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 the possible sums are 111, 141, 171. Wait, but 111, 141, 171. Are there any others?

Wait, let me check again. For example, 202: 2+0+2=4, not divisible by 3. 212: 2+1+2=5. What about 191? 1+9+1=11, nope. 181: 1+8+1=10, nope. 161: 8, nope. 151: 7, nope. 131: 5, nope. 121: 4, nope. 101: 2. So yes, only 111, 141, 171. Wait, but 202 is 202, which is less than 220. But 202 isn't divisible by 3. So only those three?

Wait, but maybe I made a mistake here. Let me check 111, 141, 171. Let me check if there are other palindromes between 100 and 220 that are divisible by 3. For example, 121 is not, 131 no, 141 yes, 151 no, 161 no, 171 yes, 181 no, 191 no, 202 no, 212 no. So only 111, 141, 171. Wait, but 212 is 212, which is less than 220. But 212 isn't divisible by 3. So only those three?

Wait, but maybe I missed some palindromes? Let me think. For example, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. That's all the three-digit palindromes less than 220. So yes, only 111, 141, 171 are divisible by 3. Therefore, the sum of the three consecutive integers must be one of these three numbers.

But wait, let me check if there are other palindromes. For example, 101 is 101, which is not divisible by 3. 111 is 111. Then 121, 131, 141, etc. So yes, only those three. Therefore, the sum can be 111, 141, or 171. Therefore, the three consecutive integers would be:

For sum = 111: 3(x + 1) = 111 ⇒ x + 1 = 37 ⇒ x = 36. Therefore, the three integers are 36, 37, 38. The largest is 38.

For sum = 141: 3(x + 1) = 141 ⇒ x + 1 = 47 ⇒ x = 46. Therefore, the integers are 46, 47, 48. Largest is 48.

For sum = 171: 3(x + 1) = 171 ⇒ x + 1 = 57 ⇒ x = 56. Therefore, integers 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". But 171 is less than 220, so that's okay. But maybe there are other palindromes that I missed?

Wait, hold on. Wait, maybe I made a mistake in assuming that the sum is 3(x + 1). Let me check again. If the three consecutive integers are x, x + 1, x + 2, then their sum is x + (x + 1) + (x + 2) = 3x + 3 = 3(x + 1). So that's correct. Therefore, the sum must be divisible by 3. Therefore, only those three palindromes. Therefore, the maximum largest integer is 58. But wait, the problem says "the greatest possible value for the largest of the three integers in the set". But maybe there are other palindromes?

Wait, maybe I need to check if there are other palindromes less than 220 that are divisible by 3. Let me think again. 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. How about 212? 212 reversed is 212, which is a palindrome. But 212 is not divisible by 3. 2 + 1 + 2 = 5. Not divisible by 3. How about 121? 1 + 2 + 1 = 4. Not divisible by 3. 131? 1 + 3 + 1 = 5. Not. 151? 1 + 5 + 1 = 7. 161? 1 + 6 + 1 = 8. 181? 1 + 8 + 1 = 10. 191? 1 + 9 + 1 = 11. 202? 2 + 0 + 2 = 4. 212? 5. So no, only 111, 141, 171. Therefore, the maximum sum is 171, leading to the largest integer being 58. But wait, the problem says "the sum is less than 220". So 171 is less than 220, so that's okay. But maybe there are other palindromes?

Wait, maybe I need to check if there are palindromes between 171 and 220 that are divisible by 3. For example, 202 is not a palindrome? Wait, 202 is a palindrome. Wait, 202 is a palindrome. But 202 is not divisible by 3. What about 212? Also not. How about 222? But that's over 220. So no. What about numbers like 121? No. Wait, maybe I need to check if there are other palindromes. For example, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. So that's all. So only 111, 141, 171. Therefore, the maximum sum is 171. 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. Suppose the three consecutive integers are n-1, n, n+1. Then their sum is 3n. So the sum is 3n. Therefore, the sum is divisible by 3. Therefore, same conclusion. So the sum is 3n, which is equal to the palindrome. Therefore, the palindrome must be divisible by 3. Therefore, same as before. Therefore, the possible palindromes are 111, 141, 171. Therefore, the maximum sum is 171, which gives n = 57. Therefore, the three numbers are 56, 57, 58. Therefore, the largest is 58. Therefore, the answer is 58? But wait, the problem says "the greatest possible value for the largest of the three integers in the set". But maybe there's a larger possible value?

Wait, maybe I need to check if there are other palindromes that I missed. Let me think again. For example, 101 is not divisible by 3. 111 is. 121 is not. 131 is not. 141 is. 151 is not. 161 is not. 171 is. 181 is not. 191 is not. 202 is not. 212 is not. So only those three. Therefore, 171 is the maximum sum. Therefore, the largest integer is 58. But wait, maybe there's another way to get a larger number?

Wait, let me think. Suppose the sum is 171. Then the numbers are 56, 57, 58. The largest is 58. But maybe if the sum is higher? But the sum has to be less than 220. Wait, 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 possible sum. Therefore, the maximum largest integer is 58. But wait, maybe there's another palindrome that I didn't consider?

Wait, let me check if there are palindromes between 171 and 220 that are divisible by 3. For example, 181 is not. 191 is not. 202 is not. 212 is not. 222 is over. How about numbers like 151? No. 161? No. 171 is the last one. So no. Therefore, 171 is the maximum sum. Therefore, the answer is 58. But wait, let me check if there's another way to get a larger number. For example, if the sum is 171, then the numbers are 56, 57, 58. But maybe if the sum is a different palindrome, but I thought only 111, 141, 171 are possible. Wait, but maybe I made a mistake in the divisibility?

Wait, let me check 111, 141, 171. Let me check if they are indeed divisible by 3. 111 divided by 3 is 37. 141 divided by 3 is 47. 171 divided by 3 is 57. Yes, correct. So those are the only ones. Therefore, the maximum sum is 171, leading to the largest integer being 58. Therefore, the answer is 58. But wait, the problem says "three consecutive positive integers". So maybe there's another way to get a larger number?

Wait, let me think again. Suppose the sum is 171. Then the numbers are 56, 57, 58. The largest is 58. But if there is a larger sum, say, 183? Wait, but 183 is not a palindrome. 183 reversed is 381, which is not the same. How about 191? 191 is a palindrome, but 191 divided by 3 is 63.666..., which is not an integer. Therefore, not possible. How about 202? 202 divided by 3 is 67.333... Not integer. 212 divided by 3 is 70.666... Not integer. So no. Therefore, no other palindromes. Therefore, 171 is the maximum sum. Therefore, the answer is 58. But wait, let me check if there are other palindromes that I might have missed. For example, numbers like 101, 111, 121, etc. Wait, 101 is not divisible by 3. 111 is. 121 is not. 131 is not. 141 is. 151 is not. 161 is not. 171 is. 181 is not. 191 is not. 202 is not. 212 is not. So yes, only those three. Therefore, 58 is the answer. But wait, let me check if there's another way to get a larger number. For example, if the sum is 171, then the numbers are 56, 57, 58. But maybe if the sum is 171, but the numbers are different? Wait, no. Because the sum is 3(x + 1). Therefore, if the sum is 171, then x + 1 = 57, so x = 56. Therefore, numbers are 56, 57, 58. So the largest is 58. Therefore, that's the maximum. Therefore, the answer is 58. But wait, let me check if there's a mistake here. Maybe I need to check if there are other palindromes that are divisible by 3. Let me think again. For example, 3-digit palindromes less than 220. Let me list them again:

