So same thing. Either way, the sum is 3 times the middle number. Therefore, it must be divisible by 3. Therefore, only those three palindromes. Therefore, the maximum possible sum is 171, leading to the largest number being 58. So maybe that is the answer? But 58 seems low. Let me check if there is a mistake here.

Wait, let me think. Suppose the sum is 171, which is 56 + 57 + 58 = 171. Correct. Suppose there is a larger sum. Let me check what is the next possible palindrome after 171 which is divisible by 3. The next palindrome after 171 would be 181, but 181 isn't divisible by 3. Then 191, also not. Then 202, not. 212, not. Then 222, which is over 220, so not allowed. Therefore, there is nothing between 171 and 220 that is a three-digit palindrome divisible by 3. Therefore, 171 is indeed the largest possible sum. Therefore, largest number is 58. Wait, but maybe there are palindromes with different digits?

Wait, for example, 121. If I tried to see if 121 can be a sum. Even though it's not divisible by 3, but let me check. Suppose someone says that maybe the three numbers aren't consecutive integers? Wait, no, the problem says "three consecutive positive integers". So they must be consecutive. Therefore, their sum must be divisible by 3. Therefore, 121 cannot be the sum.

But maybe there's another way? Wait, maybe the problem is not about the sum being a palindrome, but the three numbers themselves? Wait, no, the problem says "the sum of a particular set of three consecutive positive integers is a three-digit palindrome." So the sum is the palindrome. Therefore, we can't have the numbers themselves being palindromes. So that's not the case.

Therefore, according to this, the answer is 58. However, I feel like 58 is too small. Let me check if I messed up the calculation. Let me think again. If the sum is 171, then the numbers are 56, 57, 58. Their sum is 171. Correct. The largest is 58. If the sum is 141, then numbers are (141 / 3) - 1, so 47 -1=46? Wait, no. Wait, S = 3(x + 1). Therefore, x + 1 = S / 3. Therefore, x = (S / 3) - 1. Therefore, largest number is x + 2 = (S / 3) - 1 + 2 = (S / 3) + 1. Therefore, for S = 141, largest number is 141 / 3 + 1 = 47 + 1 = 48. Similarly for S = 111, it's 37 + 1 = 38. So 58 is the largest among those. But maybe there is another possibility? Let me think.

Wait, perhaps there is another way that the sum can be a palindrome? Let me check if there are palindromes that are divisible by 3, but I didn't consider because I thought they weren't. For example, 3-digit palindromes where the first and third digits are same. But I considered all of them. Let me check for palindromes like 151: 1+5+1=7, not divisible by 3. 161: 1+6+1=8. 171: 9. 181: 10. 191: 11. So 171 is the only one in that 1xx range. Then in the 2xx range, palindromes are 202, 212. Their digit sums are 4 and 5. Not divisible by 3. Therefore, only 111, 141, 171.

Wait, but maybe there are palindromes that are not three-digit? But the problem says the sum is a three-digit palindrome. Therefore, not possible. Therefore, only those three sums. Hence, 58 is the answer. Wait, maybe the problem is in the interpretation? Let me check the problem again:

"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?"

So, the sum is a three-digit palindrome less than 220. Therefore, the sum can be 101, 111,..., up to 212. Then, the sum must be divisible by 3. Therefore, only 111, 141, 171.

Therefore, the maximum sum is 171, leading to the largest integer being 58. But maybe there is a different way to think about this. For example, maybe considering that the sum is a palindrome, but not necessarily that the sum is 3 times the middle number. Wait, but that is the case. Let me think again. If the three numbers are consecutive integers, then their sum is 3 times the middle number, so it must be divisible by 3. Therefore, sum is 3 times something. Therefore, only palindromes divisible by 3 are allowed. So I think my reasoning is correct.

But let me check for example, take a sum of 171, which is allowed. The largest number is 58. But maybe there is another set of three numbers where the sum is a palindrome less than 220, but with a larger maximum number. How?

For example, suppose that the sum is 198. Wait, 198 is not a palindrome. But let me check. Suppose the sum is something like 202, which is a palindrome. But 202 is not divisible by 3. If I check 202 / 3 is 67.333... Not integer. So no. What about 171? Which is 171. Then, if sum is 171, the numbers are 56, 57, 58. The largest is 58. But let me think: is there a way to get a larger maximum number?

Suppose, for example, that the sum is 183, but 183 is not a palindrome. 183 reversed is 381. Not same. How about 191? Not divisible by 3. 191. 191 is a palindrome, but not divisible by 3. If I force the sum to be 191, then the numbers would be? Let me check. If sum is 191, then since 191 is not divisible by 3, there are no three consecutive integers adding up to 191. Therefore, not possible.

Wait, but maybe the problem is not that the sum is divisible by 3? Wait, no, because three consecutive integers must add up to a multiple of 3. Let me confirm with an example. Let me take numbers 1, 2, 3. Sum is 6, which is 3*2. Numbers 2, 3, 4. Sum is 9, which is 3*3. So yes, always divisible by 3. Therefore, no matter what three consecutive integers you take, the sum will be divisible by 3. Therefore, the sum must be divisible by 3. Therefore, only palindromes divisible by 3 can be considered. Therefore, only 111, 141, 171. Therefore, the answer should be 58. But let me check if I made an error in the calculation of the largest number.

