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 sequence is defined with the first four terms:

- $ x_1 = 211 $
- $ x_2 = 375 $
- $ x_3 = 420 $
- $ x_4 = 523 $

And for $ n \geq 5 $, each term is defined by the recurrence relation:

$ x_n = x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4} $

Hmm, so each term depends on the previous four terms. This seems like a linear recurrence relation. Maybe I can find a pattern or periodicity in the sequence? Because if the sequence is periodic, then I can find the period and compute the required terms modulo that period.

Let me try to compute the first few terms to see if there's a pattern or repetition.

Starting with the given terms:

$ x_1 = 211 $

$ x_2 = 375 $

$ x_3 = 420 $

$ x_4 = 523 $

Now compute $ x_5 $:

$ x_5 = x_4 - x_3 + x_2 - x_1 = 523 - 420 + 375 - 211 $

Let me calculate that step by step:

523 - 420 = 103

103 + 375 = 478

478 - 211 = 267

So $ x_5 = 267 $

Now $ x_6 = x_5 - x_4 + x_3 - x_2 $

Plugging in the numbers:

267 - 523 + 420 - 375

Compute step by step:

267 - 523 = -256

-256 + 420 = 164

164 - 375 = -211

So $ x_6 = -211 $

Next, $ x_7 = x_6 - x_5 + x_4 - x_3 $

-211 - 267 + 523 - 420

Calculating:

-211 - 267 = -478

-478 + 523 = 45

45 - 420 = -375

So $ x_7 = -375 $

Then $ x_8 = x_7 - x_6 + x_5 - x_4 $

-375 - (-211) + 267 - 523

Simplify:

-375 + 211 = -164

-164 + 267 = 103

103 - 523 = -420

So $ x_8 = -420 $

Next, $ x_9 = x_8 - x_7 + x_6 - x_5 $

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

Calculating:

-420 + 375 = -45

-45 - 211 = -256

-256 - 267 = -523

So $ x_9 = -523 $

Now $ x_{10} = x_9 - x_8 + x_7 - x_6 $

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

Simplify:

-523 + 420 = -103

-103 - 375 = -478

-478 + 211 = -267

So $ x_{10} = -267 $

Moving on to $ x_{11} = x_{10} - x_9 + x_8 - x_7 $

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

Calculating:

-267 + 523 = 256

256 - 420 = -164

-164 + 375 = 211

So $ x_{11} = 211 $

Wait a second! $ x_{11} = 211 $, which is the same as $ x_1 $. Let me check if the following terms also repeat.

Compute $ x_{12} = x_{11} - x_{10} + x_9 - x_8 $

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

Simplify:

211 + 267 = 478

478 - 523 = -45

-45 + 420 = 375

So $ x_{12} = 375 $, which is $ x_2 $.

Then $ x_{13} = x_{12} - x_{11} + x_{10} - x_9 $

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

Calculating:

375 - 211 = 164

164 - 267 = -103

-103 + 523 = 420

So $ x_{13} = 420 $, which is $ x_3 $.

Next, $ x_{14} = x_{13} - x_{12} + x_{11} - x_{10} $

420 - 375 + 211 - (-267)

Calculating:

420 - 375 = 45

45 + 211 = 256

256 + 267 = 523

So $ x_{14} = 523 $, which is $ x_4 $.

Then $ x_{15} = x_{14} - x_{13} + x_{12} - x_{11} $

523 - 420 + 375 - 211

Wait, that's exactly the same as $ x_5 $, which was 267. So $ x_{15} = 267 $, same as $ x_5 $.

So it seems like the sequence is periodic with period 10. Let me check:

Original terms:

x1=211, x2=375, x3=420, x4=523, x5=267, x6=-211, x7=-375, x8=-420, x9=-523, x10=-267, x11=211, x12=375,...

Yes! So starting from x1, the sequence repeats every 10 terms. Therefore, the period is 10.

So the sequence is periodic with period 10. Therefore, to find x_n, we can compute n modulo 10, and then find the corresponding term in the first 10 terms.

But let me confirm this periodicity. Let me check a few more terms to be sure.

x11=211 (same as x1)

x12=375 (same as x2)

x13=420 (same as x3)

x14=523 (same as x4)

x15=267 (same as x5)

x16= x15 - x14 + x13 - x12 = 267 - 523 + 420 - 375 = same as x6: -211

Yes, same as x6. So the pattern continues. Therefore, the period is indeed 10.

Therefore, the sequence repeats every 10 terms. Therefore, for any n, x_n = x_{n mod 10}, but we have to be careful with the modulus. Let me check:

For example, n=11: 11 mod 10 = 1, so x11 = x1. Similarly, n=12: 12 mod 10 = 2, so x12 = x2. However, note that when n is a multiple of 10, say n=10, 10 mod 10 = 0. But in our case, the original terms start at n=1, so perhaps we need to adjust the modulus accordingly.

Wait, let me think. If the period is 10, then x_{k} = x_{k + 10} for all k ≥ 1. Therefore, to find x_n, we can compute n modulo 10, but if the remainder is 0, then it corresponds to the 10th term. For example:

If n=10, then 10 mod 10 = 0, but we need to map that to x10. So perhaps we can define:

Let m = n mod 10. If m = 0, then the term is x10; otherwise, it's x_m.

Alternatively, since the period is 10, the sequence is:

Term number: 1 2 3 4 5 6 7 8 9 10 11 12 ...

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

So for term number n, if we compute (n - 1) mod 10, then:

For n=1: (1-1)=0 mod10=0 → index 0? But our first term is index 1. Maybe another way.

Alternatively, since the period is 10, then x_{n} = x_{n - 10} for n ≥ 11. Therefore, to find x_n, we can subtract 10 until we get into the first 10 terms.

Therefore, for any n, x_n = x_{n mod 10}, but if n mod 10 is 0, then it's x_{10}.

Alternatively, if we think of the sequence as starting at n=1, then the position within the period can be calculated as:

Let’s define k = n - 1. Then k mod 10 gives 0 to 9, which corresponds to the original terms 1 to 10. Therefore, x_n = x_{(k mod 10) + 1} where k = n - 1.

So for example, n=1: k=0, 0 mod10=0, so term is x_{0+1}=x1. Similarly, n=10: k=9, 9 mod10=9, term is x_{9+1}=x10. n=11: k=10, 10 mod10=0, term is x_{0+1}=x1. Perfect.

Therefore, in general, x_n = x_{(n - 1) mod 10 + 1}

Therefore, to compute x_n, we can compute (n - 1) mod 10, add 1, and get the position in the first 10 terms.

Therefore, for example, x_{531}:

Compute (531 - 1) mod 10 = 530 mod 10 = 0. Then add 1: 0 + 1 = 1. Therefore, x_{531} = x1 = 211.

Similarly, x_{753}:

(753 - 1) mod 10 = 752 mod 10 = 2. Add 1: 3. Therefore, x_{753} = x3 = 420.

Wait, hold on. Wait, let me check:

Wait, if (n - 1) mod 10 gives 0 to 9, then adding 1 gives 1 to 10. Therefore, if (n - 1) mod 10 = 0, then x_n = x1; if it's 1, x2; ... if it's 9, x10.

Therefore, for n=531:

(531 - 1) = 530. 530 divided by 10 is 53, remainder 0. Therefore, (n - 1) mod 10 = 0. Therefore, x_{531} = x1 = 211.

Similarly, n=753:

(753 - 1) = 752. 752 divided by 10 is 75*10=750, remainder 2. Therefore, (n - 1) mod 10 = 2. Therefore, x_{753} = x_{2 + 1} = x3? Wait, no:

Wait, if (n - 1) mod 10 = 2, then adding 1 gives 3? Wait, no:

Wait, the formula is x_n = x_{(n - 1) mod 10 + 1}

So if (n - 1) mod 10 = 0, then x_n = x_{0 + 1} = x1

If (n - 1) mod 10 = 1, then x_n = x_{1 + 1} = x2

...

If (n - 1) mod 10 = 9, then x_n = x_{9 + 1} = x10

Therefore, for n=753:

(n - 1) mod 10 = 752 mod 10 = 2. Therefore, x_{753} = x_{2 + 1} = x3? Wait, no:

Wait, (n - 1) mod 10 = 2, so the position is 2 + 1? Wait, no. Wait, the formula is:

Let me think again. Let me take n=1:

(n - 1) = 0, mod10=0, so position is 0 + 1 = 1. Correct.

n=2: (2 - 1)=1 mod10=1, position=1 + 1=2. Correct.

Wait, so the formula is x_n = x_{(n - 1) mod 10 + 1}

Therefore, if (n - 1) mod 10 = k, then x_n = x_{k + 1}

Therefore, for n=753:

k = (753 - 1) mod 10 = 752 mod 10 = 2. Therefore, x_{753} = x_{2 + 1} = x3. Which is 420. Correct.

Similarly, for n=975:

(975 - 1) = 974. 974 mod 10 = 4. Therefore, x_{975} = x_{4 + 1} = x5. Which is 267.

Wait, so then x_{531} + x_{753} + x_{975} = 211 + 420 + 267.

Let me compute that:

211 + 420 = 631; 631 + 267 = 898.

But wait, let me check again if my reasoning is correct. Because if the period is 10, then every 10 terms it repeats. Therefore, the value of x_n depends on n mod 10. However, I need to make sure that my calculation for each term is correct.

Let me check for n=10:

(n - 1) = 9, mod10=9, so x_{10} = x_{9 + 1} = x10. Correct.

n=11: (11 - 1)=10 mod10=0, so x_{11}=x_{0 + 1}=x1. Correct.

Therefore, the formula is correct.

So for n=531:

531 divided by 10 is 53*10 + 1. Therefore, 531 = 53*10 + 1. Therefore, the remainder is 1. Therefore, since the period is 10, x_{531} = x1 = 211. Correct.

Similarly, n=753:

Divide 753 by 10: 75*10 = 750, 753 - 750 = 3. Therefore, remainder is 3. Therefore, x_{753} = x3 = 420. Wait, but according to previous calculation, (n - 1) mod10 = 2. Wait, there's a discrepancy here. Wait, which is correct?

Wait, if n=753, then n mod10 is 3. But according to the formula, (n - 1) mod10 = 752 mod10=2. Therefore, x_{753} = x_{2 + 1} = x3. Which is same as n mod10=3, so x_{n} = x_{n mod10} if n mod10 ≠0, else x_{10}. Wait, but here n mod10=3, so x_{753}=x3. Which is same as the formula. So both ways, if I think of n mod10, then:

If n mod10 = 0, then x_n = x10, else x_n = x_{n mod10}

But in our case, for n=753, n mod10=3, so x_{753}=x3. Which is correct. So both methods are equivalent.

Therefore, perhaps another way to think is:

If the period is 10, then x_n = x_{n mod10} if n mod10 ≠0, else x_{10}

But in our case, since the first term is x1, so for n=10, which is 10 mod10=0, we have x_{10}.

Therefore, for n=531:

531 divided by 10 is 53*10 + 1, so remainder 1. Therefore, x_{531}=x1=211.

For n=753:

753 divided by 10 is 75*10 + 3, remainder 3. Therefore, x_{753}=x3=420.

For n=975:

975 divided by 10 is 97*10 + 5, remainder 5. Therefore, x_{975}=x5=267.

Therefore, adding them up: 211 + 420 + 267.

Let me compute that again:

211 + 420 = 631

631 + 267:

631 + 200 = 831

831 + 67 = 898

So total is 898.