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 yes, only 111, 141, 171. Therefore, the maximum sum is 171. Therefore, the largest integer is 58. Therefore, the answer is 58. But wait, the problem says "three consecutive positive integers". Let me check if there's another way to get a larger number. For example, if the sum is 171, then the numbers are 56, 57, 58. But maybe if the sum is 171, but the numbers are different? No, because the sum is fixed. If the sum is 171, then the numbers must be 56, 57, 58. Therefore, the largest is 58. Therefore, the answer is 58. But wait, let me check if there's a mistake in my calculation. For example, if the sum is 171, then 3(x + 1) = 171 ⇒ x + 1 = 57 ⇒ x = 56. Therefore, numbers are 56, 57, 58. So the largest is 58. Therefore, that's correct. Therefore, the answer is 58. But wait, I have a feeling that maybe there's a larger number. Let me think again. Maybe I need to check if there are other palindromes that are divisible by 3. Wait, for example, 3-digit palindromes can also be like 303, but that's over 220. So no. How about 111, 141, 171. Wait, 171 + 3* something? Wait, no. Wait, maybe I need to check if there are palindromes between 171 and 220 that are divisible by 3. For example, 183? But 183 is not a palindrome. 192? Not a palindrome. 201? Not a palindrome. 210? Not a palindrome. 212? Not divisible by 3. 222? Over 220. So no. Therefore, no. Therefore, 171 is the maximum. Therefore, the answer is 58. But let me check with another approach. Let me think of the three consecutive integers. Let me denote them as n-1, n, n+1. Their sum is 3n. Therefore, 3n is a palindrome less than 220. Therefore, n must be less than 220 / 3 ≈ 73.333. Therefore, n can be up to 73. Therefore, the largest integer in the set is n + 1, which would be up to 74. Wait, but if 3n is a palindrome less than 220, then n can be up to floor(219 / 3) = 73. Therefore, the maximum possible n is 73, so the largest integer is 74. But wait, but 3*73 = 219. Is 219 a palindrome? No, because 219 reversed is 912. Not the same. Therefore, 219 is not a palindrome. Therefore, the maximum n such that 3n is a palindrome less than 220. So let me check palindromes less than 220, which are divisible by 3. Which are 111, 141, 171. Therefore, 3n = 111 ⇒ n = 37. Then the largest integer is 37 + 1 = 38. 3n = 141 ⇒ n = 47 ⇒ largest integer 48. 3n = 171 ⇒ n = 57 ⇒ largest integer 58. Therefore, the maximum is 58. Therefore, the answer is 58. Therefore, I think that is correct. Therefore, the answer is 58. Wait, but let me check if there are other palindromes. For example, 3n = 201? 201 is not a palindrome. 3n = 210? Not a palindrome. 3n = 213? Not a palindrome. 3n = 216? Not a palindrome. 3n = 219? Not a palindrome. Therefore, no. Therefore, the answer is 58. Therefore, I think that is correct. Therefore, the greatest possible value for the largest of the three integers is 58. Therefore, the answer is \boxed{58}.

Wait, but hold on. Let me check if there are other palindromes that I might have missed. For example, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. Are there any other three-digit palindromes less than 220? For example, numbers like 101, 111, etc. Wait, 101 is 101, then 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. That's all. So no others. Therefore, only those three palindromes. Therefore, the answer is 58. Therefore, I think that is correct. Therefore, I think the answer is 58.

But wait, let me check once again. Suppose the sum is 171. Then the numbers are 56, 57, 58. Their sum is 56 + 57 + 58 = 171. Correct. 56 + 57 is 113, plus 58 is 171. Correct. And 171 is a palindrome. So that's correct. 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. 174 reversed is 471. Not the same. How about 58, 59, 60? Sum is 58 + 59 + 60 = 177. Not a palindrome. 59, 60, 61: 59 + 60 + 61 = 180. Not a palindrome. 60, 61, 62: 183. Not a palindrome. 61, 62, 63: 186. Not. 62, 63, 64: 189. Not. 63, 64, 65: 192. Not. 64, 65, 66: 195. Not. 65, 66, 67: 198. Not. 66, 67, 68: 201. Not a palindrome. 67, 68, 69: 204. Not. 68, 69, 70: 207. Not. 69, 70, 71: 210. Not. 70, 71, 72: 213. Not. 71, 72, 73: 216. Not. 72, 73, 74: 219. Not. 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 sum that is a palindrome between 171 and 220. Therefore, 171 is indeed the maximum sum. Therefore, the largest integer is 58. Therefore, the answer is 58. Therefore, I think that is correct.

**Final Answer**
\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. The valid palindromes less than 220 are:

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

Now, we filter these to only include those 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 3: Find the Corresponding Sets of Three Consecutive Integers

We now find the three consecutive integers whose sum equals each of these palindromes.

#### Case 1: Sum = 111

$$
3(x + 1) = 111 \Rightarrow x + 1 = 37 \Rightarrow x = 36
$$

So the integers are: **36, 37, 38**
Largest integer = **38**

#### Case 2: Sum = 141

$$
3(x + 1) = 141 \Rightarrow x + 1 = 47 \Rightarrow x = 46
$$

So the integers are: **46, 47, 48**
Largest integer = **48**

#### Case 3: Sum = 171

$$
3(x + 1) = 171 \Rightarrow x + 1 = 57 \Rightarrow x = 56
$$

So the integers are: **56, 57, 58**
Largest integer = **58**

---

### Step 4: Determine the Maximum Possible Largest Integer

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

$$
\boxed{58}
$$

This occurs when the sum is **171**, which is a valid three-digit palindrome less than 220 and divisible by 3.

---

### 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 the possible sums are 111, 141, 171. Wait, but 111, 141, 171. Are there any others?

Wait, let me check again. For example, 202: 2+0+2=4, not divisible by 3. 212: 2+1+2=5. What about 191? 1+9+1=11, nope. 181: 1+8+1=10, nope. 161: 8, nope. 151: 7, nope. 131: 5, nope. 121: 4, nope. 101: 2. So yes, only 111, 141, 171. Wait, but 202 is 202, which is less than 220. But 202 isn't divisible by 3. So only those three?

Wait, but maybe I made a mistake here. Let me check 111, 141, 171. Let me check if there are other palindromes between 100 and 220 that are divisible by 3. For example, 121 is not, 131 no, 141 yes, 151 no, 161 no, 171 yes, 181 no, 191 no, 202 no, 212 no. So only 111, 141, 171. Wait, but 212 is 212, which is less than 220. But 212 isn't divisible by 3. So only those three?

Wait, but maybe I missed some palindromes? Let me think. For example, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. That's all the three-digit palindromes less than 220. So yes, only 111, 141, 171 are divisible by 3. Therefore, the sum of the three consecutive integers must be one of these three numbers.

But wait, let me check if there are other palindromes. For example, 101 is 101, which is not divisible by 3. 111 is 111. Then 121, 131, 141, etc. So yes, only those three. Therefore, the sum can be 111, 141, or 171. Therefore, the three consecutive integers would be:

For sum = 111: 3(x + 1) = 111 ⇒ x + 1 = 37 ⇒ x = 36. Therefore, the three integers are 36, 37, 38. The largest is 38.

For sum = 141: 3(x + 1) = 141 ⇒ x + 1 = 47 ⇒ x = 46. Therefore, the integers are 46, 47, 48. Largest is 48.

For sum = 171: 3(x + 1) = 171 ⇒ x + 1 = 57 ⇒ x = 56. Therefore, integers 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". But 171 is less than 220, so that's okay. But maybe there are other palindromes that I missed?

Wait, hold on. Wait, maybe I made a mistake in assuming that the sum is 3(x + 1). Let me check again. If the three consecutive integers are x, x + 1, x + 2, then their sum is x + (x + 1) + (x + 2) = 3x + 3 = 3(x + 1). So that's correct. Therefore, the sum must be divisible by 3. Therefore, only those three palindromes. Therefore, the maximum largest integer is 58. But wait, the problem says "the greatest possible value for the largest of the three integers in the set". But maybe there are other palindromes?