Wait, the largest number is S / 3 + 1. For S = 171, that is 171 / 3 + 1 = 57 + 1 = 58. Correct. So numbers are 56, 57, 58. But wait, 56 + 57 + 58 = 171. Correct.

Wait, let me think if there is a different interpretation. Suppose the three numbers are not necessarily in the form x, x+1, x+2 but something else? But consecutive integers are always like that. So no.

Wait, another thought: maybe the problem allows the sum to be a palindrome, but not necessarily three-digit? But the problem says "the sum is a three-digit palindrome". So no. Therefore, the answer is 58. But maybe I need to check other palindromes. Wait, perhaps 191? Wait, no, not divisible by 3. 181? Also no. 202? No. 212? No. What about numbers like 3-digit palindromes with middle digit 9? For example, 191, which we already checked. Not divisible by 3. 191: 1 + 9 + 1 = 11, not divisible by 3. So no.

Wait, but hold on. What about 3-digit palindromes with first digit 1 and last digit 1, but with middle digit such that the sum is divisible by 3? For example, 111, 141, 171. Which we have already considered. So no others. Therefore, the answer is 58.

Wait, but I feel like maybe there's a mistake here. Let me check another approach. Let me suppose that the sum is a palindrome, say S, which is 3k for some integer k, and S is a three-digit palindrome less than 220. Then k = S / 3. Then the three numbers would be k - 1, k, k + 1. Therefore, the largest number is k + 1. Therefore, to maximize the largest number, we need to maximize k. Since S = 3k, and S is less than 220. Therefore, 3k < 220 => k < 220 / 3 ≈ 73.333. Therefore, maximum integer k is 73. Therefore, maximum S is 3*73 = 219. But 219 is not a palindrome. So we need to check palindromes less than 220, which are divisible by 3, and find the largest such palindrome, then compute k = S / 3, and then the largest number is k + 1.

But earlier, I thought that palindromes less than 220 and divisible by 3 are only 111, 141, 171. But 219 is not a palindrome. The next palindrome less than 220 is 212, which is not divisible by 3. Then 202, not divisible by 3. Then 191, no. 181, no. 171, yes. So 171 is the largest palindrome less than 220 divisible by 3. Therefore, k = 171 / 3 = 57, so the largest number is 57 + 1 = 58. Therefore, the answer is 58. So that's the same result as before. Therefore, 58 is correct? But let me check if there are palindromes between 171 and 220 which are divisible by 3. Let's see:

Palindromes between 171 and 220: 181, 191, 202, 212. All of these are not divisible by 3. 181: 1+8+1=10, not divisible by 3. 191: 1+9+1=11, nope. 202: 2+0+2=4, nope. 212: 2+1+2=5, nope. Therefore, 171 is the largest palindrome less than 220 divisible by 3. Therefore, answer is 58.

Wait, but hold on. What about 3-digit palindromes starting with 1, but higher than 171? Let me check. Between 171 and 220, the next palindrome after 171 would be 181, which we saw. Then 191, then 202, 212. So no. Therefore, 171 is the largest. So answer is 58. Therefore, maybe the answer is 58? But let me check once again. If I think of the largest number possible, maybe there's another set of numbers where the sum is a palindrome less than 220, but not necessarily that the sum is divisible by 3? Wait, but that's impossible, because three consecutive numbers must add to a multiple of 3. So the sum has to be divisible by 3. So no other possibilities.

Alternatively, maybe the problem allows for the numbers to be non-consecutive? But the problem says three consecutive positive integers. Therefore, they have to be consecutive. So no.

Alternatively, maybe I made a mistake in assuming that the middle number is k? Let me think again. If the three numbers are consecutive, say n-1, n, n+1, then the sum is 3n. So the sum is 3n, so S = 3n. So n = S / 3. Then the largest number is n + 1 = S / 3 + 1. Therefore, if S is 171, then the largest number is 171 / 3 + 1 = 57 + 1 = 58. Therefore, correct. Therefore, maybe 58 is correct. But why do I feel unsure? Maybe because 58 seems small. Let me check with another example. Let me check the sum of 171, which is 56 + 57 + 58 = 171. Correct. If I take the next possible palindrome divisible by 3, which is 171, that's the largest. Therefore, 58 is the answer.

Alternatively, maybe there is a mistake in considering the sum. Wait, if the sum is 183, which is not a palindrome, but 183 is 3*61. Therefore, numbers would be 60, 61, 62. Their sum is 183. Not a palindrome. But 183 is not a palindrome. Therefore, no. Similarly, 192 is 3*64, numbers 63, 64, 65. Sum is 192, which isn't a palindrome. 201 is 3*67, numbers 66, 67, 68. Sum is 201. 201 is not a palindrome. 210 is 3*70, numbers 69, 70, 71. Sum is 210. Not a palindrome. 219 is 3*73, numbers 72, 73, 74. Sum 219. Not a palindrome. Therefore, none of those sums are palindromes. Therefore, indeed, only 111, 141, 171. Therefore, the answer is 58. Therefore, maybe the answer is 58.

Wait, but hold on. Let me check if there are other palindromes in the 100s that I missed. For example, 121. Let me check if 121 is divisible by 3. 1 + 2 + 1 = 4, not divisible by 3. 131: 1+3+1=5. Not. 141: 6. 151: 7. 161: 8. 171: 9. 181: 10. 191: 11. So, no, only those three. Therefore, seems correct.