But wait, let me check if my calculation of x5 is correct. Earlier, I calculated x5 as 267. Let me check again:

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

Compute step by step:

523 - 420 = 103

103 + 375 = 478

478 - 211 = 267. Yes, correct.

So x5=267. Therefore, x_{975}=x5=267. So the sum is 211 + 420 + 267 = 898.

But wait, let me check if there's a mistake in the periodicity. Because when I calculated up to x10, then x11=x1, so the period is 10. But let me check x10 and x_{20} to be sure.

Earlier, x10=-267. Then x11=211, x12=375,..., x20 would be same as x10, which is -267. Let me check x20:

But maybe it's tedious, but since the recurrence is linear and we saw that after 10 terms it repeats, and the recurrence is of order 4, but with period 10, which is longer than the order. However, since we saw that the sequence repeats after 10 terms, and the calculation for x11 to x20 would follow the same pattern, so I think the periodicity is correct.

Therefore, the answer should be 898. However, let me check once again for n=531, 753, 975.

Alternatively, maybe I can check with another approach. Let me compute n mod 10 for each term:

For 531:

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

For 753:

753 divided by 10: 75*10=750, 753-750=3. Remainder 3. Therefore, x_{753}=x3=420.

For 975:

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

Therefore, sum is 211 + 420 + 267 = 898. So that seems correct.

But let me check if there's a possibility that the period is different? For example, maybe the period is 20 or something else? Wait, but we saw that x11=x1, x12=x2,..., so the period is 10. So unless there is a mistake in my calculation of the terms, which I checked up to x14, which repeats x4, so seems correct.

Alternatively, maybe I can check the recurrence relation for the periodicity. Suppose that the sequence is periodic with period 10, then for all n ≥ 1, x_{n+10} = x_n. Let me check for n=1:

x_{11} = x1 = 211. Correct.

Similarly, x_{12}=x2=375. Correct.

So if this holds for all n, then the period is indeed 10. Therefore, the answer is 898.

But let me check another term. For example, x_{10} = -267. Then x_{20} should be equal to x_{10} = -267. Let me compute x_{10} = -267, then x_{11}=211, x_{12}=375, x_{13}=420, x_{14}=523, x_{15}=267, x_{16}=-211, x_{17}=-375, x_{18}=-420, x_{19}=-523, x_{20}= -267. Yes, same as x_{10}. Therefore, the period is indeed 10. Therefore, the conclusion is correct.

Therefore, the answer is 211 + 420 + 267 = 898.

But wait, let me check once again the calculation of the sum:

211 + 420 = 631. Then 631 + 267:

Let me compute 631 + 200 = 831, then 831 + 67 = 898. Yes.

Alternatively, 211 + 267 = 478, then 478 + 420 = 898. Same result.

Therefore, I think the answer is 898.

But just to be thorough, let me check if there's a possibility that the problem is in the indexing. For example, sometimes sequences start at n=0, but here it's given as n=1, so I think the calculation is correct.

Alternatively, maybe I made a mistake in the initial terms? Let me check again:

x1=211, x2=375, x3=420, x4=523.

x5 = x4 - x3 + x2 - x1 = 523 - 420 + 375 - 211 = 523 - 420 is 103, 103 + 375 is 478, 478 - 211 is 267. Correct.

x6 = x5 - x4 + x3 - x2 = 267 - 523 + 420 - 375. 267 - 523 = -256, -256 + 420 = 164, 164 - 375 = -211. Correct.

x7 = x6 - x5 + x4 - x3 = -211 - 267 + 523 - 420. -211 -267 = -478, -478 + 523 = 45, 45 - 420 = -375. Correct.

x8 = x7 - x6 + x5 - x4 = -375 - (-211) + 267 - 523 = -375 + 211 = -164, -164 + 267 = 103, 103 - 523 = -420. Correct.

x9 = x8 - x7 + x6 - x5 = -420 - (-375) + (-211) - 267 = -420 + 375 = -45, -45 -211 = -256, -256 -267 = -523. Correct.

x10 = x9 - x8 + x7 - x6 = -523 - (-420) + (-375) - (-211) = -523 + 420 = -103, -103 - 375 = -478, -478 + 211 = -267. Correct.

x11 = x10 - x9 + x8 - x7 = -267 - (-523) + (-420) - (-375) = -267 + 523 = 256, 256 - 420 = -164, -164 + 375 = 211. Correct. So x11 = x1. All steps are correct.

Therefore, the periodicity is confirmed. Therefore, the answer is 898.

But just to be 100% sure, let me check another term. Let's compute x_{15} which should be equal to x5=267.

x11=211, x12=375, x13=420, x14=523, x15 = x14 - x13 + x12 - x11 = 523 - 420 + 375 - 211 = same as x5, which is 267. Correct.

Similarly, x_{16} = x15 - x14 + x13 - x12 = 267 - 523 + 420 - 375 = same as x6=-211. Correct.

Therefore, the periodicity is confirmed. Therefore, the answer is 898.

But wait, the problem says "find the value of $ x_{531} + x_{753} + x_{975} $." So according to my calculation, it's 211 + 420 + 267 = 898. However, let me check once again the values for each term:

For x_{531}:

Since 531 divided by 10 is 53*10 + 1, so remainder 1. Therefore, x_{531} = x1 = 211. Correct.

For x_{753}:

753 divided by 10 is 75*10 + 3, remainder 3. Therefore, x_{753} = x3 = 420. Correct.

For x_{975}:

975 divided by 10 is 97*10 + 5, remainder 5. Therefore, x_{975} = x5 = 267. Correct.

Therefore, adding them up: 211 + 420 + 267 = 898. So I think that's the answer.

But just to be thorough, let me check if there's an alternative way to compute this, maybe using the recurrence relation properties or matrix exponentiation or something else, but since the period is 10, it's straightforward.

Alternatively, maybe there's a mistake in assuming the period is 10? Let me check for n=10, x_{10}=-267, and then x_{10 + 10}=x_{20}=-267. Let me compute x_{15}=267, x_{16}=-211, x_{17}=-375, x_{18}=-420, x_{19}=-523, x_{20}= -267. Yes, same as x_{10}. Therefore, the period is indeed 10.

Therefore, I think my answer is correct. Therefore, the value is 898.

**Final Answer**
\boxed{898}

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

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

We are asked to compute:

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

---

### Step 1: Identify the Pattern

We begin by computing the first few terms of the sequence to look for a pattern or periodicity.

$$
\begin{align*}
x_1 &= 211 \\
x_2 &= 375 \\
x_3 &= 420 \\
x_4 &= 523 \\
x_5 &= x_4 - x_3 + x_2 - x_1 = 523 - 420 + 375 - 211 = 267 \\
x_6 &= x_5 - x_4 + x_3 - x_2 = 267 - 523 + 420 - 375 = -211 \\
x_7 &= x_6 - x_5 + x_4 - x_3 = -211 - 267 + 523 - 420 = -375 \\
x_8 &= x_7 - x_6 + x_5 - x_4 = -375 - (-211) + 267 - 523 = -420 \\
x_9 &= x_8 - x_7 + x_6 - x_5 = -420 - (-375) + (-211) - 267 = -523 \\
x_{10} &= x_9 - x_8 + x_7 - x_6 = -523 - (-420) + (-375) - (-211) = -267 \\
x_{11} &= x_{10} - x_9 + x_8 - x_7 = -267 - (-523) + (-420) - (-375) = 211 \\
x_{12} &= x_{11} - x_{10} + x_9 - x_8 = 211 - (-267) + (-523) - (-420) = 375 \\
x_{13} &= x_{12} - x_{11} + x_{10} - x_9 = 375 - 211 + (-267) - (-523) = 420 \\
x_{14} &= x_{13} - x_{12} + x_{11} - x_{10} = 420 - 375 + 211 - (-267) = 523 \\
x_{15} &= x_{14} - x_{13} + x_{12} - x_{11} = 523 - 420 + 375 - 211 = 267 \\
\end{align*}
$$

We observe that:

$$
x_{11} = x_1, \quad x_{12} = x_2, \quad x_{13} = x_3, \quad x_{14} = x_4, \quad x_{15} = x_5
$$

This confirms that the sequence is **periodic with period 10**.

---

### Step 2: Use the Periodicity to Compute the Required Terms

Since the sequence is periodic with period 10, we can compute any term $ x_n $ by finding the position of $ n $ within the first 10 terms.

To do this, we compute:

$$
n \mod 10
$$

- If the remainder is 0, then $ x_n = x_{10} $
- Otherwise, $ x_n = x_{\text{remainder}} $

Let’s compute each term:

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

---

### Step 3: Compute the Final 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 sequence is defined with the first four terms:

- $ x_1 = 211 $
- $ x_2 = 375 $
- $ x_3 = 420 $
- $ x_4 = 523 $

And for $ n \geq 5 $, each term is defined by the recurrence relation:

$ x_n = x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4} $

Hmm, so each term depends on the previous four terms. This seems like a linear recurrence relation. Maybe I can find a pattern or periodicity in the sequence? Because if the sequence is periodic, then I can find the period and compute the required terms modulo that period.

Let me try to compute the first few terms to see if there's a pattern or repetition.

Starting with the given terms:

$ x_1 = 211 $

$ x_2 = 375 $

$ x_3 = 420 $

$ x_4 = 523 $

Now compute $ x_5 $:

$ x_5 = x_4 - x_3 + x_2 - x_1 = 523 - 420 + 375 - 211 $

Let me calculate that step by step:

523 - 420 = 103

103 + 375 = 478

478 - 211 = 267

So $ x_5 = 267 $

Now $ x_6 = x_5 - x_4 + x_3 - x_2 $

Plugging in the numbers:

267 - 523 + 420 - 375

Compute step by step:

267 - 523 = -256

-256 + 420 = 164

164 - 375 = -211

So $ x_6 = -211 $

Next, $ x_7 = x_6 - x_5 + x_4 - x_3 $

-211 - 267 + 523 - 420

Calculating:

-211 - 267 = -478

-478 + 523 = 45

45 - 420 = -375

So $ x_7 = -375 $

Then $ x_8 = x_7 - x_6 + x_5 - x_4 $

-375 - (-211) + 267 - 523

Simplify:

-375 + 211 = -164

-164 + 267 = 103

103 - 523 = -420

So $ x_8 = -420 $

Next, $ x_9 = x_8 - x_7 + x_6 - x_5 $

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

Calculating:

-420 + 375 = -45

-45 - 211 = -256

-256 - 267 = -523

So $ x_9 = -523 $

Now $ x_{10} = x_9 - x_8 + x_7 - x_6 $

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

Simplify:

-523 + 420 = -103

-103 - 375 = -478

-478 + 211 = -267

So $ x_{10} = -267 $

Moving on to $ x_{11} = x_{10} - x_9 + x_8 - x_7 $

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

Calculating:

-267 + 523 = 256

256 - 420 = -164

-164 + 375 = 211

So $ x_{11} = 211 $

Wait a second! $ x_{11} = 211 $, which is the same as $ x_1 $. Let me check if the following terms also repeat.

Compute $ x_{12} = x_{11} - x_{10} + x_9 - x_8 $

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

Simplify:

211 + 267 = 478

478 - 523 = -45

-45 + 420 = 375

So $ x_{12} = 375 $, which is $ x_2 $.

Then $ x_{13} = x_{12} - x_{11} + x_{10} - x_9 $

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

Calculating:

375 - 211 = 164

164 - 267 = -103

-103 + 523 = 420

So $ x_{13} = 420 $, which is $ x_3 $.