Wait, maybe I need to check if there are other palindromes less than 220 that are divisible by 3. Let me think again. 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. How about 212? 212 reversed is 212, which is a palindrome. But 212 is not divisible by 3. 2 + 1 + 2 = 5. Not divisible by 3. How about 121? 1 + 2 + 1 = 4. Not divisible by 3. 131? 1 + 3 + 1 = 5. Not. 151? 1 + 5 + 1 = 7. 161? 1 + 6 + 1 = 8. 181? 1 + 8 + 1 = 10. 191? 1 + 9 + 1 = 11. 202? 2 + 0 + 2 = 4. 212? 5. So no, only 111, 141, 171. Therefore, the maximum sum is 171, leading to the largest integer being 58. But wait, the problem says "the sum is less than 220". So 171 is less than 220, so that's okay. But maybe there are other palindromes?

Wait, maybe I need to check if there are palindromes between 171 and 220 that are divisible by 3. For example, 202 is not a palindrome? Wait, 202 is a palindrome. Wait, 202 is a palindrome. But 202 is not divisible by 3. What about 212? Also not. How about 222? But that's over 220. So no. What about numbers like 121? No. Wait, maybe I need to check if there are other palindromes. For example, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. So that's all. So only 111, 141, 171. Therefore, the maximum sum is 171. 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. Suppose the three consecutive integers are n-1, n, n+1. Then their sum is 3n. So the sum is 3n. Therefore, the sum is divisible by 3. Therefore, same conclusion. So the sum is 3n, which is equal to the palindrome. Therefore, the palindrome must be divisible by 3. Therefore, same as before. Therefore, the possible palindromes are 111, 141, 171. Therefore, the maximum sum is 171, which gives n = 57. Therefore, the three numbers are 56, 57, 58. Therefore, the largest is 58. Therefore, the answer is 58? But wait, the problem says "the greatest possible value for the largest of the three integers in the set". But maybe there's a larger possible value?

Wait, maybe I need to check if there are other palindromes that I missed. Let me think again. For example, 101 is not divisible by 3. 111 is. 121 is not. 131 is not. 141 is. 151 is not. 161 is not. 171 is. 181 is not. 191 is not. 202 is not. 212 is not. So only those three. Therefore, 171 is the maximum sum. Therefore, the largest integer is 58. But wait, maybe there's another way to get a larger number?

Wait, let me think. Suppose the sum is 171. Then the numbers are 56, 57, 58. The largest is 58. But maybe if the sum is higher? But the sum has to be less than 220. Wait, 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 possible sum. Therefore, the maximum largest integer is 58. But wait, maybe there's another palindrome that I didn't consider?

Wait, let me check if there are palindromes between 171 and 220 that are divisible by 3. For example, 181 is not. 191 is not. 202 is not. 212 is not. 222 is over. How about numbers like 151? No. 161? No. 171 is the last one. So no. Therefore, 171 is the maximum sum. Therefore, the answer is 58. But wait, let me check if there's another way to get a larger number. For example, if the sum is 171, then the numbers are 56, 57, 58. But maybe if the sum is a different palindrome, but I thought only 111, 141, 171 are possible. Wait, but maybe I made a mistake in the divisibility?

Wait, let me check 111, 141, 171. Let me check if they are indeed divisible by 3. 111 divided by 3 is 37. 141 divided by 3 is 47. 171 divided by 3 is 57. Yes, correct. So those are the only ones. Therefore, the maximum sum is 171, leading to the largest integer being 58. Therefore, the answer is 58. But wait, the problem says "three consecutive positive integers". So maybe there's another way to get a larger number?

Wait, let me think again. Suppose the sum is 171. Then the numbers are 56, 57, 58. The largest is 58. But if there is a larger sum, say, 183? Wait, but 183 is not a palindrome. 183 reversed is 381, which is not the same. How about 191? 191 is a palindrome, but 191 divided by 3 is 63.666..., which is not an integer. Therefore, not possible. How about 202? 202 divided by 3 is 67.333... Not integer. 212 divided by 3 is 70.666... Not integer. So no. Therefore, no other palindromes. Therefore, 171 is the maximum sum. Therefore, the answer is 58. But wait, let me check if there are other palindromes that I might have missed. For example, numbers like 101, 111, 121, etc. Wait, 101 is not divisible by 3. 111 is. 121 is not. 131 is not. 141 is. 151 is not. 161 is not. 171 is. 181 is not. 191 is not. 202 is not. 212 is not. So yes, only those three. Therefore, 58 is the answer. But wait, let me check if there's another way to get a larger number. For example, if the sum is 171, then the numbers are 56, 57, 58. But maybe if the sum is 171, but the numbers are different? Wait, no. Because the sum is 3(x + 1). Therefore, if the sum is 171, then x + 1 = 57, so x = 56. Therefore, numbers are 56, 57, 58. So the largest is 58. Therefore, that's the maximum. Therefore, the answer is 58. But wait, let me check if there's a mistake here. Maybe I need to check if there are other palindromes that are divisible by 3. Let me think again. For example, 3-digit palindromes less than 220. Let me list them again:

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 yes, only 111, 141, 171. Therefore, the maximum sum is 171. Therefore, the largest integer is 58. Therefore, the answer is 58. But wait, the problem says "three consecutive positive integers". Let me check if there's another way to get a larger number. For example, if the sum is 171, then the numbers are 56, 57, 58. But maybe if the sum is 171, but the numbers are different? No, because the sum is fixed. If the sum is 171, then the numbers must be 56, 57, 58. Therefore, the largest is 58. Therefore, the answer is 58. But wait, let me check if there's a mistake in my calculation. For example, if the sum is 171, then 3(x + 1) = 171 ⇒ x + 1 = 57 ⇒ x = 56. Therefore, numbers are 56, 57, 58. So the largest is 58. Therefore, that's correct. Therefore, the answer is 58. But wait, I have a feeling that maybe there's a larger number. Let me think again. Maybe I need to check if there are other palindromes that are divisible by 3. Wait, for example, 3-digit palindromes can also be like 303, but that's over 220. So no. How about 111, 141, 171. Wait, 171 + 3* something? Wait, no. Wait, maybe I need to check if there are palindromes between 171 and 220 that are divisible by 3. For example, 183? But 183 is not a palindrome. 192? Not a palindrome. 201? Not a palindrome. 210? Not a palindrome. 212? Not divisible by 3. 222? Over 220. So no. Therefore, no. Therefore, 171 is the maximum. Therefore, the answer is 58. But let me check with another approach. Let me think of the three consecutive integers. Let me denote them as n-1, n, n+1. Their sum is 3n. Therefore, 3n is a palindrome less than 220. Therefore, n must be less than 220 / 3 ≈ 73.333. Therefore, n can be up to 73. Therefore, the largest integer in the set is n + 1, which would be up to 74. Wait, but if 3n is a palindrome less than 220, then n can be up to floor(219 / 3) = 73. Therefore, the maximum possible n is 73, so the largest integer is 74. But wait, but 3*73 = 219. Is 219 a palindrome? No, because 219 reversed is 912. Not the same. Therefore, 219 is not a palindrome. Therefore, the maximum n such that 3n is a palindrome less than 220. So let me check palindromes less than 220, which are divisible by 3. Which are 111, 141, 171. Therefore, 3n = 111 ⇒ n = 37. Then the largest integer is 37 + 1 = 38. 3n = 141 ⇒ n = 47 ⇒ largest integer 48. 3n = 171 ⇒ n = 57 ⇒ largest integer 58. Therefore, the maximum is 58. Therefore, the answer is 58. Therefore, I think that is correct. Therefore, the answer is 58. Wait, but let me check if there are other palindromes. For example, 3n = 201? 201 is not a palindrome. 3n = 210? Not a palindrome. 3n = 213? Not a palindrome. 3n = 216? Not a palindrome. 3n = 219? Not a palindrome. Therefore, no. Therefore, the answer is 58. Therefore, I think that is correct. Therefore, the greatest possible value for the largest of the three integers is 58. Therefore, the answer is \boxed{58}.