But wait, what about 3-digit palindromes that start with 1, middle digit 8, which is 181. Not divisible by 3. So, yes. Therefore, I think that the answer is 58. Maybe I was overthinking.

Wait, but let me check with S = 171, which gives the numbers 56, 57, 58. Largest number is 58. But maybe there's another set of three consecutive numbers where the sum is a palindrome less than 220, but the numbers are larger? For example, suppose that the sum is 198. Wait, 198 is not a palindrome. But if there is a palindrome like 181 or something else. Wait, but again, 181 is not divisible by 3. So no.

Alternatively, suppose the sum is 183, not a palindrome. 192 isn't. 198 isn't. 201 isn't. 207 isn't. 210 isn't. 213 isn't. 216 isn't. 219 isn't. So none of these are palindromes. So no. Therefore, no other possibilities. Therefore, the answer is 58. Therefore, maybe 58 is correct. Therefore, I think the answer is 58.

Wait, but to be thorough, let me check if there's a palindrome with leading zeros? But no, palindromes are three-digit numbers, which can't have leading zeros. So 010 is just 10, which is two-digit. Therefore, not allowed. Therefore, no.

Therefore, after all that thinking, I think the answer is 58. Therefore, the greatest possible value for the largest of the three integers is 58.

But wait, hold on. Let me check once again for palindromes. Let me check 3-digit palindromes divisible by 3 less than 220. So 111, 141, 171. What about 3-digit palindromes starting with 2? Like 202, 212. 2+0+2=4, not divisible by 3. 2+1+2=5, not. So no. What about 222? It's 222, which is 222. But the problem says the sum is less than 220, so 222 is excluded. So no. Therefore, no other palindromes.

Therefore, my conclusion is that the answer is 58. Therefore, I think I should go with 58. However, just to make sure, let me check if there's an alternative interpretation of the problem.

Wait, maybe the problem isn't saying that the sum is a palindrome, but that the three numbers are palindromes? But the wording says: "The sum of a particular set of three consecutive positive integers is a three-digit palindrome." So it's the sum that's the palindrome, not the numbers themselves. Therefore, my original interpretation is correct.

Therefore, after thorough checking, I think the answer is 58. Therefore, the greatest possible value is \boxed{58}.

Wait, hold on, but I just had a thought. Let me check with another approach. Let me suppose that the three numbers are n, n+1, n+2. Their sum is 3n + 3 = 3(n + 1). Therefore, the sum S = 3(n + 1). Therefore, S must be divisible by 3. So we need S to be a three-digit palindrome less than 220, so S ∈ {101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212}, and S must be divisible by 3. As established before, only 111, 141, 171. Therefore, the maximum S is 171, which gives n + 1 = 57, so n = 56, so the numbers are 56, 57, 58. Therefore, the largest is 58. Therefore, answer is 58. Therefore, I think this is correct.

Alternatively, maybe the problem is from a source where the answer is different? But I don't think so. Unless there is an error in my logic. Let me think again: the sum is a three-digit palindrome. Therefore, it's between 100 and 219. The sum must be divisible by 3. Therefore, the palindromes divisible by 3 are 111, 141, 171. Therefore, the maximum sum is 171. Therefore, the maximum largest integer is 58. Yes. Therefore, I think the answer is 58. Therefore, I will go with that.

**Final Answer**
\boxed{58}

To solve this problem, we are given the following:

- We are to find three **consecutive positive integers** whose **sum is a three-digit palindrome**, and the **sum is less than 220**.
- We are to determine the **greatest possible value** for the **largest of the three integers** in the set.

---

### Step 1: Understanding the Sum of Three Consecutive Integers

Let the three consecutive integers be:

$$
n, \, n+1, \, n+2
$$

Their sum is:

$$
n + (n+1) + (n+2) = 3n + 3 = 3(n + 1)
$$

This tells us that the **sum must be divisible by 3**, making it crucial to consider **three-digit palindromes divisible by 3**, less than 220.

---

### Step 2: List of Valid Palindromes

A **three-digit palindrome** reads the same forwards and backwards. So the number must be of the form $ ABA $, where $ A $ is a digit from 1 to 9, and $ B $ is any digit from 0 to 9.

So, all three-digit palindromes less than 220 are:

$$
101, \, 111, \, 121, \, 131, \, 141, \, 151, \, 161, \, 171, \, 181, \, 191, \, 202, \, 212
$$

We now check which of these are divisible by 3 by adding their digits:

- 101 → 1+0+1 = 2 → Not divisible by 3
- 111 → 3 → ✅
- 121 → 4 → ❌
- 131 → 5 → ❌
- 141 → 6 → ✅
- 151 → 7 → ❌
- 161 → 8 → ❌
- 171 → 9 → ✅
- 181 → 10 → ❌
- 191 → 11 → ❌
- 202 → 4 → ❌
- 212 → 5 → ❌

✅ **Valid Sums**: 111, 141, 171

---

### Step 3: Determine the Largest Integer in the Set

We now compute the corresponding three consecutive integers for each valid sum (which is of the form $ S = 3(n + 1) $):

- **If $ S = 111 $:**
  $$
  n + 1 = \frac{111}{3} = 37,\quad n = 36
  $$
  The integers are: 36, 37, 38
  **Largest = 38**

- **If $ S = 141 $:**
  $$
  n + 1 = \frac{141}{3} = 47,\quad n = 46
  $$
  The integers are: 46, 47, 48
  **Largest = 48**

- **If $ S = 171 $:**
  $$
  n + 1 = \frac{171}{3} = 57,\quad n = 56
  $$
  The integers are: 56, 57, 58
  **Largest = 58**

---

### Step 4: Conclusion

Among all valid sets, the largest value of the **largest integer** occurs when the **sum is 171**, and the corresponding integers are **56, 57, 58**.