Next, $ x_{14} = x_{13} - x_{12} + x_{11} - x_{10} $

420 - 375 + 211 - (-267)

Calculating:

420 - 375 = 45

45 + 211 = 256

256 + 267 = 523

So $ x_{14} = 523 $, which is $ x_4 $.

Then $ x_{15} = x_{14} - x_{13} + x_{12} - x_{11} $

523 - 420 + 375 - 211

Wait, that's exactly the same as $ x_5 $, which was 267. So $ x_{15} = 267 $, same as $ x_5 $.

So it seems like the sequence is periodic with period 10. Let me check:

Original terms:

x1=211, x2=375, x3=420, x4=523, x5=267, x6=-211, x7=-375, x8=-420, x9=-523, x10=-267, x11=211, x12=375,...

Yes! So starting from x1, the sequence repeats every 10 terms. Therefore, the period is 10.

So the sequence is periodic with period 10. Therefore, to find x_n, we can compute n modulo 10, and then find the corresponding term in the first 10 terms.

But let me confirm this periodicity. Let me check a few more terms to be sure.

x11=211 (same as x1)

x12=375 (same as x2)

x13=420 (same as x3)

x14=523 (same as x4)

x15=267 (same as x5)

x16= x15 - x14 + x13 - x12 = 267 - 523 + 420 - 375 = same as x6: -211

Yes, same as x6. So the pattern continues. Therefore, the period is indeed 10.

Therefore, the sequence repeats every 10 terms. Therefore, for any n, x_n = x_{n mod 10}, but we have to be careful with the modulus. Let me check:

For example, n=11: 11 mod 10 = 1, so x11 = x1. Similarly, n=12: 12 mod 10 = 2, so x12 = x2. However, note that when n is a multiple of 10, say n=10, 10 mod 10 = 0. But in our case, the original terms start at n=1, so perhaps we need to adjust the modulus accordingly.

Wait, let me think. If the period is 10, then x_{k} = x_{k + 10} for all k ≥ 1. Therefore, to find x_n, we can compute n modulo 10, but if the remainder is 0, then it corresponds to the 10th term. For example:

If n=10, then 10 mod 10 = 0, but we need to map that to x10. So perhaps we can define:

Let m = n mod 10. If m = 0, then the term is x10; otherwise, it's x_m.

Alternatively, since the period is 10, the sequence is:

Term number: 1 2 3 4 5 6 7 8 9 10 11 12 ...

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

So for term number n, if we compute (n - 1) mod 10, then:

For n=1: (1-1)=0 mod10=0 → index 0? But our first term is index 1. Maybe another way.

Alternatively, since the period is 10, then x_{n} = x_{n - 10} for n ≥ 11. Therefore, to find x_n, we can subtract 10 until we get into the first 10 terms.

Therefore, for any n, x_n = x_{n mod 10}, but if n mod 10 is 0, then it's x_{10}.

Alternatively, if we think of the sequence as starting at n=1, then the position within the period can be calculated as:

Let’s define k = n - 1. Then k mod 10 gives 0 to 9, which corresponds to the original terms 1 to 10. Therefore, x_n = x_{(k mod 10) + 1} where k = n - 1.

So for example, n=1: k=0, 0 mod10=0, so term is x_{0+1}=x1. Similarly, n=10: k=9, 9 mod10=9, term is x_{9+1}=x10. n=11: k=10, 10 mod10=0, term is x_{0+1}=x1. Perfect.

Therefore, in general, x_n = x_{(n - 1) mod 10 + 1}

Therefore, to compute x_n, we can compute (n - 1) mod 10, add 1, and get the position in the first 10 terms.

Therefore, for example, x_{531}:

Compute (531 - 1) mod 10 = 530 mod 10 = 0. Then add 1: 0 + 1 = 1. Therefore, x_{531} = x1 = 211.

Similarly, x_{753}:

(753 - 1) mod 10 = 752 mod 10 = 2. Add 1: 3. Therefore, x_{753} = x3 = 420.

Wait, hold on. Wait, let me check:

Wait, if (n - 1) mod 10 gives 0 to 9, then adding 1 gives 1 to 10. Therefore, if (n - 1) mod 10 = 0, then x_n = x1; if it's 1, x2; ... if it's 9, x10.

Therefore, for n=531:

(531 - 1) = 530. 530 divided by 10 is 53, remainder 0. Therefore, (n - 1) mod 10 = 0. Therefore, x_{531} = x1 = 211.

Similarly, n=753:

(753 - 1) = 752. 752 divided by 10 is 75*10=750, remainder 2. Therefore, (n - 1) mod 10 = 2. Therefore, x_{753} = x_{2 + 1} = x3? Wait, no:

Wait, if (n - 1) mod 10 = 2, then adding 1 gives 3? Wait, no:

Wait, the formula is x_n = x_{(n - 1) mod 10 + 1}

So if (n - 1) mod 10 = 0, then x_n = x_{0 + 1} = x1

If (n - 1) mod 10 = 1, then x_n = x_{1 + 1} = x2

...

If (n - 1) mod 10 = 9, then x_n = x_{9 + 1} = x10

Therefore, for n=753:

(n - 1) mod 10 = 752 mod 10 = 2. Therefore, x_{753} = x_{2 + 1} = x3? Wait, no:

Wait, (n - 1) mod 10 = 2, so the position is 2 + 1? Wait, no. Wait, the formula is:

Let me think again. Let me take n=1:

(n - 1) = 0, mod10=0, so position is 0 + 1 = 1. Correct.

n=2: (2 - 1)=1 mod10=1, position=1 + 1=2. Correct.

Wait, so the formula is x_n = x_{(n - 1) mod 10 + 1}

Therefore, if (n - 1) mod 10 = k, then x_n = x_{k + 1}

Therefore, for n=753:

k = (753 - 1) mod 10 = 752 mod 10 = 2. Therefore, x_{753} = x_{2 + 1} = x3. Which is 420. Correct.

Similarly, for n=975:

(975 - 1) = 974. 974 mod 10 = 4. Therefore, x_{975} = x_{4 + 1} = x5. Which is 267.

Wait, so then x_{531} + x_{753} + x_{975} = 211 + 420 + 267.

Let me compute that:

211 + 420 = 631; 631 + 267 = 898.

But wait, let me check again if my reasoning is correct. Because if the period is 10, then every 10 terms it repeats. Therefore, the value of x_n depends on n mod 10. However, I need to make sure that my calculation for each term is correct.

Let me check for n=10:

(n - 1) = 9, mod10=9, so x_{10} = x_{9 + 1} = x10. Correct.

n=11: (11 - 1)=10 mod10=0, so x_{11}=x_{0 + 1}=x1. Correct.

Therefore, the formula is correct.

So for n=531:

531 divided by 10 is 53*10 + 1. Therefore, 531 = 53*10 + 1. Therefore, the remainder is 1. Therefore, since the period is 10, x_{531} = x1 = 211. Correct.

Similarly, n=753:

Divide 753 by 10: 75*10 = 750, 753 - 750 = 3. Therefore, remainder is 3. Therefore, x_{753} = x3 = 420. Wait, but according to previous calculation, (n - 1) mod10 = 2. Wait, there's a discrepancy here. Wait, which is correct?

Wait, if n=753, then n mod10 is 3. But according to the formula, (n - 1) mod10 = 752 mod10=2. Therefore, x_{753} = x_{2 + 1} = x3. Which is same as n mod10=3, so x_{n} = x_{n mod10} if n mod10 ≠0, else x_{10}. Wait, but here n mod10=3, so x_{753}=x3. Which is same as the formula. So both ways, if I think of n mod10, then:

If n mod10 = 0, then x_n = x10, else x_n = x_{n mod10}

But in our case, for n=753, n mod10=3, so x_{753}=x3. Which is correct. So both methods are equivalent.

Therefore, perhaps another way to think is:

If the period is 10, then x_n = x_{n mod10} if n mod10 ≠0, else x_{10}

But in our case, since the first term is x1, so for n=10, which is 10 mod10=0, we have x_{10}.

Therefore, for n=531:

531 divided by 10 is 53*10 + 1, so remainder 1. Therefore, x_{531}=x1=211.

For n=753:

753 divided by 10 is 75*10 + 3, remainder 3. Therefore, x_{753}=x3=420.

For n=975:

975 divided by 10 is 97*10 + 5, remainder 5. Therefore, x_{975}=x5=267.

Therefore, adding them up: 211 + 420 + 267.

Let me compute that again:

211 + 420 = 631

631 + 267:

631 + 200 = 831

831 + 67 = 898

So total is 898.

But wait, let me check if my calculation of x5 is correct. Earlier, I calculated x5 as 267. Let me check again:

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

Compute step by step:

523 - 420 = 103

103 + 375 = 478

478 - 211 = 267. Yes, correct.

So x5=267. Therefore, x_{975}=x5=267. So the sum is 211 + 420 + 267 = 898.

But wait, let me check if there's a mistake in the periodicity. Because when I calculated up to x10, then x11=x1, so the period is 10. But let me check x10 and x_{20} to be sure.

Earlier, x10=-267. Then x11=211, x12=375,..., x20 would be same as x10, which is -267. Let me check x20:

But maybe it's tedious, but since the recurrence is linear and we saw that after 10 terms it repeats, and the recurrence is of order 4, but with period 10, which is longer than the order. However, since we saw that the sequence repeats after 10 terms, and the calculation for x11 to x20 would follow the same pattern, so I think the periodicity is correct.

Therefore, the answer should be 898. However, let me check once again for n=531, 753, 975.

Alternatively, maybe I can check with another approach. Let me compute n mod 10 for each term:

For 531:

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

For 753:

753 divided by 10: 75*10=750, 753-750=3. Remainder 3. Therefore, x_{753}=x3=420.

For 975:

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

Therefore, sum is 211 + 420 + 267 = 898. So that seems correct.

But let me check if there's a possibility that the period is different? For example, maybe the period is 20 or something else? Wait, but we saw that x11=x1, x12=x2,..., so the period is 10. So unless there is a mistake in my calculation of the terms, which I checked up to x14, which repeats x4, so seems correct.

Alternatively, maybe I can check the recurrence relation for the periodicity. Suppose that the sequence is periodic with period 10, then for all n ≥ 1, x_{n+10} = x_n. Let me check for n=1:

x_{11} = x1 = 211. Correct.

Similarly, x_{12}=x2=375. Correct.

So if this holds for all n, then the period is indeed 10. Therefore, the answer is 898.

But let me check another term. For example, x_{10} = -267. Then x_{20} should be equal to x_{10} = -267. Let me compute x_{10} = -267, then x_{11}=211, x_{12}=375, x_{13}=420, x_{14}=523, x_{15}=267, x_{16}=-211, x_{17}=-375, x_{18}=-420, x_{19}=-523, x_{20}= -267. Yes, same as x_{10}. Therefore, the period is indeed 10. Therefore, the conclusion is correct.

Therefore, the answer is 211 + 420 + 267 = 898.

But wait, let me check once again the calculation of the sum:

211 + 420 = 631. Then 631 + 267:

Let me compute 631 + 200 = 831, then 831 + 67 = 898. Yes.

Alternatively, 211 + 267 = 478, then 478 + 420 = 898. Same result.

Therefore, I think the answer is 898.

But just to be thorough, let me check if there's a possibility that the problem is in the indexing. For example, sometimes sequences start at n=0, but here it's given as n=1, so I think the calculation is correct.