Wait, but hold on. Let me check if there are other palindromes that I might have missed. For example, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. Are there any other three-digit palindromes less than 220? For example, numbers like 101, 111, etc. Wait, 101 is 101, then 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212. That's all. So no others. Therefore, only those three palindromes. Therefore, the answer is 58. Therefore, I think that is correct. Therefore, I think the answer is 58.

But wait, let me check once again. Suppose the sum is 171. Then the numbers are 56, 57, 58. Their sum is 56 + 57 + 58 = 171. Correct. 56 + 57 is 113, plus 58 is 171. Correct. And 171 is a palindrome. So that's correct. 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. 174 reversed is 471. Not the same. How about 58, 59, 60? Sum is 58 + 59 + 60 = 177. Not a palindrome. 59, 60, 61: 59 + 60 + 61 = 180. Not a palindrome. 60, 61, 62: 183. Not a palindrome. 61, 62, 63: 186. Not. 62, 63, 64: 189. Not. 63, 64, 65: 192. Not. 64, 65, 66: 195. Not. 65, 66, 67: 198. Not. 66, 67, 68: 201. Not a palindrome. 67, 68, 69: 204. Not. 68, 69, 70: 207. Not. 69, 70, 71: 210. Not. 70, 71, 72: 213. Not. 71, 72, 73: 216. Not. 72, 73, 74: 219. Not. 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 sum that is a palindrome between 171 and 220. Therefore, 171 is indeed the maximum sum. Therefore, the largest integer is 58. Therefore, the answer is 58. Therefore, I think that is correct.

**Final Answer**
\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. The valid palindromes less than 220 are:

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

Now, we filter these to only include those 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 3: Find the Corresponding Sets of Three Consecutive Integers

We now find the three consecutive integers whose sum equals each of these palindromes.

#### Case 1: Sum = 111

$$
3(x + 1) = 111 \Rightarrow x + 1 = 37 \Rightarrow x = 36
$$

So the integers are: **36, 37, 38**
Largest integer = **38**

#### Case 2: Sum = 141

$$
3(x + 1) = 141 \Rightarrow x + 1 = 47 \Rightarrow x = 46
$$

So the integers are: **46, 47, 48**
Largest integer = **48**

#### Case 3: Sum = 171

$$
3(x + 1) = 171 \Rightarrow x + 1 = 57 \Rightarrow x = 56
$$

So the integers are: **56, 57, 58**
Largest integer = **58**

---

### Step 4: Determine the Maximum Possible Largest Integer

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

$$
\boxed{58}
$$

This occurs when the sum is **171**, which is a valid three-digit palindrome less than 220 and divisible by 3.

---

### 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 books, leaves, and pages. Let me start by recalling what I know.

First, the problem says that a book has n leaves, which means it's made up of n pieces of paper. But then it mentions that 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, each leaf has two pages, right? So, total pages would be 2n. That makes sense. So pages = 2 * leaves. Got that.

Now, the problem states that the number of pages is 3 more than a multiple of 7. So, mathematically, that would be pages ≡ 3 mod 7. Since pages = 2n, then 2n ≡ 3 mod 7. Also, the number of leaves is greater than 100, so n > 100. We need to find the smallest possible n (number of leaves) that satisfies these conditions.

So, let me write down the equations:

1. 2n ≡ 3 mod 7
2. n > 100
3. Find the smallest n satisfying 1 and 2.

Alright, so first, let me solve the congruence equation 2n ≡ 3 mod 7. To solve for n, I need to find the multiplicative inverse of 2 modulo 7. The inverse of 2 mod 7 is a number x such that 2x ≡ 1 mod 7. Let me compute that.

Trying x=4: 2*4=8≡1 mod7. Yes, because 8-7=1. So inverse of 2 mod7 is 4. Therefore, multiplying both sides of 2n ≡3 mod7 by 4:

n ≡ 4*3 mod7 => n ≡12 mod7. 12 divided by 7 is 1 with remainder 5. So n ≡5 mod7. Therefore, n can be written as n=7k +5 for some integer k.

So, n must be of the form 7k +5. Also, n must be greater than 100. So, we need to find the smallest integer k such that 7k +5 >100. Let me solve for k:

7k +5 >100 => 7k >95 => k >95/7. Let me compute 95 divided by 7. 7*13=91, 7*14=98. So 95/7 is 13 + 2/7. So k must be greater than 13 + 2/7. Since k must be integer, the smallest integer k is 14. Therefore, the smallest n is 7*14 +5. Let me compute that: 7*14 is 98, plus 5 is 103. Therefore, n=103. Is that the answer? Wait, but let me check if this satisfies the original condition.

So, pages = 2n = 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. Therefore, 206 ≡3 mod7. Perfect. Also, n=103 is greater than 100. So, seems like 103 is the answer. But wait, let me check if there is a smaller n>100 that satisfies n≡5 mod7. Because maybe k=13 gives n=7*13 +5=91 +5=96, which is less than 100. So, k=14 gives 103. So, 103 is the first n>100. Therefore, 103 is the minimal number of leaves. So, the answer is 103?

Wait, but let me check again. Maybe I made a mistake in solving the congruence? Let me check again.

Original equation: 2n ≡3 mod7. Multiply both sides by inverse of 2 mod7, which is 4. So 4*2n ≡4*3 mod7 => 8n≡12 mod7. But 8≡1 mod7, so 1*n≡12 mod7. 12 mod7 is 5. Therefore, n≡5 mod7. So that's correct. So n=7k+5. So yes, the solutions are numbers congruent to 5 mod7. So the numbers are 5,12,19,... up to 103. So the first one above 100 is 103. Therefore, 103 is the answer. But let me check if there is a smaller n>100. Let me check n=103-7=96, which is less than 100. So next one is 103. So yes, 103 is the minimal. Therefore, the answer is 103. Wait, but let me check if there is a mistake in the problem statement interpretation.

Wait, 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, each leaf is a piece of paper, which has two pages. So, if there are n leaves, then there are 2n pages. So that's correct. So pages=2n. So the problem says pages is 3 more than a multiple of 7. So 2n ≡3 mod7. So solving that gives n≡5 mod7. So n=7k+5. So n>100. So the minimal n is 103. Therefore, answer is 103. But let me check with n=103. Let me check the number of pages: 206. 206 divided by 7 is 29*7=203, remainder 3. So yes, 206=7*29 +3. So that's correct. So 206 is 3 more than a multiple of 7. So that's correct. So seems like 103 is the answer. But let me check if there is a smaller n>100. For example, maybe n=103 is the first one. Let me check n=100. Let me check if n=100 is possible. For n=100, pages=200. 200 divided by 7 is 28*7=196, remainder 4. So 200≡4 mod7. Not 3. n=101: pages=202. 202-196=6, so 202≡6 mod7. Not 3. n=102: pages=204. 204-196=8, which is 1 mod7. 204≡1 mod7. Not 3. n=103: pages=206≡3 mod7. So yes, that's the first one. Therefore, 103 is the answer. Therefore, the smallest possible number of leaves is 103. So I think that's the answer. Wait, but let me check if there's another way to approach this problem. Maybe I made a mistake in the congruence?