Thus, the **greatest possible value for the largest of the three integers** is:

$$
\boxed{58}
$$

[Assistant]
Explanation:
Okay, so I need to find three consecutive positive integers where their sum is a three-digit palindrome, and the sum is less than 220. Then, I have to find the greatest possible value for the largest of those three integers. Alright, let me start breaking this down.

First, let me recall what a palindrome is. A palindrome is a number that reads the same forwards and backwards. For example, 121 or 1331. Since the problem mentions a three-digit palindrome, that means the sum has to be between 100 and 999, but specifically less than 220. So the possible sums are three-digit palindromes from 101 up to 219. Wait, but palindromes have the first and last digits the same. So for three-digit numbers, the first digit is the same as the third. Let me list all three-digit palindromes less than 220.

Three-digit palindromes less than 220 would start with 1 or 2. Let's see:

- Starting with 1: The first digit is 1, the third digit is also 1. The middle digit can be anything from 0-9. So these are 101, 111, 121, 131, 141, 151, 161, 171, 181, 191. Then starting with 2, the first digit is 2, so the third digit must be 2 as well. But the number has to be less than 220. So the first digit is 2, the second digit must be less than 2, because 220 is excluded. Wait, the number is less than 220. So if the palindrome starts with 2, then the number is between 200 and 219. Therefore, the second digit must be 0 or 1. Let me check:

- If first digit is 2, third digit is 2. The middle digit can be 0 or 1? Because 202 is okay, but 212 is 212 which is less than 220. Wait, 212 is less than 220? Yes, 212 is 212, and 220 is excluded. So palindromes starting with 2 can be 202, 212. Then 222 would be over 220, so not included. So the three-digit palindromes less than 220 are:

101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212.

Okay, so that's my list of possible sums. Now, the problem is about three consecutive positive integers. Let me denote them as n-1, n, n+1. Wait, but actually, they are three consecutive integers. Let me think. Let me denote the three numbers as x, x+1, x+2. Then their sum is x + (x+1) + (x+2) = 3x + 3 = 3(x + 1). Therefore, the sum of three consecutive integers is always a multiple of 3. Therefore, the sum S = 3(x + 1). Therefore, S must be divisible by 3.

Therefore, among the palindromes I listed above, which ones are divisible by 3? Let me check each of them:

First, 101: 1+0+1=2, not divisible by 3. 101 divided by 3 is 33.666... Not integer. So not divisible by 3.

111: 1+1+1=3, which is divisible by 3. So 111 is divisible by 3.

121: 1+2+1=4, not divisible by 3.

131: 1+3+1=5, nope.

141: 1+4+1=6, yes, divisible by 3.

151: 1+5+1=7, nope.

161: 1+6+1=8, nope.

171: 1+7+1=9, yes.

181: 1+8+1=10, nope.

191: 1+9+1=11, nope.

202: 2+0+2=4, nope.

212: 2+1+2=5, nope.

So the palindromes that are divisible by 3 are 111, 141, 171. Wait, let me check if there are others? Wait, 111, 141, 171. Then 202? 2+0+2=4, not. 212? 2+1+2=5. No. So only those three? Wait, maybe I made a mistake. Let me check again:

Original palindromes less than 220:

101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212.

Sum digits:

101: 1+0+1=2

111: 1+1+1=3 ✔

121: 1+2+1=4

131: 1+3+1=5

141: 1+4+1=6 ✔

151: 1+5+1=7

161: 1+6+1=8

171: 1+7+1=9 ✔

181: 1+8+1=10

191: 1+9+1=11

202: 2+0+2=4

212: 2+1+2=5

Yes. So only 111, 141, 171. So those are the possible sums. Therefore, the sum can be 111, 141, or 171. Wait, but wait, what about numbers like 121? Wait, 121 is not divisible by 3. Let me check 121 divided by 3. 121 / 3 is 40.333... So no, not an integer. So only those three.

Wait, but hold on, is that all? Let me check if there are other palindromes that are divisible by 3. For example, let me check 212. 212 / 3 is 70.666. No. 202 / 3 is 67.333. No. So yeah, only 111, 141, 171. Wait, but maybe I missed some palindromes? Let me check again.

Wait, three-digit palindromes less than 220: starting with 1, the middle digit varies from 0 to 9, so 101, 111, 121,..., up to 191. Then starting with 2, the middle digit can be 0 or 1, so 202 and 212. So that's all. So only 111, 141, 171 are divisible by 3. Therefore, the possible sums are 111, 141, 171. Wait, but why are there only three? Because only those three palindromes have digit sums divisible by 3. So that seems correct.

But let me check if there's something wrong here. Let me check 111: sum is 111. Then 141, 171. So, the three possible sums. Therefore, for each of these sums, we can find the three consecutive numbers. Let me do that.