Alternatively, maybe I made a mistake in the initial terms? Let me check again:

x1=211, x2=375, x3=420, x4=523.

x5 = x4 - x3 + x2 - x1 = 523 - 420 + 375 - 211 = 523 - 420 is 103, 103 + 375 is 478, 478 - 211 is 267. Correct.

x6 = x5 - x4 + x3 - x2 = 267 - 523 + 420 - 375. 267 - 523 = -256, -256 + 420 = 164, 164 - 375 = -211. Correct.

x7 = x6 - x5 + x4 - x3 = -211 - 267 + 523 - 420. -211 -267 = -478, -478 + 523 = 45, 45 - 420 = -375. Correct.

x8 = x7 - x6 + x5 - x4 = -375 - (-211) + 267 - 523 = -375 + 211 = -164, -164 + 267 = 103, 103 - 523 = -420. Correct.

x9 = x8 - x7 + x6 - x5 = -420 - (-375) + (-211) - 267 = -420 + 375 = -45, -45 -211 = -256, -256 -267 = -523. Correct.

x10 = x9 - x8 + x7 - x6 = -523 - (-420) + (-375) - (-211) = -523 + 420 = -103, -103 - 375 = -478, -478 + 211 = -267. Correct.

x11 = x10 - x9 + x8 - x7 = -267 - (-523) + (-420) - (-375) = -267 + 523 = 256, 256 - 420 = -164, -164 + 375 = 211. Correct. So x11 = x1. All steps are correct.

Therefore, the periodicity is confirmed. Therefore, the answer is 898.

But just to be 100% sure, let me check another term. Let's compute x_{15} which should be equal to x5=267.

x11=211, x12=375, x13=420, x14=523, x15 = x14 - x13 + x12 - x11 = 523 - 420 + 375 - 211 = same as x5, which is 267. Correct.

Similarly, x_{16} = x15 - x14 + x13 - x12 = 267 - 523 + 420 - 375 = same as x6=-211. Correct.

Therefore, the periodicity is confirmed. Therefore, the answer is 898.

But wait, the problem says "find the value of $ x_{531} + x_{753} + x_{975} $." So according to my calculation, it's 211 + 420 + 267 = 898. However, let me check once again the values for each term:

For x_{531}:

Since 531 divided by 10 is 53*10 + 1, so remainder 1. Therefore, x_{531} = x1 = 211. Correct.

For x_{753}:

753 divided by 10 is 75*10 + 3, remainder 3. Therefore, x_{753} = x3 = 420. Correct.

For x_{975}:

975 divided by 10 is 97*10 + 5, remainder 5. Therefore, x_{975} = x5 = 267. Correct.

Therefore, adding them up: 211 + 420 + 267 = 898. So I think that's the answer.

But just to be thorough, let me check if there's an alternative way to compute this, maybe using the recurrence relation properties or matrix exponentiation or something else, but since the period is 10, it's straightforward.

Alternatively, maybe there's a mistake in assuming the period is 10? Let me check for n=10, x_{10}=-267, and then x_{10 + 10}=x_{20}=-267. Let me compute x_{15}=267, x_{16}=-211, x_{17}=-375, x_{18}=-420, x_{19}=-523, x_{20}= -267. Yes, same as x_{10}. Therefore, the period is indeed 10.

Therefore, I think my answer is correct. Therefore, the value is 898.

**Final Answer**
\boxed{898}

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

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

We are asked to compute:

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

---

### Step 1: Identify the Pattern

We begin by computing the first few terms of the sequence to look for a pattern or periodicity.

$$
\begin{align*}
x_1 &= 211 \\
x_2 &= 375 \\
x_3 &= 420 \\
x_4 &= 523 \\
x_5 &= x_4 - x_3 + x_2 - x_1 = 523 - 420 + 375 - 211 = 267 \\
x_6 &= x_5 - x_4 + x_3 - x_2 = 267 - 523 + 420 - 375 = -211 \\
x_7 &= x_6 - x_5 + x_4 - x_3 = -211 - 267 + 523 - 420 = -375 \\
x_8 &= x_7 - x_6 + x_5 - x_4 = -375 - (-211) + 267 - 523 = -420 \\
x_9 &= x_8 - x_7 + x_6 - x_5 = -420 - (-375) + (-211) - 267 = -523 \\
x_{10} &= x_9 - x_8 + x_7 - x_6 = -523 - (-420) + (-375) - (-211) = -267 \\
x_{11} &= x_{10} - x_9 + x_8 - x_7 = -267 - (-523) + (-420) - (-375) = 211 \\
x_{12} &= x_{11} - x_{10} + x_9 - x_8 = 211 - (-267) + (-523) - (-420) = 375 \\
x_{13} &= x_{12} - x_{11} + x_{10} - x_9 = 375 - 211 + (-267) - (-523) = 420 \\
x_{14} &= x_{13} - x_{12} + x_{11} - x_{10} = 420 - 375 + 211 - (-267) = 523 \\
x_{15} &= x_{14} - x_{13} + x_{12} - x_{11} = 523 - 420 + 375 - 211 = 267 \\
\end{align*}
$$

We observe that:

$$
x_{11} = x_1, \quad x_{12} = x_2, \quad x_{13} = x_3, \quad x_{14} = x_4, \quad x_{15} = x_5
$$

This confirms that the sequence is **periodic with period 10**.

---

### Step 2: Use the Periodicity to Compute the Required Terms

Since the sequence is periodic with period 10, we can compute any term $ x_n $ by finding the position of $ n $ within the first 10 terms.

To do this, we compute:

$$
n \mod 10
$$

- If the remainder is 0, then $ x_n = x_{10} $
- Otherwise, $ x_n = x_{\text{remainder}} $

Let’s compute each term:

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

---

### Step 3: Compute the Final 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 this sum is less than 220. Then, among all such possible sets, I have to find the greatest possible value for the largest of the three integers. Hmm, let me break this down step by step.

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 sum is a three-digit palindrome, it must be between 100 and 999, but in this case, it's specified to be less than 220. So the sum can be from 101 up to 219, but only palindromic numbers in that range.

Let me note that three-digit palindromes have the form ABA, where A is a digit from 1 to 9 (since it's a three-digit number) and B is a digit from 0 to 9. So the numbers would be 101, 111, 121, ..., 191, 202, 212, etc. But since the sum is less than 220, the maximum possible palindrome would be 212, because 222 is over 220. Wait, 212 is less than 220, right? Yes. So the possible palindromes are from 101 up to 212, but only those that are palindromic.

So first, maybe I should list all three-digit palindromes less than 220. Let me do that.

Starting with 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. Wait, 202 is 202, which is less than 220, and 212 is also less than 220. The next one would be 222, which is over 220, so we stop there. So the possible palindromes are 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. That's 12 numbers.

Now, the problem says that the sum of three consecutive positive integers is one of these palindromes. Let me denote the three consecutive integers as n-1, n, n+1. Wait, but maybe it's easier to think of them as x, x+1, x+2. Let me check.

If I take three consecutive integers, their sum is x + (x+1) + (x+2) = 3x + 3 = 3(x + 1). So the sum is three times (x + 1). Therefore, the sum must be divisible by 3. Therefore, the palindrome we are looking for must be divisible by 3. So that's a key point. Therefore, among the palindromes listed above, I need to find those that are divisible by 3.

So let me check which of these palindromes are divisible by 3. Let me recall that a number is divisible by 3 if the sum of its digits is divisible by 3.

So let's check each palindrome:

101: 1 + 0 + 1 = 2 → not divisible by 3.

111: 1 + 1 + 1 = 3 → divisible by 3. Okay.

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, that's only three numbers? Let me check again.

Wait, 111: sum digits 3, yes. 141: 1+4+1=6, yes. 171: 1+7+1=9, yes. Then 202: 2+0+2=4, no. 212: 2+1+2=5, no. So only those three. Wait, but 101, 121, etc., are not divisible by 3. So that's only three possible sums? That seems low. Wait, but maybe I made a mistake here?

Wait, let me check again. For example, 111 is 111, which is 3*37. 141 is 3*47, 171 is 3*57. So yes, those are the only ones. So the sum must be one of these three numbers. Therefore, the sum is either 111, 141, or 171. Wait, but 171 is less than 220, so that's okay. So then, if the sum is 3(x + 1), then x + 1 is equal to sum divided by 3. Therefore, x is (sum / 3) - 1. Therefore, the three numbers are:

For sum = 111: x = (111 / 3) - 1 = 37 - 1 = 36. Therefore, the numbers are 36, 37, 38. The largest is 38.

For sum = 141: x = 141 / 3 - 1 = 47 - 1 = 46. Therefore, numbers are 46, 47, 48. Largest is 48.

For sum = 171: x = 171 / 3 - 1 = 57 - 1 = 56. Therefore, numbers are 56, 57, 58. Largest is 58.

So according to this, the largest possible value for the largest integer is 58. But wait, the problem says "the sum is less than 220", so 171 is less than 220, so that's okay. But is there a possibility that there are other palindromes that I missed?

Wait, maybe I made a mistake in considering the palindromes. Let me check again. The three-digit palindromes less than 220. Let me think again. The first digit is 1 or 2. If first digit is 1, then the last digit is 1, so numbers like 101, 111, 121,..., 191. Then if first digit is 2, the last digit is 2, so numbers like 202, 212. But 222 is over 220. So that's correct. So the palindromes are as I listed before. So only 111, 141, 171 are divisible by 3. Therefore, the maximum sum is 171, leading to the largest integer being 58. But wait, the problem says "the greatest possible value for the largest of the three integers in the set". So 58? But maybe there's a mistake here. Let me check again.

Wait, perhaps I need to check if there are other palindromes that I missed? For example, maybe numbers like 121? Wait, 121 is not divisible by 3. Let me check 121 divided by 3: 121 / 3 is 40.333... So no. Similarly, 131 is 131 / 3 ≈ 43.666... Not integer. 151 / 3 ≈ 50.333... Not integer. 161 / 3 ≈ 53.666... Not integer. 181 / 3 ≈ 60.333... Not integer. 191 / 3 ≈ 63.666... Not integer. 202 / 3 ≈ 67.333... Not integer. 212 / 3 ≈ 70.666... Not integer. So only 111, 141, 171. Therefore, only those three sums. Therefore, the maximum largest integer is 58. But wait, maybe there's another way?

Wait, maybe I made a mistake in assuming that the three consecutive integers are x, x+1, x+2. Let me check again. The sum is 3x + 3, which is 3(x + 1). Therefore, the sum must be divisible by 3. Therefore, the palindrome must be divisible by 3. Therefore, only those three palindromes. Therefore, the maximum sum is 171, leading to the largest integer being 58. But maybe there's a larger sum? Wait, but the sum is less than 220. The next palindrome after 171 would be 202, but 202 isn't divisible by 3. Then 212, which also isn't divisible by 3. So no. Therefore, 171 is the maximum sum. Therefore, the largest integer is 58. But wait, let me check if there are other palindromes that I might have missed. For example, maybe numbers like 101, but they aren't divisible by 3. Wait, but maybe there's another way to get the sum?

Wait, hold on. Let me think again. Maybe I need to check if there are other palindromes that are divisible by 3. Wait, for example, 111, 141, 171, 201? Wait, 201 is a palindrome? No, 201 reversed is 102, which is not the same. So 201 is not a palindrome. Similarly, 212 is a palindrome, but not divisible by 3. 222 is over 220. So no. So only 111, 141, 171. Therefore, the maximum sum is 171. Therefore, the largest integer is 58. But wait, let me check if there's another way to get a larger sum. For example, if the sum is 171, then the numbers are 56, 57, 58. The largest is 58. But maybe if there is a larger sum, but not a palindrome? But the problem says the sum must be a palindrome. So no. Therefore, 58 is the answer? But I feel like maybe I missed something here. Let me check again.