Alternatively, maybe I can think of 2n ≡3 mod7. So 2n -3 is divisible by 7. So 2n -3 =7m, for some integer m. Then 2n=7m +3. Therefore, n=(7m +3)/2. Since n must be integer, 7m +3 must be even. Therefore, 7m must be odd, since 7m +3 is even. 7m is odd implies that m is odd. Because 7 is odd, odd times odd is odd, odd times even is even. Therefore, m must be odd. Let me write m=2k +1, where k is integer. Then substituting back:

n=(7*(2k +1) +3)/2=(14k +7 +3)/2=(14k +10)/2=7k +5. So same result as before. Therefore, n=7k +5. Therefore, same conclusion. Therefore, n must be 5,12,19,... So the first n>100 is 103. Therefore, answer is 103. Therefore, I think that's correct. So I think the answer is 103. Wait, but let me check if there's a possible mistake in the problem statement. For example, sometimes in books, the term "leaf" can refer to a single sheet, which has two pages, but sometimes people might count leaves differently? Wait, the problem says "a book is said to have n leaves if it is composed of n pieces of paper." So each leaf is a piece of paper. So each leaf has two pages. So pages=2n. So that's correct. So 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 that's consistent. So no problem here. Therefore, I think my answer is correct. Therefore, the smallest possible number of leaves 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** $ n $ in a book such that:

1. The **number of pages** is **twice the number of leaves**, i.e., $ \text{pages} = 2n $.
2. The **number of pages** is **3 more than a multiple of 7**, i.e., $ 2n \equiv 3 \pmod{7} $.
3. The **number of leaves** is **greater than 100**, i.e., $ n > 100 $.

---

### Step 1: Solve the Congruence

We are given:

$$
2n \equiv 3 \pmod{7}
$$

To solve for $ n $, we need the **multiplicative inverse** of 2 modulo 7. That is, a number $ x $ such that:

$$
2x \equiv 1 \pmod{7}
$$

Trying values, we find:

$$
2 \times 4 = 8 \equiv 1 \pmod{7}
$$

So, the inverse of 2 modulo 7 is 4. Multiply both sides of the original congruence by 4:

$$
4 \cdot 2n \equiv 4 \cdot 3 \pmod{7} \Rightarrow 8n \equiv 12 \pmod{7}
$$

Since $ 8 \equiv 1 \pmod{7} $, this simplifies to:

$$
n \equiv 12 \pmod{7}
$$

Now, reduce 12 modulo 7:

$$
12 \div 7 = 1 \text{ remainder } 5 \Rightarrow n \equiv 5 \pmod{7}
$$

So, the general solution is:

$$
n = 7k + 5 \quad \text{for some integer } k
$$

---

### Step 2: Find the Smallest $ n > 100 $

We now find the smallest integer $ k $ such that:

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

So the smallest integer $ k $ is 14. Substituting into the formula:

$$
n = 7 \cdot 14 + 5 = 98 + 5 = 103
$$

---

### Step 3: Verify the Solution

- Number of leaves: $ n = 103 $
- Number of pages: $ 2n = 206 $
- Check if $ 206 \equiv 3 \pmod{7} $:

$$
206 \div 7 = 29 \text{ remainder } 3 \Rightarrow 206 \equiv 3 \pmod{7}
$$

✅ This satisfies the condition.

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

---

### Final Answer

$$
\boxed{103}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, the question is about books, leaves, and pages. Let me start by recalling what I know.

First, the problem says that a book has n leaves, which means it's made up of n pieces of paper. But then it mentions that 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, each leaf has two pages, right? So, total pages would be 2n. That makes sense. So pages = 2 * leaves. Got that.

Now, the problem states that the number of pages is 3 more than a multiple of 7. So, mathematically, that would be pages ≡ 3 mod 7. Since pages = 2n, then 2n ≡ 3 mod 7. Also, the number of leaves is greater than 100, so n > 100. We need to find the smallest possible n (number of leaves) that satisfies these conditions.

So, let me write down the equations:

1. 2n ≡ 3 mod 7
2. n > 100
3. Find the smallest n satisfying 1 and 2.

Alright, so first, let me solve the congruence equation 2n ≡ 3 mod 7. To solve for n, I need to find the multiplicative inverse of 2 modulo 7. The inverse of 2 mod 7 is a number x such that 2x ≡ 1 mod 7. Let me compute that.

Trying x=4: 2*4=8≡1 mod7. Yes, because 8-7=1. So inverse of 2 mod7 is 4. Therefore, multiplying both sides of 2n ≡3 mod7 by 4:

n ≡ 4*3 mod7 => n ≡12 mod7. 12 divided by 7 is 1 with remainder 5. So n ≡5 mod7. Therefore, n can be written as n=7k +5 for some integer k.

So, n must be of the form 7k +5. Also, n must be greater than 100. So, we need to find the smallest integer k such that 7k +5 >100. Let me solve for k:

7k +5 >100 => 7k >95 => k >95/7. Let me compute 95 divided by 7. 7*13=91, 7*14=98. So 95/7 is 13 + 2/7. So k must be greater than 13 + 2/7. Since k must be integer, the smallest integer k is 14. Therefore, the smallest n is 7*14 +5. Let me compute that: 7*14 is 98, plus 5 is 103. Therefore, n=103. Is that the answer? Wait, but let me check if this satisfies the original condition.

So, pages = 2n = 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. Therefore, 206 ≡3 mod7. Perfect. Also, n=103 is greater than 100. So, seems like 103 is the answer. But wait, let me check if there is a smaller n>100 that satisfies n≡5 mod7. Because maybe k=13 gives n=7*13 +5=91 +5=96, which is less than 100. So, k=14 gives 103. So, 103 is the first n>100. Therefore, 103 is the minimal number of leaves. So, the answer is 103?

Wait, but let me check again. Maybe I made a mistake in solving the congruence? Let me check again.

Original equation: 2n ≡3 mod7. Multiply both sides by inverse of 2 mod7, which is 4. So 4*2n ≡4*3 mod7 => 8n≡12 mod7. But 8≡1 mod7, so 1*n≡12 mod7. 12 mod7 is 5. Therefore, n≡5 mod7. So that's correct. So n=7k+5. So yes, the solutions are numbers congruent to 5 mod7. So the numbers are 5,12,19,... up to 103. So the first one above 100 is 103. Therefore, 103 is the answer. But let me check if there is a smaller n>100. Let me check n=103-7=96, which is less than 100. So next one is 103. So yes, 103 is the minimal. Therefore, the answer is 103. Wait, but let me check if there is a mistake in the problem statement interpretation.

Wait, 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, each leaf is a piece of paper, which has two pages. So, if there are n leaves, then there are 2n pages. So that's correct. So pages=2n. So the problem says pages is 3 more than a multiple of 7. So 2n ≡3 mod7. So solving that gives n≡5 mod7. So n=7k+5. So n>100. So the minimal n is 103. Therefore, answer is 103. But let me check with n=103. Let me check the number of pages: 206. 206 divided by 7 is 29*7=203, remainder 3. So yes, 206=7*29 +3. So that's correct. So 206 is 3 more than a multiple of 7. So that's correct. So seems like 103 is the answer. But let me check if there is a smaller n>100. For example, maybe n=103 is the first one. Let me check n=100. Let me check if n=100 is possible. For n=100, pages=200. 200 divided by 7 is 28*7=196, remainder 4. So 200≡4 mod7. Not 3. n=101: pages=202. 202-196=6, so 202≡6 mod7. Not 3. n=102: pages=204. 204-196=8, which is 1 mod7. 204≡1 mod7. Not 3. n=103: pages=206≡3 mod7. So yes, that's the first one. Therefore, 103 is the answer. Therefore, the smallest possible number of leaves is 103. So I think that's the answer. Wait, but let me check if there's another way to approach this problem. Maybe I made a mistake in the congruence?