First, if the sum is S = 3(x + 1). Therefore, x + 1 = S / 3. Therefore, x = (S / 3) - 1. Then the three numbers are x, x + 1, x + 2. Therefore, the largest number is x + 2 = (S / 3) - 1 + 2 = (S / 3) + 1.

Therefore, for each sum S, the largest number is (S / 3) + 1. Therefore, if I want the greatest possible value for the largest of the three integers, I should take the largest possible sum S, which is 171. Therefore, (171 / 3) + 1 = 57 + 1 = 58. Therefore, 58? Wait, but that seems low. Wait, let me check.

Wait, if the sum is 171, then the three numbers would be?

Since sum is 3(x + 1) = 171, so x + 1 = 57, so x = 56. Therefore, numbers are 56, 57, 58. Their sum is 56 + 57 + 58 = 171. Correct. So the largest number is 58. But maybe there's a larger possible sum? Wait, but we thought only 171 is the largest possible sum. Wait, is that right? Let me check if there are any other palindromes that I missed which are divisible by 3. Wait, maybe I missed some palindromes? Let me think.

Wait, three-digit palindromes less than 220. Let me think again. For example, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. So that's all. And only 111, 141, 171 are divisible by 3. So 171 is the largest. Therefore, according to this, the largest possible value of the largest integer is 58. But wait, that seems low. The problem says "the sum is less than 220", but maybe we need to check other possibilities?

Wait, perhaps there is a mistake here. Let me think again. Wait, maybe there are other palindromes which are divisible by 3? Let me check for example 212. 212 divided by 3 is 70.666... Not integer. 202 / 3 is 67.333... Also not. 191 / 3? 191 divided by 3 is 63.666... Not integer. 181 / 3? 181 / 3 = 60.333... Not. 171 / 3 = 57, which is okay. 161 / 3? 53.666... 151 / 3? 50.333... 141 / 3 = 47. 131 / 3? 43.666... 121 / 3? 40.333... 111 / 3 = 37. 101 / 3? 33.666... So yeah, only 111, 141, 171. So that seems correct.

But 58 seems low. Let me think again. Wait, maybe there is a mistake in how I set up the problem. The problem says "three consecutive positive integers". So if the sum is S, then S = 3x + 3, where x is the first integer. Therefore, the largest integer is x + 2. So the largest integer is (S / 3 - 1) + 2 = S / 3 + 1. So for S = 171, largest integer is 171 / 3 + 1 = 57 + 1 = 58. So that seems right. But 58? Let me think, is there a way to get a larger number?

Wait, maybe there's another approach. Let me check if there are other palindromic sums that I might have missed. Wait, suppose that I didn't consider that the sum could be something else? Wait, but the problem says the sum is a three-digit palindrome less than 220, so only those numbers. But maybe the problem allows the sum to be a three-digit palindrome, but not necessarily that the sum is 3(x + 1). Wait, no, the sum of three consecutive integers is always 3 times the middle number. So if we have three consecutive integers, the middle one is (x + 1), so sum is 3(x + 1). Therefore, the sum must be divisible by 3. Therefore, only those three sums. Therefore, the maximum sum is 171, leading to the largest number being 58. But maybe there's a mistake here? Let me check with the sum of 171.

Let me check if there is another set of numbers where the sum is a palindrome less than 220, but maybe not necessarily following that structure? Wait, no, the sum of three consecutive numbers is always three times the middle number. Therefore, it must be divisible by 3. Therefore, only those three sums. Therefore, the answer would be 58. But maybe I need to check if there's a higher possible sum. Wait, what if there's a palindrome that is not divisible by 3, but the sum is equal to that palindrome? Wait, but if the sum is not divisible by 3, then there can't be three consecutive integers adding up to that sum. Because sum of three consecutive numbers is always divisible by 3. So that's a key point. Therefore, only palindromes divisible by 3 can be sums. Therefore, only 111, 141, 171. Hence, maximum sum is 171. Hence, largest number is 58.

But wait, maybe there is another way. Wait, maybe the problem is not about three consecutive integers starting from x, x+1, x+2? Wait, but consecutive integers must be like that. For example, if the numbers are n-1, n, n+1, their sum is 3n. So same thing. Either way, the sum is 3 times the middle number. Therefore, it must be divisible by 3. Therefore, only those three palindromes. Therefore, the maximum possible sum is 171, leading to the largest number being 58. So maybe that is the answer? But 58 seems low. Let me check if there is a mistake here.

Wait, let me think. Suppose the sum is 171, which is 56 + 57 + 58 = 171. Correct. Suppose there is a larger sum. Let me check what is the next possible palindrome after 171 which is divisible by 3. The next palindrome after 171 would be 181, but 181 isn't divisible by 3. Then 191, also not. Then 202, not. 212, not. Then 222, which is over 220, so not allowed. Therefore, there is nothing between 171 and 220 that is a three-digit palindrome divisible by 3. Therefore, 171 is indeed the largest possible sum. Therefore, largest number is 58. Wait, but maybe there are palindromes with different digits?

Wait, for example, 121. If I tried to see if 121 can be a sum. Even though it's not divisible by 3, but let me check. Suppose someone says that maybe the three numbers aren't consecutive integers? Wait, no, the problem says "three consecutive positive integers". So they must be consecutive. Therefore, their sum must be divisible by 3. Therefore, 121 cannot be the sum.