Alternatively, maybe I need to consider that the three consecutive integers could be something else? Wait, no. The sum of three consecutive integers is always 3 times the middle number. So if the three numbers are n-1, n, n+1, then the sum is 3n. Therefore, the sum is 3n, so the sum must be divisible by 3. Therefore, the palindrome must be divisible by 3. Therefore, only those three palindromes. Therefore, the maximum sum is 171, leading to the middle number being 57, so the largest number is 58. Therefore, the answer is 58? But let me check if there's a larger possible sum. Wait, maybe there's another palindrome that I didn't consider?

Wait, let me think again. For example, 191 is a palindrome. 191 divided by 3 is 63.666... Not integer. 181 divided by 3 is 60.333... Not integer. 161 divided by 3 is 53.666... Not integer. 151 divided by 3 is 50.333... Not integer. 141 is okay. 131 is not. 121 is not. 111 is okay. 101 is not. Then 202 is not. 212 is not. So only 111, 141, 171. Therefore, 171 is the maximum. Therefore, the answer is 58. But wait, let me check if there's another way to get a larger sum. For example, if the sum is 171, then the numbers are 56, 57, 58. But maybe if the sum is higher, but still a palindrome? But the next palindrome after 171 is 202, which is not divisible by 3. Then 212, which is not. Then 222, which is over 220. So no. Therefore, 171 is the maximum. Therefore, the answer is 58. But wait, let me check if there's a mistake in my calculation. Let me check the sum for 56, 57, 58. 56 + 57 + 58 = 56 + 57 is 113, plus 58 is 171. Yes, that's correct. And 171 is a palindrome. So that works. So then, the largest number is 58. But wait, the problem says "the greatest possible value for the largest of the three integers in the set". Is there a way to get a larger number?

Wait, maybe I need to check if there are other palindromes that I missed. Let me think again. For example, maybe numbers like 1001? No, that's four digits. Wait, the sum is a three-digit palindrome. So three digits. So only up to 999. But the sum is less than 220, so up to 219. So palindromes are as I listed. So maybe there's another way? Wait, perhaps I made a mistake in considering the three consecutive integers. Let me think again. Suppose the three numbers are not starting from x, x+1, x+2, but maybe something else? No, three consecutive integers must be like that. The sum is always 3 times the middle number. Therefore, the sum must be divisible by 3. Therefore, only those three palindromes. Therefore, 58 is the answer. But maybe I need to check if there are other palindromes that are divisible by 3. Let me check 111, 141, 171, 201? Wait, 201 is not a palindrome. 212 is not divisible by 3. 222 is over 220. 191? No. 181? No. 161? No. 151? No. 131? No. 121? No. 101? No. So only 111, 141, 171. Therefore, 171 is the maximum. Therefore, the answer is 58. But wait, let me check if there's another way. For example, maybe the three numbers are not starting from x, x+1, x+2, but maybe different? No, three consecutive integers must be like that. So the sum is 3x + 3. Therefore, the sum is 3(x + 1). Therefore, the sum must be divisible by 3. Therefore, only those three palindromes. Therefore, 58 is the answer. But maybe I need to check if there are other palindromes that are divisible by 3. Let me check 111, 141, 171, 201, 231... but 231 is over 220. So no. Therefore, 171 is the maximum. Therefore, the answer is 58. Wait, but let me check if there's a larger sum. For example, if the sum is 171, then the numbers are 56, 57, 58. If I take a larger sum, say 180, which is not a palindrome. 181 is a palindrome, but 181 divided by 3 is 60.333... Not integer. 191 is a palindrome, but not divisible by 3. 200 is not a palindrome. 201 is not a palindrome. 210 is not a palindrome. 211 is not a palindrome. 212 is a palindrome, but not divisible by 3. So no. Therefore, 171 is the maximum. Therefore, the answer is 58. But wait, let me check if there's another palindrome between 171 and 220 that is divisible by 3. Let me think. For example, 181 is not. 191 is not. 202 is not. 212 is not. 222 is over. So no. Therefore, 171 is the maximum. Therefore, the answer is 58. But wait, let me check if there's another way. Suppose the three numbers are not starting from x, x+1, x+2. Wait, no, they have to be consecutive. So the sum is 3x + 3. Therefore, the sum must be divisible by 3. Therefore, only those three palindromes. Therefore, 58 is the answer. But I have a feeling that maybe there's a mistake here. Let me check with another approach.

Alternatively, maybe I can think of the sum as a palindrome less than 220, and then check if it's divisible by 3, and then compute the numbers. So let me list all three-digit palindromes less than 220:

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

Now, check which of these are divisible by 3:

- 101: 1+0+1=2 → not divisible by 3
- 111: 1+1+1=3 → yes
- 121: 1+2+1=4 → no
- 131: 1+3+1=5 → no
- 141: 1+4+1=6 → yes
- 151: 1+5+1=7 → no
- 161: 1+6+1=8 → no
- 171: 1+7+1=9 → yes
- 181: 1+8+1=10 → no
- 191: 1+9+1=11 → no
- 202: 2+0+2=4 → no
- 212: 2+1+2=5 → no

So only 111, 141, 171. Therefore, the possible sums are 111, 141, 171. Therefore, the corresponding middle numbers are 111/3=37, 141/3=47, 171/3=57. Therefore, the three numbers are:

For 111: 36, 37, 38 → largest is 38

For 141: 46, 47, 48 → largest is 48

For 171: 56, 57, 58 → largest is 58

Therefore, the maximum possible value for the largest integer is 58. Therefore, the answer is 58. But wait, let me check if there's a possibility that the sum is a palindrome which is not divisible by 3, but the three numbers still add up to it? Wait, no. Because the sum of three consecutive integers is always divisible by 3. Therefore, the sum must be divisible by 3. Therefore, only those three palindromes. Therefore, 58 is the answer. Therefore, I think that is correct.

But let me check once again. Suppose someone says that maybe the three numbers are not in the form x, x+1, x+2. Wait, but they are three consecutive integers. So they have to be like that. For example, if they are n-1, n, n+1, then the sum is 3n. Therefore, the sum is divisible by 3. Therefore, the palindrome must be divisible by 3. Therefore, only those three. Therefore, 58 is the answer. Therefore, I think that is correct.

But just to be thorough, let me check if there are other palindromes that I might have missed. For example, numbers like 1001? No, that's four digits. 111 is three digits. 121, etc. So no. Therefore, I think my answer is correct. Therefore, the greatest possible value for the largest integer is 58.

Wait, but let me check if there's a larger sum. For example, if the sum is 171, which is 56 + 57 + 58. What if there is another set of three numbers where the sum is a palindrome less than 220, but not necessarily following the 3(x + 1) formula? Wait, no. The sum of three consecutive integers is always 3 times the middle number. Therefore, it must be divisible by 3. Therefore, the sum must be divisible by 3. Therefore, only those three palindromes. Therefore, 58 is the answer. Therefore, I think that is correct.

But let me check with another example. Suppose the sum is 171. Then the numbers are 56, 57, 58. The largest is 58. If I try to get a larger number, say 59, then the three numbers would be 57, 58, 59. Their sum is 57 + 58 + 59 = 174. Is 174 a palindrome? No, because 174 reversed is 471. Not the same. So not a palindrome. Similarly, if I take numbers 58, 59, 60. Sum is 58 + 59 + 60 = 177. 177 reversed is 771. Not a palindrome. 59, 60, 61: sum is 180. 180 reversed is 081, which is 81, not same. Not a palindrome. 60, 61, 62: sum is 183. Not a palindrome. 61, 62, 63: 186. Not a palindrome. 62, 63, 64: 189. Not a palindrome. 63, 64, 65: 192. Not a palindrome. 64, 65, 66: 195. Not a palindrome. 65, 66, 67: 198. Not a palindrome. 66, 67, 68: 201. Not a palindrome. 67, 68, 69: 204. Not a palindrome. 68, 69, 70: 207. Not a palindrome. 69, 70, 71: 210. Not a palindrome. 70, 71, 72: 213. Not a palindrome. 71, 72, 73: 216. Not a palindrome. 72, 73, 74: 219. Not a palindrome. 73, 74, 75: 222. Which is over 220. So none of these sums are palindromes. Therefore, there is no set of three consecutive integers with a larger largest number than 58 where the sum is a palindrome less than 220. Therefore, 58 is indeed the correct answer.

Therefore, I think the answer is 58. Wait, but let me check if there's another palindrome that I might have missed. For example, 191. If the sum is 191, which is not divisible by 3. But if I try to see if there are three consecutive numbers that add up to 191. Let me check. If the sum is 191, then the middle number would be 191 / 3 ≈ 63.666... which is not an integer. Therefore, there are no such integers. Therefore, no. Similarly, for 202, which is not divisible by 3. 202 / 3 ≈ 67.333... Not integer. Therefore, no. Therefore, no other possibilities. Therefore, 58 is the answer.

But just to make sure, let me check if there's a palindrome between 171 and 220 that is divisible by 3. Let me think. The next palindrome after 171 is 202, which is not divisible by 3. Then 212, which is not. Then 222, which is over. So no. Therefore, 171 is the maximum. Therefore, the answer is 58. Therefore, I think that is correct.

**Final Answer**
The greatest possible value for the largest of the three integers is \boxed{58}.

To solve the problem, we are given that the **sum of three consecutive positive integers** is a **three-digit palindrome**, and this sum is **less than 220**. We are to find the **greatest possible value for the largest of the three integers** in such a set.

---

### Step 1: Understanding the Structure of the Sum

Let the three consecutive integers be:

$$
x, \quad x+1, \quad x+2
$$

Their sum is:

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

So, the **sum is always divisible by 3**. Therefore, the **palindrome must also be divisible by 3**.

---

### Step 2: List All Three-Digit Palindromes Less Than 220

Three-digit palindromes have the form **ABA**, where A and B are digits. Since the sum is less than 220, the first digit A can only be **1 or 2**.

So, the possible palindromes are:

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

---

### Step 3: Filter Palindromes Divisible by 3

We now check which of these are divisible by 3. A number is divisible by 3 if the **sum of its digits is divisible by 3**.

- 101 → 1 + 0 + 1 = 2 → Not divisible by 3
- 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 → ❌

So, the valid palindromes divisible by 3 are:

$$
111, \quad 141, \quad 171
$$

---

### Step 4: Find the Corresponding Sets of Three Consecutive Integers

We now compute the three consecutive integers for each valid sum:

- **Sum = 111**
  $$
  x + 1 = \frac{111}{3} = 37 \Rightarrow x = 36
  $$
  The integers are: **36, 37, 38** → Largest is **38**

- **Sum = 141**
  $$
  x + 1 = \frac{141}{3} = 47 \Rightarrow x = 46
  $$
  The integers are: **46, 47, 48** → Largest is **48**

- **Sum = 171**
  $$
  x + 1 = \frac{171}{3} = 57 \Rightarrow x = 56
  $$
  The integers are: **56, 57, 58** → Largest is **58**

---

### Step 5: Determine the Maximum Possible Largest Integer

From the above, the largest possible value for the **largest of the three integers** is:

$$
\boxed{58}
$$

---

### Final Answer

$$
\boxed{58}
$$

[Assistant]
Explanation:
Okay, so I need to find three consecutive positive integers where their sum is a three-digit palindrome, and this sum is less than 220. Then, among all such possible sets, I have to find the greatest possible value for the largest of the three integers. Hmm, let me break this down step by step.

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 sum is a three-digit palindrome, it must be between 100 and 999, but in this case, it's specified to be less than 220. So the sum can be from 101 up to 219, but only palindromic numbers in that range.