Alternatively, maybe I can think of 2n ≡3 mod7. So 2n -3 is divisible by 7. So 2n -3 =7m, for some integer m. Then 2n=7m +3. Therefore, n=(7m +3)/2. Since n must be integer, 7m +3 must be even. Therefore, 7m must be odd, since 7m +3 is even. 7m is odd implies that m is odd. Because 7 is odd, odd times odd is odd, odd times even is even. Therefore, m must be odd. Let me write m=2k +1, where k is integer. Then substituting back:

n=(7*(2k +1) +3)/2=(14k +7 +3)/2=(14k +10)/2=7k +5. So same result as before. Therefore, n=7k +5. Therefore, same conclusion. Therefore, n must be 5,12,19,... So the first n>100 is 103. Therefore, answer is 103. Therefore, I think that's correct. So I think the answer is 103. Wait, but let me check if there's a possible mistake in the problem statement. For example, sometimes in books, the term "leaf" can refer to a single sheet, which has two pages, but sometimes people might count leaves differently? Wait, the problem says "a book is said to have n leaves if it is composed of n pieces of paper." So each leaf is a piece of paper. So each leaf has two pages. So pages=2n. So that's correct. So 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 that's consistent. So no problem here. Therefore, I think my answer is correct. Therefore, the smallest possible number of leaves 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** $ n $ in a book such that:

1. The **number of pages** is **twice the number of leaves**, i.e., $ \text{pages} = 2n $.
2. The **number of pages** is **3 more than a multiple of 7**, i.e., $ 2n \equiv 3 \pmod{7} $.
3. The **number of leaves** is **greater than 100**, i.e., $ n > 100 $.

---

### Step 1: Solve the Congruence

We are given:

$$
2n \equiv 3 \pmod{7}
$$

To solve for $ n $, we need the **multiplicative inverse** of 2 modulo 7. That is, a number $ x $ such that:

$$
2x \equiv 1 \pmod{7}
$$

Trying values, we find:

$$
2 \times 4 = 8 \equiv 1 \pmod{7}
$$

So, the inverse of 2 modulo 7 is 4. Multiply both sides of the original congruence by 4:

$$
4 \cdot 2n \equiv 4 \cdot 3 \pmod{7} \Rightarrow 8n \equiv 12 \pmod{7}
$$

Since $ 8 \equiv 1 \pmod{7} $, this simplifies to:

$$
n \equiv 12 \pmod{7}
$$

Now, reduce 12 modulo 7:

$$
12 \div 7 = 1 \text{ remainder } 5 \Rightarrow n \equiv 5 \pmod{7}
$$

So, the general solution is:

$$
n = 7k + 5 \quad \text{for some integer } k
$$

---

### Step 2: Find the Smallest $ n > 100 $

We now find the smallest integer $ k $ such that:

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

So the smallest integer $ k $ is 14. Substituting into the formula:

$$
n = 7 \cdot 14 + 5 = 98 + 5 = 103
$$

---

### Step 3: Verify the Solution

- Number of leaves: $ n = 103 $
- Number of pages: $ 2n = 206 $
- Check if $ 206 \equiv 3 \pmod{7} $:

$$
206 \div 7 = 29 \text{ remainder } 3 \Rightarrow 206 \equiv 3 \pmod{7}
$$

✅ This satisfies the condition.

Also, $ n = 103 > 100 $, so it meets 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 $5+7\cdot 14=\boxed{103}$.


Explanation:
Okay, so I need to compute the cross product (2b - a) × (3c + a). The problem gives me the cross products of a×b, a×c, and b×c. Let me recall some cross product properties to see how I can use these given vectors.

First, cross product is distributive over vector addition, right? So, maybe I can expand the expression (2b - a) × (3c + a) using distributive property. Let me write that out:

(2b - a) × (3c + a) = 2b × 3c + 2b × a - a × 3c - a × a.

Simplifying each term:

First term: 2b × 3c = 6(b × c)
Second term: 2b × a = 2(b × a)
Third term: -a × 3c = -3(a × c)
Fourth term: -a × a = - (a × a)

Now, I remember that the cross product of any vector with itself is zero, so a × a = 0. So the fourth term is zero. That simplifies things.

So now, the expression becomes:

6(b × c) + 2(b × a) - 3(a × c)

But wait, cross product is anti-commutative, meaning that b × a = -a × b. So let me rewrite the second term:

2(b × a) = -2(a × b)

So substituting back:

6(b × c) - 2(a × b) - 3(a × c)

Now, the problem gives me the values of a×b, a×c, and b×c. Let me note them down:

Given:
a × b = [6, -7, 3]^T
a × c = [4, 7, 2]^T
b × c = [1, -7, 18]^T

So substituting these into the expression:

First term: 6*(b × c) = 6*[1, -7, 18]^T = [6*1, 6*(-7), 6*18]^T = [6, -42, 108]^T

Second term: -2*(a × b) = -2*[6, -7, 3]^T = [-12, 14, -6]^T

Third term: -3*(a × c) = -3*[4, 7, 2]^T = [-12, -21, -6]^T

Now, I need to add these three vectors together:

First term: [6, -42, 108]
Second term: [-12, 14, -6]
Third term: [-12, -21, -6]

Let me add them component-wise.

For the first component:

6 + (-12) + (-12) = 6 - 12 -12 = 6 - 24 = -18

Second component:

-42 + 14 + (-21) = -42 + 14 -21 = (-42 -21) +14 = -63 +14 = -49

Third component:

108 + (-6) + (-6) = 108 -6 -6 = 108 -12 = 96

So putting it all together, the result is [-18, -49, 96]^T.

Wait, let me check my calculations again to make sure I didn't make any arithmetic errors.

First component:

6 (from first term) -12 (second term) -12 (third term). 6 -12 is -6, -6 -12 is -18. Correct.

Second component:

-42 (from first term) +14 (second term) -21 (third term). Let me compute step by step:

Start with -42 +14 = -28. Then -28 -21 = -49. Correct.

Third component:

108 (from first term) -6 (second term) -6 (third term). 108 -6 is 102, 102 -6 is 96. Correct.

So the final result is the vector (-18, -49, 96). Let me check if there's another way to approach this problem or if I missed any cross product properties.

Alternatively, maybe I can check if there's a different expansion or if I made a mistake in the sign during the expansion.

Original expansion:

(2b - a) × (3c + a) = 2b × 3c + 2b × a - a × 3c - a × a

Yes, that's correct. Then simplifying:

6(b × c) + 2(b × a) - 3(a × c) - 0

Then using anti-commutativity: b × a = -a × b, so 2(b × a) = -2(a × b). So substituting:

6(b × c) - 2(a × b) - 3(a × c). Correct.

Then substituting the given vectors:

Yes, that's exactly what I did. So substituting each term:

First term: 6*(b × c) = 6*[1, -7, 18] = [6, -42, 108]

Second term: -2*(a × b) = -2*[6, -7, 3] = [-12, 14, -6]

Third term: -3*(a × c) = -3*[4, 7, 2] = [-12, -21, -6]

Adding them up:

First component: 6 -12 -12 = -18

Second component: -42 +14 -21 = -49

Third component: 108 -6 -6 = 96

So the result is (-18, -49, 96). Let me check if the problem expects the answer in a specific format. The problem says "put your final answer within \boxed{}", so likely as a column vector. But since the given cross products are written as column vectors, I can present it similarly.

Alternatively, maybe I can check if there's a different approach. For example, if I can find vectors a, b, c, then compute the cross product directly. But since we don't have the actual vectors, only their cross products, that might not be straightforward. However, maybe there's a way to relate them?

Alternatively, perhaps I can check if my answer is consistent with some other properties. For example, if I compute the magnitude or something else? But since the problem is purely algebraic, and my steps seem correct, maybe it's okay.