But maybe there's another way? Wait, maybe the problem is not about the sum being a palindrome, but the three numbers themselves? Wait, no, the problem says "the sum of a particular set of three consecutive positive integers is a three-digit palindrome." So the sum is the palindrome. Therefore, we can't have the numbers themselves being palindromes. So that's not the case.

Therefore, according to this, the answer is 58. However, I feel like 58 is too small. Let me check if I messed up the calculation. Let me think again. If the sum is 171, then the numbers are 56, 57, 58. Their sum is 171. Correct. The largest is 58. If the sum is 141, then numbers are (141 / 3) - 1, so 47 -1=46? Wait, no. Wait, S = 3(x + 1). Therefore, x + 1 = S / 3. Therefore, x = (S / 3) - 1. Therefore, largest number is x + 2 = (S / 3) - 1 + 2 = (S / 3) + 1. Therefore, for S = 141, largest number is 141 / 3 + 1 = 47 + 1 = 48. Similarly for S = 111, it's 37 + 1 = 38. So 58 is the largest among those. But maybe there is another possibility? Let me think.

Wait, perhaps there is another way that the sum can be a palindrome? Let me check if there are palindromes that are divisible by 3, but I didn't consider because I thought they weren't. For example, 3-digit palindromes where the first and third digits are same. But I considered all of them. Let me check for palindromes like 151: 1+5+1=7, not divisible by 3. 161: 1+6+1=8. 171: 9. 181: 10. 191: 11. So 171 is the only one in that 1xx range. Then in the 2xx range, palindromes are 202, 212. Their digit sums are 4 and 5. Not divisible by 3. Therefore, only 111, 141, 171.

Wait, but maybe there are palindromes that are not three-digit? But the problem says the sum is a three-digit palindrome. Therefore, not possible. Therefore, only those three sums. Hence, 58 is the answer. Wait, maybe the problem is in the interpretation? Let me check the problem again:

"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?"

So, the sum is a three-digit palindrome less than 220. Therefore, the sum can be 101, 111,..., up to 212. Then, the sum must be divisible by 3. Therefore, only 111, 141, 171.

Therefore, the maximum sum is 171, leading to the largest integer being 58. But maybe there is a different way to think about this. For example, maybe considering that the sum is a palindrome, but not necessarily that the sum is 3 times the middle number. Wait, but that is the case. Let me think again. If the three numbers are consecutive integers, then their sum is 3 times the middle number, so it must be divisible by 3. Therefore, sum is 3 times something. Therefore, only palindromes divisible by 3 are allowed. So I think my reasoning is correct.

But let me check for example, take a sum of 171, which is allowed. The largest number is 58. But maybe there is another set of three numbers where the sum is a palindrome less than 220, but with a larger maximum number. How?

For example, suppose that the sum is 198. Wait, 198 is not a palindrome. But let me check. Suppose the sum is something like 202, which is a palindrome. But 202 is not divisible by 3. If I check 202 / 3 is 67.333... Not integer. So no. What about 171? Which is 171. Then, if sum is 171, the numbers are 56, 57, 58. The largest is 58. But let me think: is there a way to get a larger maximum number?

Suppose, for example, that the sum is 183, but 183 is not a palindrome. 183 reversed is 381. Not same. How about 191? Not divisible by 3. 191. 191 is a palindrome, but not divisible by 3. If I force the sum to be 191, then the numbers would be? Let me check. If sum is 191, then since 191 is not divisible by 3, there are no three consecutive integers adding up to 191. Therefore, not possible.

Wait, but maybe the problem is not that the sum is divisible by 3? Wait, no, because three consecutive integers must add up to a multiple of 3. Let me confirm with an example. Let me take numbers 1, 2, 3. Sum is 6, which is 3*2. Numbers 2, 3, 4. Sum is 9, which is 3*3. So yes, always divisible by 3. Therefore, no matter what three consecutive integers you take, the sum will be divisible by 3. Therefore, the sum must be divisible by 3. Therefore, only palindromes divisible by 3 can be considered. Therefore, only 111, 141, 171. Therefore, the answer should be 58. But let me check if I made an error in the calculation of the largest number.

Wait, the largest number is S / 3 + 1. For S = 171, that is 171 / 3 + 1 = 57 + 1 = 58. Correct. So numbers are 56, 57, 58. But wait, 56 + 57 + 58 = 171. Correct.

Wait, let me think if there is a different interpretation. Suppose the three numbers are not necessarily in the form x, x+1, x+2 but something else? But consecutive integers are always like that. So no.

Wait, another thought: maybe the problem allows the sum to be a palindrome, but not necessarily three-digit? But the problem says "the sum is a three-digit palindrome". So no. Therefore, the answer is 58. But maybe I need to check other palindromes. Wait, perhaps 191? Wait, no, not divisible by 3. 181? Also no. 202? No. 212? No. What about numbers like 3-digit palindromes with middle digit 9? For example, 191, which we already checked. Not divisible by 3. 191: 1 + 9 + 1 = 11, not divisible by 3. So no.

Wait, but hold on. What about 3-digit palindromes with first digit 1 and last digit 1, but with middle digit such that the sum is divisible by 3? For example, 111, 141, 171. Which we have already considered. So no others. Therefore, the answer is 58.