Let me note that three-digit palindromes have the form ABA, where A is a digit from 1 to 9 (since it's a three-digit number) and B is a digit from 0 to 9. So the numbers would be 101, 111, 121, ..., 191, 202, 212, etc. But since the sum is less than 220, the maximum possible palindrome would be 212, because 222 is over 220. Wait, 212 is less than 220, right? Yes. So the possible palindromes are from 101 up to 212, but only those that are palindromic.

So first, maybe I should list all three-digit palindromes less than 220. Let me do that.

Starting with 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. Wait, 202 is 202, which is less than 220, and 212 is also less than 220. The next one would be 222, which is over 220, so we stop there. So the possible palindromes are 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. That's 12 numbers.

Now, the problem says that the sum of three consecutive positive integers is one of these palindromes. Let me denote the three consecutive integers as n-1, n, n+1. Wait, but maybe it's easier to think of them as x, x+1, x+2. Let me check.

If I take three consecutive integers, their sum is x + (x+1) + (x+2) = 3x + 3 = 3(x + 1). So the sum is three times (x + 1). Therefore, the sum must be divisible by 3. Therefore, the palindrome we are looking for must be divisible by 3. So that's a key point. Therefore, among the palindromes listed above, I need to find those that are divisible by 3.

So let me check which of these palindromes are divisible by 3. Let me recall that a number is divisible by 3 if the sum of its digits is divisible by 3.

So let's check each palindrome:

101: 1 + 0 + 1 = 2 → not divisible by 3.

111: 1 + 1 + 1 = 3 → divisible by 3. Okay.

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, that's only three numbers? Let me check again.

Wait, 111: sum digits 3, yes. 141: 1+4+1=6, yes. 171: 1+7+1=9, yes. Then 202: 2+0+2=4, no. 212: 2+1+2=5, no. So only those three. Wait, but 101, 121, etc., are not divisible by 3. So that's only three possible sums? That seems low. Wait, but maybe I made a mistake here?

Wait, let me check again. For example, 111 is 111, which is 3*37. 141 is 3*47, 171 is 3*57. So yes, those are the only ones. So the sum must be one of these three numbers. Therefore, the sum is either 111, 141, or 171. Wait, but 171 is less than 220, so that's okay. So then, if the sum is 3(x + 1), then x + 1 is equal to sum divided by 3. Therefore, x is (sum / 3) - 1. Therefore, the three numbers are:

For sum = 111: x = (111 / 3) - 1 = 37 - 1 = 36. Therefore, the numbers are 36, 37, 38. The largest is 38.

For sum = 141: x = 141 / 3 - 1 = 47 - 1 = 46. Therefore, numbers are 46, 47, 48. Largest is 48.

For sum = 171: x = 171 / 3 - 1 = 57 - 1 = 56. Therefore, numbers are 56, 57, 58. Largest is 58.

So according to this, the largest possible value for the largest integer is 58. But wait, the problem says "the sum is less than 220", so 171 is less than 220, so that's okay. But is there a possibility that there are other palindromes that I missed?

Wait, maybe I made a mistake in considering the palindromes. Let me check again. The three-digit palindromes less than 220. Let me think again. The first digit is 1 or 2. If first digit is 1, then the last digit is 1, so numbers like 101, 111, 121,..., 191. Then if first digit is 2, the last digit is 2, so numbers like 202, 212. But 222 is over 220. So that's correct. So the palindromes are as I listed before. So only 111, 141, 171 are divisible by 3. Therefore, the maximum sum is 171, leading to the largest integer being 58. But wait, the problem says "the greatest possible value for the largest of the three integers in the set". So 58? But maybe there's a mistake here. Let me check again.

Wait, perhaps I need to check if there are other palindromes that I missed? For example, maybe numbers like 121? Wait, 121 is not divisible by 3. Let me check 121 divided by 3: 121 / 3 is 40.333... So no. Similarly, 131 is 131 / 3 ≈ 43.666... Not integer. 151 / 3 ≈ 50.333... Not integer. 161 / 3 ≈ 53.666... Not integer. 181 / 3 ≈ 60.333... Not integer. 191 / 3 ≈ 63.666... Not integer. 202 / 3 ≈ 67.333... Not integer. 212 / 3 ≈ 70.666... Not integer. So only 111, 141, 171. Therefore, only those three sums. Therefore, the maximum largest integer is 58. But wait, maybe there's another way?

Wait, maybe I made a mistake in assuming that the three consecutive integers are x, x+1, x+2. Let me check again. The sum is 3x + 3, which is 3(x + 1). Therefore, the sum must be divisible by 3. Therefore, the palindrome must be divisible by 3. Therefore, only those three palindromes. Therefore, the maximum sum is 171, leading to the largest integer being 58. But maybe there's a larger sum? Wait, but the sum is less than 220. The next palindrome after 171 would be 202, but 202 isn't divisible by 3. Then 212, which also isn't divisible by 3. So no. Therefore, 171 is the maximum sum. Therefore, the largest integer is 58. But wait, let me check if there are other palindromes that I might have missed. For example, maybe numbers like 101, but they aren't divisible by 3. Wait, but maybe there's another way to get the sum?

Wait, hold on. Let me think again. Maybe I need to check if there are other palindromes that are divisible by 3. Wait, for example, 111, 141, 171, 201? Wait, 201 is a palindrome? No, 201 reversed is 102, which is not the same. So 201 is not a palindrome. Similarly, 212 is a palindrome, but not divisible by 3. 222 is over 220. So no. So only 111, 141, 171. Therefore, the maximum sum is 171. Therefore, the largest integer is 58. But wait, let me check if there's another way to get a larger sum. For example, if the sum is 171, then the numbers are 56, 57, 58. The largest is 58. But maybe if there is a larger sum, but not a palindrome? But the problem says the sum must be a palindrome. So no. Therefore, 58 is the answer? But I feel like maybe I missed something here. Let me check again.

Alternatively, maybe I need to consider that the three consecutive integers could be something else? Wait, no. The sum of three consecutive integers is always 3 times the middle number. So if the three numbers are n-1, n, n+1, then the sum is 3n. Therefore, the sum is 3n, so the sum must be divisible by 3. Therefore, the palindrome must be divisible by 3. Therefore, only those three palindromes. Therefore, the maximum sum is 171, leading to the middle number being 57, so the largest number is 58. Therefore, the answer is 58? But let me check if there's a larger possible sum. Wait, maybe there's another palindrome that I didn't consider?

Wait, let me think again. For example, 191 is a palindrome. 191 divided by 3 is 63.666... Not integer. 181 divided by 3 is 60.333... Not integer. 161 divided by 3 is 53.666... Not integer. 151 divided by 3 is 50.333... Not integer. 141 is okay. 131 is not. 121 is not. 111 is okay. 101 is not. Then 202 is not. 212 is not. So only 111, 141, 171. Therefore, 171 is the maximum. Therefore, the answer is 58. But wait, let me check if there's another way to get a larger sum. For example, if the sum is 171, then the numbers are 56, 57, 58. But maybe if the sum is higher, but still a palindrome? But the next palindrome after 171 is 202, which is not divisible by 3. Then 212, which is not. Then 222, which is over 220. So no. Therefore, 171 is the maximum. Therefore, the answer is 58. But wait, let me check if there's a mistake in my calculation. Let me check the sum for 56, 57, 58. 56 + 57 + 58 = 56 + 57 is 113, plus 58 is 171. Yes, that's correct. And 171 is a palindrome. So that works. So then, the largest number is 58. But wait, the problem says "the greatest possible value for the largest of the three integers in the set". Is there a way to get a larger number?

Wait, maybe I need to check if there are other palindromes that I missed. Let me think again. For example, maybe numbers like 1001? No, that's four digits. Wait, the sum is a three-digit palindrome. So three digits. So only up to 999. But the sum is less than 220, so up to 219. So palindromes are as I listed. So maybe there's another way? Wait, perhaps I made a mistake in considering the three consecutive integers. Let me think again. Suppose the three numbers are not starting from x, x+1, x+2, but maybe something else? No, three consecutive integers must be like that. The sum is always 3 times the middle number. Therefore, the sum must be divisible by 3. Therefore, only those three palindromes. Therefore, 58 is the answer. But maybe I need to check if there are other palindromes that are divisible by 3. Let me check 111, 141, 171, 201? Wait, 201 is not a palindrome. 212 is not divisible by 3. 222 is over 220. 191? No. 181? No. 161? No. 151? No. 131? No. 121? No. 101? No. So only 111, 141, 171. Therefore, 171 is the maximum. Therefore, the answer is 58. But wait, let me check if there's another way. For example, maybe the three numbers are not starting from x, x+1, x+2, but maybe different? No, three consecutive integers must be like that. So the sum is 3x + 3. Therefore, the sum is 3(x + 1). Therefore, the sum must be divisible by 3. Therefore, only those three palindromes. Therefore, 58 is the answer. But maybe I need to check if there are other palindromes that are divisible by 3. Let me check 111, 141, 171, 201, 231... but 231 is over 220. So no. Therefore, 171 is the maximum. Therefore, the answer is 58. Wait, but let me check if there's a larger sum. For example, if the sum is 171, then the numbers are 56, 57, 58. If I take a larger sum, say 180, which is not a palindrome. 181 is a palindrome, but 181 divided by 3 is 60.333... Not integer. 191 is a palindrome, but not divisible by 3. 200 is not a palindrome. 201 is not a palindrome. 210 is not a palindrome. 211 is not a palindrome. 212 is a palindrome, but not divisible by 3. So no. Therefore, 171 is the maximum. Therefore, the answer is 58. But wait, let me check if there's another palindrome between 171 and 220 that is divisible by 3. Let me think. For example, 181 is not. 191 is not. 202 is not. 212 is not. 222 is over. So no. Therefore, 171 is the maximum. Therefore, the answer is 58. But wait, let me check if there's another way. Suppose the three numbers are not starting from x, x+1, x+2. Wait, no, they have to be consecutive. So the sum is 3x + 3. Therefore, the sum must be divisible by 3. Therefore, only those three palindromes. Therefore, 58 is the answer. But I have a feeling that maybe there's a mistake here. Let me check with another approach.

Alternatively, maybe I can think of the sum as a palindrome less than 220, and then check if it's divisible by 3, and then compute the numbers. So let me list all three-digit palindromes less than 220:

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

Now, check which of these are divisible by 3:

- 101: 1+0+1=2 → not divisible by 3
- 111: 1+1+1=3 → yes
- 121: 1+2+1=4 → no
- 131: 1+3+1=5 → no
- 141: 1+4+1=6 → yes
- 151: 1+5+1=7 → no
- 161: 1+6+1=8 → no
- 171: 1+7+1=9 → yes
- 181: 1+8+1=10 → no
- 191: 1+9+1=11 → no
- 202: 2+0+2=4 → no
- 212: 2+1+2=5 → no

So only 111, 141, 171. Therefore, the possible sums are 111, 141, 171. Therefore, the corresponding middle numbers are 111/3=37, 141/3=47, 171/3=57. Therefore, the three numbers are:

For 111: 36, 37, 38 → largest is 38

For 141: 46, 47, 48 → largest is 48

For 171: 56, 57, 58 → largest is 58

Therefore, the maximum possible value for the largest integer is 58. Therefore, the answer is 58. But wait, let me check if there's a possibility that the sum is a palindrome which is not divisible by 3, but the three numbers still add up to it? Wait, no. Because the sum of three consecutive integers is always divisible by 3. Therefore, the sum must be divisible by 3. Therefore, only those three palindromes. Therefore, 58 is the answer. Therefore, I think that is correct.