Wait, let me check once again the cross product expansion. Let me recall that cross product is distributive over addition, so:

(2b - a) × (3c + a) = 2b × 3c + 2b × a - a × 3c - a × a. Yes, that's correct. Because cross product distributes over addition, so:

First, expand (2b - a) × (3c + a) = (2b - a) × 3c + (2b - a) × a = 2b × 3c - a × 3c + 2b × a - a × a. Which is same as before. So that's correct.

Alternatively, maybe I can think of the cross product as a linear operator? But I think my approach is correct.

Alternatively, maybe I can check if the answer is correct by considering some specific vectors. But since we don't have the actual vectors, that might be difficult. However, maybe I can check if the cross products given are consistent with each other?

For example, if I suppose that vectors a, b, c are in 3D space, then the cross products should satisfy certain relations. For example, the scalar triple product [a, b, c] = a ⋅ (b × c). But since we have a×b, a×c, b×c, maybe we can find some relations?

Alternatively, maybe not necessary here. Since my calculation seems consistent, and I double-checked the arithmetic, I think my answer is correct.

Therefore, the final answer is the vector (-18, -49, 96). So in boxed form, it would be \boxed{\begin{pmatrix} -18 \\ -49 \\ 96 \end{pmatrix}}.

**Final Answer**
\boxed{\begin{pmatrix} -18 \\ -49 \\ 96 \end{pmatrix}}

To compute the cross product
$$
(2\mathbf{b} - \mathbf{a}) \times (3\mathbf{c} + \mathbf{a}),
$$
we begin by applying the **distributive property** of the cross product over vector addition:

$$
(2\mathbf{b} - \mathbf{a}) \times (3\mathbf{c} + \mathbf{a}) = 2\mathbf{b} \times 3\mathbf{c} + 2\mathbf{b} \times \mathbf{a} - \mathbf{a} \times 3\mathbf{c} - \mathbf{a} \times \mathbf{a}.
$$

---

### Step 1: Simplify each term

- $ 2\mathbf{b} \times 3\mathbf{c} = 6(\mathbf{b} \times \mathbf{c}) $
- $ 2\mathbf{b} \times \mathbf{a} = 2(\mathbf{b} \times \mathbf{a}) = -2(\mathbf{a} \times \mathbf{b}) $ (using anti-commutativity: $ \mathbf{b} \times \mathbf{a} = -\mathbf{a} \times \mathbf{b} $)
- $ -\mathbf{a} \times 3\mathbf{c} = -3(\mathbf{a} \times \mathbf{c}) $
- $ -\mathbf{a} \times \mathbf{a} = 0 $ (since the cross product of any vector with itself is zero)

So the expression becomes:

$$
6(\mathbf{b} \times \mathbf{c}) - 2(\mathbf{a} \times \mathbf{b}) - 3(\mathbf{a} \times \mathbf{c}).
$$

---

### Step 2: Substitute known values

We are given:

- $ \mathbf{a} \times \mathbf{b} = \begin{pmatrix} 6 \\ -7 \\ 3 \end{pmatrix} $
- $ \mathbf{a} \times \mathbf{c} = \begin{pmatrix} 4 \\ 7 \\ 2 \end{pmatrix} $
- $ \mathbf{b} \times \mathbf{c} = \begin{pmatrix} 1 \\ -7 \\ 18 \end{pmatrix} $

Now compute each term:

- $ 6(\mathbf{b} \times \mathbf{c}) = 6 \begin{pmatrix} 1 \\ -7 \\ 18 \end{pmatrix} = \begin{pmatrix} 6 \\ -42 \\ 108 \end{pmatrix} $
- $ -2(\mathbf{a} \times \mathbf{b}) = -2 \begin{pmatrix} 6 \\ -7 \\ 3 \end{pmatrix} = \begin{pmatrix} -12 \\ 14 \\ -6 \end{pmatrix} $
- $ -3(\mathbf{a} \times \mathbf{c}) = -3 \begin{pmatrix} 4 \\ 7 \\ 2 \end{pmatrix} = \begin{pmatrix} -12 \\ -21 \\ -6 \end{pmatrix} $

---

### Step 3: Add the resulting vectors

Now sum the three vectors:

$$
\begin{pmatrix} 6 \\ -42 \\ 108 \end{pmatrix} + \begin{pmatrix} -12 \\ 14 \\ -6 \end{pmatrix} + \begin{pmatrix} -12 \\ -21 \\ -6 \end{pmatrix}
$$

Compute component-wise:

- First component: $ 6 - 12 - 12 = -18 $
- Second component: $ -42 + 14 - 21 = -49 $
- Third component: $ 108 - 6 - 6 = 96 $

---

### Final Answer

$$
\boxed{\begin{pmatrix} -18 \\ -49 \\ 96 \end{pmatrix}}
$$

[Assistant]
Explanation:
Okay, so I need to compute the cross product (2b - a) × (3c + a). The problem gives me the cross products of a×b, a×c, and b×c. Let me recall some cross product properties to see how I can use these given vectors.

First, cross product is distributive over vector addition, right? So, maybe I can expand the expression (2b - a) × (3c + a) using distributive property. Let me write that out:

(2b - a) × (3c + a) = 2b × 3c + 2b × a - a × 3c - a × a.

Simplifying each term:

First term: 2b × 3c = 6(b × c)
Second term: 2b × a = 2(b × a)
Third term: -a × 3c = -3(a × c)
Fourth term: -a × a = - (a × a)

Now, I remember that the cross product of any vector with itself is zero, so a × a = 0. So the fourth term is zero. That simplifies things.

So now, the expression becomes:

6(b × c) + 2(b × a) - 3(a × c)

But wait, cross product is anti-commutative, meaning that b × a = -a × b. So let me rewrite the second term:

2(b × a) = -2(a × b)

So substituting back:

6(b × c) - 2(a × b) - 3(a × c)

Now, the problem gives me the values of a×b, a×c, and b×c. Let me note them down:

Given:
a × b = [6, -7, 3]^T
a × c = [4, 7, 2]^T
b × c = [1, -7, 18]^T

So substituting these into the expression:

First term: 6*(b × c) = 6*[1, -7, 18]^T = [6*1, 6*(-7), 6*18]^T = [6, -42, 108]^T

Second term: -2*(a × b) = -2*[6, -7, 3]^T = [-12, 14, -6]^T

Third term: -3*(a × c) = -3*[4, 7, 2]^T = [-12, -21, -6]^T

Now, I need to add these three vectors together:

First term: [6, -42, 108]
Second term: [-12, 14, -6]
Third term: [-12, -21, -6]

Let me add them component-wise.

For the first component:

6 + (-12) + (-12) = 6 - 12 -12 = 6 - 24 = -18

Second component:

-42 + 14 + (-21) = -42 + 14 -21 = (-42 -21) +14 = -63 +14 = -49

Third component:

108 + (-6) + (-6) = 108 -6 -6 = 108 -12 = 96

So putting it all together, the result is [-18, -49, 96]^T.

Wait, let me check my calculations again to make sure I didn't make any arithmetic errors.

First component:

6 (from first term) -12 (second term) -12 (third term). 6 -12 is -6, -6 -12 is -18. Correct.

Second component:

-42 (from first term) +14 (second term) -21 (third term). Let me compute step by step:

Start with -42 +14 = -28. Then -28 -21 = -49. Correct.

Third component:

108 (from first term) -6 (second term) -6 (third term). 108 -6 is 102, 102 -6 is 96. Correct.

So the final result is the vector (-18, -49, 96). Let me check if there's another way to approach this problem or if I missed any cross product properties.

Alternatively, maybe I can check if there's a different expansion or if I made a mistake in the sign during the expansion.