Wait, but I feel like maybe there's a mistake here. Let me check another approach. Let me suppose that the sum is a palindrome, say S, which is 3k for some integer k, and S is a three-digit palindrome less than 220. Then k = S / 3. Then the three numbers would be k - 1, k, k + 1. Therefore, the largest number is k + 1. Therefore, to maximize the largest number, we need to maximize k. Since S = 3k, and S is less than 220. Therefore, 3k < 220 => k < 220 / 3 ≈ 73.333. Therefore, maximum integer k is 73. Therefore, maximum S is 3*73 = 219. But 219 is not a palindrome. So we need to check palindromes less than 220, which are divisible by 3, and find the largest such palindrome, then compute k = S / 3, and then the largest number is k + 1.

But earlier, I thought that palindromes less than 220 and divisible by 3 are only 111, 141, 171. But 219 is not a palindrome. The next palindrome less than 220 is 212, which is not divisible by 3. Then 202, not divisible by 3. Then 191, no. 181, no. 171, yes. So 171 is the largest palindrome less than 220 divisible by 3. Therefore, k = 171 / 3 = 57, so the largest number is 57 + 1 = 58. Therefore, the answer is 58. So that's the same result as before. Therefore, 58 is correct? But let me check if there are palindromes between 171 and 220 which are divisible by 3. Let's see:

Palindromes between 171 and 220: 181, 191, 202, 212. All of these are not divisible by 3. 181: 1+8+1=10, not divisible by 3. 191: 1+9+1=11, nope. 202: 2+0+2=4, nope. 212: 2+1+2=5, nope. Therefore, 171 is the largest palindrome less than 220 divisible by 3. Therefore, answer is 58.

Wait, but hold on. What about 3-digit palindromes starting with 1, but higher than 171? Let me check. Between 171 and 220, the next palindrome after 171 would be 181, which we saw. Then 191, then 202, 212. So no. Therefore, 171 is the largest. So answer is 58. Therefore, maybe the answer is 58? But let me check once again. If I think of the largest number possible, maybe there's another set of numbers where the sum is a palindrome less than 220, but not necessarily that the sum is divisible by 3? Wait, but that's impossible, because three consecutive numbers must add to a multiple of 3. So the sum has to be divisible by 3. So no other possibilities.

Alternatively, maybe the problem allows for the numbers to be non-consecutive? But the problem says three consecutive positive integers. Therefore, they have to be consecutive. So no.

Alternatively, maybe I made a mistake in assuming that the middle number is k? Let me think again. If the three numbers are consecutive, say n-1, n, n+1, then the sum is 3n. So the sum is 3n, so S = 3n. So n = S / 3. Then the largest number is n + 1 = S / 3 + 1. Therefore, if S is 171, then the largest number is 171 / 3 + 1 = 57 + 1 = 58. Therefore, correct. Therefore, maybe 58 is correct. But why do I feel unsure? Maybe because 58 seems small. Let me check with another example. Let me check the sum of 171, which is 56 + 57 + 58 = 171. Correct. If I take the next possible palindrome divisible by 3, which is 171, that's the largest. Therefore, 58 is the answer.

Alternatively, maybe there is a mistake in considering the sum. Wait, if the sum is 183, which is not a palindrome, but 183 is 3*61. Therefore, numbers would be 60, 61, 62. Their sum is 183. Not a palindrome. But 183 is not a palindrome. Therefore, no. Similarly, 192 is 3*64, numbers 63, 64, 65. Sum is 192, which isn't a palindrome. 201 is 3*67, numbers 66, 67, 68. Sum is 201. 201 is not a palindrome. 210 is 3*70, numbers 69, 70, 71. Sum is 210. Not a palindrome. 219 is 3*73, numbers 72, 73, 74. Sum 219. Not a palindrome. Therefore, none of those sums are palindromes. Therefore, indeed, only 111, 141, 171. Therefore, the answer is 58. Therefore, maybe the answer is 58.

Wait, but hold on. Let me check if there are other palindromes in the 100s that I missed. For example, 121. Let me check if 121 is divisible by 3. 1 + 2 + 1 = 4, not divisible by 3. 131: 1+3+1=5. Not. 141: 6. 151: 7. 161: 8. 171: 9. 181: 10. 191: 11. So, no, only those three. Therefore, seems correct.

But wait, what about 3-digit palindromes that start with 1, middle digit 8, which is 181. Not divisible by 3. So, yes. Therefore, I think that the answer is 58. Maybe I was overthinking.

Wait, but let me check with S = 171, which gives the numbers 56, 57, 58. Largest number is 58. But maybe there's another set of three consecutive numbers where the sum is a palindrome less than 220, but the numbers are larger? For example, suppose that the sum is 198. Wait, 198 is not a palindrome. But if there is a palindrome like 181 or something else. Wait, but again, 181 is not divisible by 3. So no.

Alternatively, suppose the sum is 183, not a palindrome. 192 isn't. 198 isn't. 201 isn't. 207 isn't. 210 isn't. 213 isn't. 216 isn't. 219 isn't. So none of these are palindromes. So no. Therefore, no other possibilities. Therefore, the answer is 58. Therefore, maybe 58 is correct. Therefore, I think the answer is 58.

Wait, but to be thorough, let me check if there's a palindrome with leading zeros? But no, palindromes are three-digit numbers, which can't have leading zeros. So 010 is just 10, which is two-digit. Therefore, not allowed. Therefore, no.