But let me check once again. Suppose someone says that maybe the three numbers are not in the form x, x+1, x+2. Wait, but they are three consecutive integers. So they have to be like that. For example, if they are n-1, n, n+1, then the sum is 3n. Therefore, the sum is divisible by 3. Therefore, the palindrome must be divisible by 3. Therefore, only those three. Therefore, 58 is the answer. Therefore, I think that is correct.

But just to be thorough, let me check if there are other palindromes that I might have missed. For example, numbers like 1001? No, that's four digits. 111 is three digits. 121, etc. So no. Therefore, I think my answer is correct. Therefore, the greatest possible value for the largest integer is 58.

Wait, but let me check if there's a larger sum. For example, if the sum is 171, which is 56 + 57 + 58. What if there is another set of three numbers where the sum is a palindrome less than 220, but not necessarily following the 3(x + 1) formula? Wait, no. The sum of three consecutive integers is always 3 times the middle number. Therefore, it must be divisible by 3. Therefore, the sum must be divisible by 3. Therefore, only those three palindromes. Therefore, 58 is the answer. Therefore, I think that is correct.

But let me check with another example. Suppose the sum is 171. Then the numbers are 56, 57, 58. The largest is 58. If I try to get a larger number, say 59, then the three numbers would be 57, 58, 59. Their sum is 57 + 58 + 59 = 174. Is 174 a palindrome? No, because 174 reversed is 471. Not the same. So not a palindrome. Similarly, if I take numbers 58, 59, 60. Sum is 58 + 59 + 60 = 177. 177 reversed is 771. Not a palindrome. 59, 60, 61: sum is 180. 180 reversed is 081, which is 81, not same. Not a palindrome. 60, 61, 62: sum is 183. Not a palindrome. 61, 62, 63: 186. Not a palindrome. 62, 63, 64: 189. Not a palindrome. 63, 64, 65: 192. Not a palindrome. 64, 65, 66: 195. Not a palindrome. 65, 66, 67: 198. Not a palindrome. 66, 67, 68: 201. Not a palindrome. 67, 68, 69: 204. Not a palindrome. 68, 69, 70: 207. Not a palindrome. 69, 70, 71: 210. Not a palindrome. 70, 71, 72: 213. Not a palindrome. 71, 72, 73: 216. Not a palindrome. 72, 73, 74: 219. Not a palindrome. 73, 74, 75: 222. Which is over 220. So none of these sums are palindromes. Therefore, there is no set of three consecutive integers with a larger largest number than 58 where the sum is a palindrome less than 220. Therefore, 58 is indeed the correct answer.

Therefore, I think the answer is 58. Wait, but let me check if there's another palindrome that I might have missed. For example, 191. If the sum is 191, which is not divisible by 3. But if I try to see if there are three consecutive numbers that add up to 191. Let me check. If the sum is 191, then the middle number would be 191 / 3 ≈ 63.666... which is not an integer. Therefore, there are no such integers. Therefore, no. Similarly, for 202, which is not divisible by 3. 202 / 3 ≈ 67.333... Not integer. Therefore, no. Therefore, no other possibilities. Therefore, 58 is the answer.

But just to make sure, let me check if there's a palindrome between 171 and 220 that is divisible by 3. Let me think. The next palindrome after 171 is 202, which is not divisible by 3. Then 212, which is not. Then 222, which is over. So no. Therefore, 171 is the maximum. Therefore, the answer is 58. Therefore, I think that is correct.

**Final Answer**
The greatest possible value for the largest of the three integers is \boxed{58}.

To solve the problem, we are given that the **sum of three consecutive positive integers** is a **three-digit palindrome**, and this sum is **less than 220**. We are to find the **greatest possible value for the largest of the three integers** in such a set.

---

### Step 1: Understanding the Structure of the Sum

Let the three consecutive integers be:

$$
x, \quad x+1, \quad x+2
$$

Their sum is:

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

So, the **sum is always divisible by 3**. Therefore, the **palindrome must also be divisible by 3**.

---

### Step 2: List All Three-Digit Palindromes Less Than 220

Three-digit palindromes have the form **ABA**, where A and B are digits. Since the sum is less than 220, the first digit A can only be **1 or 2**.

So, the possible palindromes are:

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

---

### Step 3: Filter Palindromes Divisible by 3

We now check which of these are divisible by 3. A number is divisible by 3 if the **sum of its digits is divisible by 3**.

- 101 → 1 + 0 + 1 = 2 → Not divisible by 3
- 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 → ❌

So, the valid palindromes divisible by 3 are:

$$
111, \quad 141, \quad 171
$$

---

### Step 4: Find the Corresponding Sets of Three Consecutive Integers

We now compute the three consecutive integers for each valid sum:

- **Sum = 111**
  $$
  x + 1 = \frac{111}{3} = 37 \Rightarrow x = 36
  $$
  The integers are: **36, 37, 38** → Largest is **38**

- **Sum = 141**
  $$
  x + 1 = \frac{141}{3} = 47 \Rightarrow x = 46
  $$
  The integers are: **46, 47, 48** → Largest is **48**

- **Sum = 171**
  $$
  x + 1 = \frac{171}{3} = 57 \Rightarrow x = 56
  $$
  The integers are: **56, 57, 58** → Largest is **58**

---

### Step 5: Determine the Maximum Possible Largest Integer

From the above, the largest possible value for the **largest of the three integers** is:

$$
\boxed{58}
$$

---

### Final Answer

$$
\boxed{58}
$$

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 $\boxed{58}$.


Explanation:
Okay, let me try to figure out this problem. So, the question is about a book's leaves and pages. Let me start by recalling what the problem says.

First, it says a book has n leaves if it's composed of n pieces of paper. Then, the number of pages is twice the number of leaves because each side of a piece of paper is a page. So, if there are n leaves (pieces of paper), then there are 2n pages. That makes sense because each leaf has two pages: front and back, right?

Now, the problem states that the number of pages is 3 more than a multiple of 7. So, mathematically, the number of pages can be written as 7k + 3, where k is some integer. Since the number of pages is 2n, we can write that 2n = 7k + 3. Therefore, n = (7k + 3)/2.

But n has to be an integer because you can't have a fraction of a leaf. So, (7k + 3) must be even. Let me think about that. 7k is either even or odd depending on k. Since 7 is odd, multiplying it by k will result in even if k is even, and odd if k is odd. Then adding 3: if 7k is even, adding 3 makes it odd. If 7k is odd, adding 3 makes it even. Therefore, for (7k + 3) to be even, 7k must be odd, which means k must be odd.

So, k is an odd integer. Let me write that down. Let k = 2m + 1, where m is a non-negative integer. Then substituting back into the equation for n:

n = (7*(2m + 1) + 3)/2 = (14m + 7 + 3)/2 = (14m + 10)/2 = 7m + 5.

Therefore, n = 7m + 5. So, the number of leaves is 7m + 5.

Now, the problem also says that the number of leaves is greater than 100. So, 7m + 5 > 100. Let me solve for m:

7m + 5 > 100
7m > 95
m > 95/7
Calculating 95 divided by 7: 7*13 = 91, so 95 - 91 = 4. So, 95/7 = 13 + 4/7 ≈ 13.571. Therefore, m must be greater than approximately 13.571. Since m is an integer (because k is an integer and k = 2m + 1), m must be at least 14.

Therefore, the smallest possible m is 14. Then, substituting back into n:

n = 7*14 + 5 = 98 + 5 = 103.

Wait, so the smallest number of leaves is 103? Let me check if this satisfies all the conditions.

First, number of leaves is 103. Then, number of pages is 2*103 = 206. Now, check if 206 is 3 more than a multiple of 7. Let me divide 206 by 7.

7*29 = 203. 206 - 203 = 3. Yes, so 206 = 7*29 + 3. So that works. Also, the number of leaves is 103, which is greater than 100. So that seems to satisfy all the conditions.

But wait, let me check if there's a smaller m that I might have missed. Because m has to be greater than 13.571, so m=14 is the first integer. But let me check m=13. If m=13, then n=7*13 +5= 91 +5=96. But 96 is less than 100, so that's not allowed. Therefore, m=14 gives n=103, which is the first one above 100.

But let me think again. Maybe there's another way to approach this problem? Let me verify.

Alternatively, since the number of pages is 2n, and 2n ≡ 3 mod 7. So, 2n ≡ 3 mod 7. Let me solve this congruence for n.

Multiplying both sides by the inverse of 2 mod 7. Since 2*4=8≡1 mod7, so inverse of 2 mod7 is 4. Therefore, multiplying both sides by 4:

n ≡ 3*4 mod7
n ≡ 12 mod7
12 divided by 7 is 1 with remainder 5, so n ≡5 mod7.

Therefore, n=7m +5, which matches what I had earlier. Therefore, n must be congruent to 5 mod7. So, the smallest n>100 such that n≡5 mod7.

So, let me find the smallest n>100 where n ≡5 mod7.

First, find the smallest number greater than 100 that is congruent to 5 mod7.

Let me compute 100 divided by 7. 7*14=98, so 100-98=2. Therefore, 100 ≡2 mod7.

We need n ≡5 mod7. So, starting from 100, let's find how much we need to add to 100 to get to 5 mod7.

Current remainder: 2. We need remainder 5. The difference is 5 - 2 = 3. So, adding 3 to 100 would give 103, which is 100 +3=103. Let me check 103 mod7. 7*14=98, 103-98=5. Yes, 103≡5 mod7. Therefore, the smallest n>100 is 103.

Therefore, the answer is 103.

But let me check again if there's a smaller n. For example, if n=103 is the first one above 100. Let me check n=103-7=96. Which is less than 100, so not allowed. Therefore, 103 is indeed the smallest.

Therefore, the answer is 103.

But let me check once more with the original problem.

Number of leaves is 103. Number of pages is 206. 206 divided by 7 is 29*7=203, remainder 3. So 206=29*7 +3. Therefore, 206 is 3 more than a multiple of 7. And leaves are 103>100. So all conditions are satisfied.

Is there a possibility that there is a smaller number of leaves? For example, if there is a number between 100 and 103 that satisfies the condition? Let me check.

Numbers greater than 100 and less than 103 are 101 and 102. Let me check if they satisfy n ≡5 mod7.

For n=101: 101 divided by 7 is 14*7=98, 101-98=3. So 101≡3 mod7. Not 5.

n=102: 102-98=4, so 102≡4 mod7. Not 5.

Therefore, the next one is 103≡5 mod7. So yes, 103 is the first one.

Therefore, the answer is 103.

But let me think again if there's another way this could be approached. For example, maybe considering that the number of pages is 2n, which is 3 mod7. So 2n ≡3 mod7. Therefore, n≡(3/2) mod7. Which is same as multiplying by inverse of 2 mod7, which is 4, so 3*4=12≡5 mod7. So n≡5 mod7. So same result.

Therefore, the number of leaves must be congruent to 5 mod7, and greater than 100. So the minimal such n is 103.

Therefore, I think the answer is 103.

But just to make sure, let me check if there's a mistake in the initial assumption.

The problem says "the number of pages is twice the number of leaves because each side of a piece of paper is defined as a page." So, if there are n leaves (pieces of paper), then each leaf has two pages, so total pages is 2n. That seems correct.

So, for example, if you have 1 leaf, that's one piece of paper, which has two pages. So pages=2*1=2. So that's correct.

Therefore, the logic is sound.

Another check: if n=5, then pages=10. 10 mod7 is 3, since 7*1=7, 10-7=3. So 10 is 3 more than a multiple of 7. So n=5 would satisfy the condition. But since the problem says leaves must be greater than 100, so n=5 is too small.

So, the general solution is n=7m +5. So, the numbers are 5, 12, 19,... up to 103.

So, the answer is 103.

Therefore, I think the answer is 103.

**Final Answer**
The smallest possible number of leaves is \boxed{103}.

To solve the problem, we need to determine the **smallest number of leaves** in a book that satisfies two conditions:

1. The **number of pages** is **3 more than a multiple of 7**.
2. The **number of leaves** is **greater than 100**.

---

### Step 1: Understanding the Relationship Between Leaves and Pages

- A **leaf** is a single sheet of paper.
- Each leaf has **two pages** (front and back).
- Therefore, if a book has **n leaves**, it has **2n pages**.

---

### Step 2: Translating the Page Condition

We are told that the number of pages is **3 more than a multiple of 7**. That is:

$$
\text{Pages} = 7k + 3 \quad \text{for some integer } k
$$

But since the number of pages is also equal to $ 2n $, we can write:

$$
2n = 7k + 3
$$

Solving for $ n $:

$$
n = \frac{7k + 3}{2}
$$

For $ n $ to be an integer, the numerator $ 7k + 3 $ must be even. Since 7 is odd, $ 7k $ is even if and only if $ k $ is even. Therefore, $ 7k + 3 $ is even **only when $ k $ is odd**.

Let’s define $ k = 2m + 1 $, where $ m $ is a non-negative integer. Then:

$$
n = \frac{7(2m + 1) + 3}{2} = \frac{14m + 10}{2} = 7m + 5
$$

So, the number of leaves $ n $ must be of the form:

$$
n = 7m + 5
$$

---

### Step 3: Applying the Leaf Condition

We are told that the number of leaves must be **greater than 100**:

$$
7m + 5 > 100 \Rightarrow 7m > 95 \Rightarrow m > \frac{95}{7} \approx 13.57
$$

Since $ m $ must be an integer, the smallest possible value is $ m = 14 $.

Substituting back:

$$
n = 7(14) + 5 = 98 + 5 = 103
$$

---

### Step 4: Verifying the Solution

- Number of leaves: $ n = 103 $
- Number of pages: $ 2n = 206 $
- Check if 206 is 3 more than a multiple of 7:

$$
206 \div 7 = 29 \text{ remainder } 3 \Rightarrow 206 = 7 \cdot 29 + 3
$$

✅ This satisfies the condition.

- Also, $ n = 103 > 100 $, so it satisfies the second condition.

---

### Final Answer

$$
\boxed{103}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, the question is about a book's leaves and pages. Let me start by recalling what the problem says.

First, it says a book has n leaves if it's composed of n pieces of paper. Then, the number of pages is twice the number of leaves because each side of a piece of paper is a page. So, if there are n leaves (pieces of paper), then there are 2n pages. That makes sense because each leaf has two pages: front and back, right?

Now, the problem states that the number of pages is 3 more than a multiple of 7. So, mathematically, the number of pages can be written as 7k + 3, where k is some integer. Since the number of pages is 2n, we can write that 2n = 7k + 3. Therefore, n = (7k + 3)/2.

But n has to be an integer because you can't have a fraction of a leaf. So, (7k + 3) must be even. Let me think about that. 7k is either even or odd depending on k. Since 7 is odd, multiplying it by k will result in even if k is even, and odd if k is odd. Then adding 3: if 7k is even, adding 3 makes it odd. If 7k is odd, adding 3 makes it even. Therefore, for (7k + 3) to be even, 7k must be odd, which means k must be odd.

So, k is an odd integer. Let me write that down. Let k = 2m + 1, where m is a non-negative integer. Then substituting back into the equation for n:

n = (7*(2m + 1) + 3)/2 = (14m + 7 + 3)/2 = (14m + 10)/2 = 7m + 5.

Therefore, n = 7m + 5. So, the number of leaves is 7m + 5.

Now, the problem also says that the number of leaves is greater than 100. So, 7m + 5 > 100. Let me solve for m:

7m + 5 > 100
7m > 95
m > 95/7
Calculating 95 divided by 7: 7*13 = 91, so 95 - 91 = 4. So, 95/7 = 13 + 4/7 ≈ 13.571. Therefore, m must be greater than approximately 13.571. Since m is an integer (because k is an integer and k = 2m + 1), m must be at least 14.

Therefore, the smallest possible m is 14. Then, substituting back into n:

n = 7*14 + 5 = 98 + 5 = 103.

Wait, so the smallest number of leaves is 103? Let me check if this satisfies all the conditions.

First, number of leaves is 103. Then, number of pages is 2*103 = 206. Now, check if 206 is 3 more than a multiple of 7. Let me divide 206 by 7.

7*29 = 203. 206 - 203 = 3. Yes, so 206 = 7*29 + 3. So that works. Also, the number of leaves is 103, which is greater than 100. So that seems to satisfy all the conditions.

But wait, let me check if there's a smaller m that I might have missed. Because m has to be greater than 13.571, so m=14 is the first integer. But let me check m=13. If m=13, then n=7*13 +5= 91 +5=96. But 96 is less than 100, so that's not allowed. Therefore, m=14 gives n=103, which is the first one above 100.

But let me think again. Maybe there's another way to approach this problem? Let me verify.

Alternatively, since the number of pages is 2n, and 2n ≡ 3 mod 7. So, 2n ≡ 3 mod 7. Let me solve this congruence for n.

Multiplying both sides by the inverse of 2 mod 7. Since 2*4=8≡1 mod7, so inverse of 2 mod7 is 4. Therefore, multiplying both sides by 4:

n ≡ 3*4 mod7
n ≡ 12 mod7
12 divided by 7 is 1 with remainder 5, so n ≡5 mod7.

Therefore, n=7m +5, which matches what I had earlier. Therefore, n must be congruent to 5 mod7. So, the smallest n>100 such that n≡5 mod7.

So, let me find the smallest n>100 where n ≡5 mod7.

First, find the smallest number greater than 100 that is congruent to 5 mod7.

Let me compute 100 divided by 7. 7*14=98, so 100-98=2. Therefore, 100 ≡2 mod7.

We need n ≡5 mod7. So, starting from 100, let's find how much we need to add to 100 to get to 5 mod7.

Current remainder: 2. We need remainder 5. The difference is 5 - 2 = 3. So, adding 3 to 100 would give 103, which is 100 +3=103. Let me check 103 mod7. 7*14=98, 103-98=5. Yes, 103≡5 mod7. Therefore, the smallest n>100 is 103.

Therefore, the answer is 103.

But let me check again if there's a smaller n. For example, if n=103 is the first one above 100. Let me check n=103-7=96. Which is less than 100, so not allowed. Therefore, 103 is indeed the smallest.

Therefore, the answer is 103.

But let me check once more with the original problem.

Number of leaves is 103. Number of pages is 206. 206 divided by 7 is 29*7=203, remainder 3. So 206=29*7 +3. Therefore, 206 is 3 more than a multiple of 7. And leaves are 103>100. So all conditions are satisfied.

Is there a possibility that there is a smaller number of leaves? For example, if there is a number between 100 and 103 that satisfies the condition? Let me check.

Numbers greater than 100 and less than 103 are 101 and 102. Let me check if they satisfy n ≡5 mod7.

For n=101: 101 divided by 7 is 14*7=98, 101-98=3. So 101≡3 mod7. Not 5.

n=102: 102-98=4, so 102≡4 mod7. Not 5.

Therefore, the next one is 103≡5 mod7. So yes, 103 is the first one.

Therefore, the answer is 103.

But let me think again if there's another way this could be approached. For example, maybe considering that the number of pages is 2n, which is 3 mod7. So 2n ≡3 mod7. Therefore, n≡(3/2) mod7. Which is same as multiplying by inverse of 2 mod7, which is 4, so 3*4=12≡5 mod7. So n≡5 mod7. So same result.

Therefore, the number of leaves must be congruent to 5 mod7, and greater than 100. So the minimal such n is 103.

Therefore, I think the answer is 103.

But just to make sure, let me check if there's a mistake in the initial assumption.

The problem says "the number of pages is twice the number of leaves because each side of a piece of paper is defined as a page." So, if there are n leaves (pieces of paper), then each leaf has two pages, so total pages is 2n. That seems correct.

So, for example, if you have 1 leaf, that's one piece of paper, which has two pages. So pages=2*1=2. So that's correct.

Therefore, the logic is sound.

Another check: if n=5, then pages=10. 10 mod7 is 3, since 7*1=7, 10-7=3. So 10 is 3 more than a multiple of 7. So n=5 would satisfy the condition. But since the problem says leaves must be greater than 100, so n=5 is too small.

So, the general solution is n=7m +5. So, the numbers are 5, 12, 19,... up to 103.

So, the answer is 103.

Therefore, I think the answer is 103.

**Final Answer**
The smallest possible number of leaves is \boxed{103}.

To solve the problem, we need to determine the **smallest number of leaves** in a book that satisfies two conditions:

1. The **number of pages** is **3 more than a multiple of 7**.
2. The **number of leaves** is **greater than 100**.

---

### Step 1: Understanding the Relationship Between Leaves and Pages

- A **leaf** is a single sheet of paper.
- Each leaf has **two pages** (front and back).
- Therefore, if a book has **n leaves**, it has **2n pages**.

---

### Step 2: Translating the Page Condition

We are told that the number of pages is **3 more than a multiple of 7**. That is:

$$
\text{Pages} = 7k + 3 \quad \text{for some integer } k
$$

But since the number of pages is also equal to $ 2n $, we can write:

$$
2n = 7k + 3
$$

Solving for $ n $:

$$
n = \frac{7k + 3}{2}
$$

For $ n $ to be an integer, the numerator $ 7k + 3 $ must be even. Since 7 is odd, $ 7k $ is even if and only if $ k $ is even. Therefore, $ 7k + 3 $ is even **only when $ k $ is odd**.

Let’s define $ k = 2m + 1 $, where $ m $ is a non-negative integer. Then:

$$
n = \frac{7(2m + 1) + 3}{2} = \frac{14m + 10}{2} = 7m + 5
$$

So, the number of leaves $ n $ must be of the form:

$$
n = 7m + 5
$$

---

### Step 3: Applying the Leaf Condition

We are told that the number of leaves must be **greater than 100**:

$$
7m + 5 > 100 \Rightarrow 7m > 95 \Rightarrow m > \frac{95}{7} \approx 13.57
$$

Since $ m $ must be an integer, the smallest possible value is $ m = 14 $.

Substituting back:

$$
n = 7(14) + 5 = 98 + 5 = 103
$$

---

### Step 4: Verifying the Solution

- Number of leaves: $ n = 103 $
- Number of pages: $ 2n = 206 $
- Check if 206 is 3 more than a multiple of 7:

$$
206 \div 7 = 29 \text{ remainder } 3 \Rightarrow 206 = 7 \cdot 29 + 3
$$

✅ This satisfies the condition.

- Also, $ n = 103 > 100 $, so it satisfies the second condition.

---

### Final Answer

$$
\boxed{103}
$$

Solution: Let $m$ be the least possible number of leaves. Then $2m$ is the least possible number of pages. We know that $2m\equiv 3\pmod 7\implies 8m \equiv 3\cdot 4\pmod 7\implies m\equiv 12\equiv 5\pmod 7$. So $m=5+7a$ for some positive integer $a$. The smallest such number greater than $100$ is $5+7\cdot 14=\boxed{103}$.