Therefore, after all that thinking, I think the answer is 58. Therefore, the greatest possible value for the largest of the three integers is 58.

But wait, hold on. Let me check once again for palindromes. Let me check 3-digit palindromes divisible by 3 less than 220. So 111, 141, 171. What about 3-digit palindromes starting with 2? Like 202, 212. 2+0+2=4, not divisible by 3. 2+1+2=5, not. So no. What about 222? It's 222, which is 222. But the problem says the sum is less than 220, so 222 is excluded. So no. Therefore, no other palindromes.

Therefore, my conclusion is that the answer is 58. Therefore, I think I should go with 58. However, just to make sure, let me check if there's an alternative interpretation of the problem.

Wait, maybe the problem isn't saying that the sum is a palindrome, but that the three numbers are palindromes? But the wording says: "The sum of a particular set of three consecutive positive integers is a three-digit palindrome." So it's the sum that's the palindrome, not the numbers themselves. Therefore, my original interpretation is correct.

Therefore, after thorough checking, I think the answer is 58. Therefore, the greatest possible value is \boxed{58}.

Wait, hold on, but I just had a thought. Let me check with another approach. Let me suppose that the three numbers are n, n+1, n+2. Their sum is 3n + 3 = 3(n + 1). Therefore, the sum S = 3(n + 1). Therefore, S must be divisible by 3. So we need S to be a three-digit palindrome less than 220, so S ∈ {101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212}, and S must be divisible by 3. As established before, only 111, 141, 171. Therefore, the maximum S is 171, which gives n + 1 = 57, so n = 56, so the numbers are 56, 57, 58. Therefore, the largest is 58. Therefore, answer is 58. Therefore, I think this is correct.

Alternatively, maybe the problem is from a source where the answer is different? But I don't think so. Unless there is an error in my logic. Let me think again: the sum is a three-digit palindrome. Therefore, it's between 100 and 219. The sum must be divisible by 3. Therefore, the palindromes divisible by 3 are 111, 141, 171. Therefore, the maximum sum is 171. Therefore, the maximum largest integer is 58. Yes. Therefore, I think the answer is 58. Therefore, I will go with that.

**Final Answer**
\boxed{58}

To solve this problem, we are given the following:

- We are to find three **consecutive positive integers** whose **sum is a three-digit palindrome**, and the **sum is less than 220**.
- We are to determine the **greatest possible value** for the **largest of the three integers** in the set.

---

### Step 1: Understanding the Sum of Three Consecutive Integers

Let the three consecutive integers be:

$$
n, \, n+1, \, n+2
$$

Their sum is:

$$
n + (n+1) + (n+2) = 3n + 3 = 3(n + 1)
$$

This tells us that the **sum must be divisible by 3**, making it crucial to consider **three-digit palindromes divisible by 3**, less than 220.

---

### Step 2: List of Valid Palindromes

A **three-digit palindrome** reads the same forwards and backwards. So the number must be of the form $ ABA $, where $ A $ is a digit from 1 to 9, and $ B $ is any digit from 0 to 9.

So, all three-digit palindromes less than 220 are:

$$
101, \, 111, \, 121, \, 131, \, 141, \, 151, \, 161, \, 171, \, 181, \, 191, \, 202, \, 212
$$

We now check which of these are divisible by 3 by adding their digits:

- 101 → 1+0+1 = 2 → Not divisible by 3
- 111 → 3 → ✅
- 121 → 4 → ❌
- 131 → 5 → ❌
- 141 → 6 → ✅
- 151 → 7 → ❌
- 161 → 8 → ❌
- 171 → 9 → ✅
- 181 → 10 → ❌
- 191 → 11 → ❌
- 202 → 4 → ❌
- 212 → 5 → ❌

✅ **Valid Sums**: 111, 141, 171

---

### Step 3: Determine the Largest Integer in the Set

We now compute the corresponding three consecutive integers for each valid sum (which is of the form $ S = 3(n + 1) $):

- **If $ S = 111 $:**
  $$
  n + 1 = \frac{111}{3} = 37,\quad n = 36
  $$
  The integers are: 36, 37, 38
  **Largest = 38**

- **If $ S = 141 $:**
  $$
  n + 1 = \frac{141}{3} = 47,\quad n = 46
  $$
  The integers are: 46, 47, 48
  **Largest = 48**

- **If $ S = 171 $:**
  $$
  n + 1 = \frac{171}{3} = 57,\quad n = 56
  $$
  The integers are: 56, 57, 58
  **Largest = 58**

---

### Step 4: Conclusion

Among all valid sets, the largest value of the **largest integer** occurs when the **sum is 171**, and the corresponding integers are **56, 57, 58**.

Thus, the **greatest possible value for the largest of the three integers** is:

$$
\boxed{58}
$$

Solution: The sum of the set $\{n-1, n, n+1\}$ of three consecutive integers is $3n$.  Therefore, we are looking for the largest three-digit palindromic multiple of $3$ less than $220$.  Checking through $212, 202, 191, 181$, and $171$, we find that $171$ is the greatest palindrome which is also a multiple of $3$.  Solving $3n=171$ for $n$ we find $n=57$. The three integers are $56,57,58$, and the greatest is $\boxed{58}$.