Original expansion:

(2b - a) × (3c + a) = 2b × 3c + 2b × a - a × 3c - a × a

Yes, that's correct. Then simplifying:

6(b × c) + 2(b × a) - 3(a × c) - 0

Then using anti-commutativity: b × a = -a × b, so 2(b × a) = -2(a × b). So substituting:

6(b × c) - 2(a × b) - 3(a × c). Correct.

Then substituting the given vectors:

Yes, that's exactly what I did. So substituting each term:

First term: 6*(b × c) = 6*[1, -7, 18] = [6, -42, 108]

Second term: -2*(a × b) = -2*[6, -7, 3] = [-12, 14, -6]

Third term: -3*(a × c) = -3*[4, 7, 2] = [-12, -21, -6]

Adding them up:

First component: 6 -12 -12 = -18

Second component: -42 +14 -21 = -49

Third component: 108 -6 -6 = 96

So the result is (-18, -49, 96). Let me check if the problem expects the answer in a specific format. The problem says "put your final answer within \boxed{}", so likely as a column vector. But since the given cross products are written as column vectors, I can present it similarly.

Alternatively, maybe I can check if there's a different approach. For example, if I can find vectors a, b, c, then compute the cross product directly. But since we don't have the actual vectors, only their cross products, that might not be straightforward. However, maybe there's a way to relate them?

Alternatively, perhaps I can check if my answer is consistent with some other properties. For example, if I compute the magnitude or something else? But since the problem is purely algebraic, and my steps seem correct, maybe it's okay.

Wait, let me check once again the cross product expansion. Let me recall that cross product is distributive over addition, so:

(2b - a) × (3c + a) = 2b × 3c + 2b × a - a × 3c - a × a. Yes, that's correct. Because cross product distributes over addition, so:

First, expand (2b - a) × (3c + a) = (2b - a) × 3c + (2b - a) × a = 2b × 3c - a × 3c + 2b × a - a × a. Which is same as before. So that's correct.

Alternatively, maybe I can think of the cross product as a linear operator? But I think my approach is correct.

Alternatively, maybe I can check if the answer is correct by considering some specific vectors. But since we don't have the actual vectors, that might be difficult. However, maybe I can check if the cross products given are consistent with each other?

For example, if I suppose that vectors a, b, c are in 3D space, then the cross products should satisfy certain relations. For example, the scalar triple product [a, b, c] = a ⋅ (b × c). But since we have a×b, a×c, b×c, maybe we can find some relations?

Alternatively, maybe not necessary here. Since my calculation seems consistent, and I double-checked the arithmetic, I think my answer is correct.

Therefore, the final answer is the vector (-18, -49, 96). So in boxed form, it would be \boxed{\begin{pmatrix} -18 \\ -49 \\ 96 \end{pmatrix}}.

**Final Answer**
\boxed{\begin{pmatrix} -18 \\ -49 \\ 96 \end{pmatrix}}

To compute the cross product
$$
(2\mathbf{b} - \mathbf{a}) \times (3\mathbf{c} + \mathbf{a}),
$$
we begin by applying the **distributive property** of the cross product over vector addition:

$$
(2\mathbf{b} - \mathbf{a}) \times (3\mathbf{c} + \mathbf{a}) = 2\mathbf{b} \times 3\mathbf{c} + 2\mathbf{b} \times \mathbf{a} - \mathbf{a} \times 3\mathbf{c} - \mathbf{a} \times \mathbf{a}.
$$

---

### Step 1: Simplify each term

- $ 2\mathbf{b} \times 3\mathbf{c} = 6(\mathbf{b} \times \mathbf{c}) $
- $ 2\mathbf{b} \times \mathbf{a} = 2(\mathbf{b} \times \mathbf{a}) = -2(\mathbf{a} \times \mathbf{b}) $ (using anti-commutativity: $ \mathbf{b} \times \mathbf{a} = -\mathbf{a} \times \mathbf{b} $)
- $ -\mathbf{a} \times 3\mathbf{c} = -3(\mathbf{a} \times \mathbf{c}) $
- $ -\mathbf{a} \times \mathbf{a} = 0 $ (since the cross product of any vector with itself is zero)

So the expression becomes:

$$
6(\mathbf{b} \times \mathbf{c}) - 2(\mathbf{a} \times \mathbf{b}) - 3(\mathbf{a} \times \mathbf{c}).
$$

---

### Step 2: Substitute known values

We are given:

- $ \mathbf{a} \times \mathbf{b} = \begin{pmatrix} 6 \\ -7 \\ 3 \end{pmatrix} $
- $ \mathbf{a} \times \mathbf{c} = \begin{pmatrix} 4 \\ 7 \\ 2 \end{pmatrix} $
- $ \mathbf{b} \times \mathbf{c} = \begin{pmatrix} 1 \\ -7 \\ 18 \end{pmatrix} $

Now compute each term:

- $ 6(\mathbf{b} \times \mathbf{c}) = 6 \begin{pmatrix} 1 \\ -7 \\ 18 \end{pmatrix} = \begin{pmatrix} 6 \\ -42 \\ 108 \end{pmatrix} $
- $ -2(\mathbf{a} \times \mathbf{b}) = -2 \begin{pmatrix} 6 \\ -7 \\ 3 \end{pmatrix} = \begin{pmatrix} -12 \\ 14 \\ -6 \end{pmatrix} $
- $ -3(\mathbf{a} \times \mathbf{c}) = -3 \begin{pmatrix} 4 \\ 7 \\ 2 \end{pmatrix} = \begin{pmatrix} -12 \\ -21 \\ -6 \end{pmatrix} $

---

### Step 3: Add the resulting vectors

Now sum the three vectors:

$$
\begin{pmatrix} 6 \\ -42 \\ 108 \end{pmatrix} + \begin{pmatrix} -12 \\ 14 \\ -6 \end{pmatrix} + \begin{pmatrix} -12 \\ -21 \\ -6 \end{pmatrix}
$$

Compute component-wise:

- First component: $ 6 - 12 - 12 = -18 $
- Second component: $ -42 + 14 - 21 = -49 $
- Third component: $ 108 - 6 - 6 = 96 $

---

### Final Answer

$$
\boxed{\begin{pmatrix} -18 \\ -49 \\ 96 \end{pmatrix}}
$$

Solution:
Expanding, we get
\begin{align*}
(2 \mathbf{b} - \mathbf{a}) \times (3 \mathbf{c} + \mathbf{a}) &= 6 \mathbf{b} \times \mathbf{c} + 2 \mathbf{b} \times \mathbf{a} - 3 \mathbf{a} \times \mathbf{c} - \mathbf{a} \times \mathbf{a} \\
&= 6 \mathbf{b} \times \mathbf{c} - 2 \mathbf{a} \times \mathbf{b} - 3 \mathbf{a} \times \mathbf{c} - \mathbf{0} \\
&= 6 \begin{pmatrix} 1 \\ - 7 \\ 18 \end{pmatrix} - 2 \begin{pmatrix} 6 \\ - 7 \\ 3 \end{pmatrix} - 3 \begin{pmatrix} 4 \\ 7 \\ 2 \end{pmatrix} \\
&= \boxed{\begin{pmatrix} -18 \\ -49 \\ 96 \end{pmatrix}}.
\end{align*}


Explanation:
Okay, so I need to figure out how many integer coordinates are strictly inside this rectangle with vertices at (5,4), (5,-4), (-5,4), and (-5,-4). Let me start by visualizing the rectangle. The rectangle is centered at the origin, right? Because the x-coordinates go from -5 to 5 and the y-coordinates go from -4 to 4. So, it's a rectangle that's 10 units wide (from -5 to 5 on the x-axis) and 8 units tall (from -4 to 4 on the y-axis).

But the question is about integer coordinates strictly inside the rectangle. That means I need to count all the points (x, y) where x and y are integers, and the point is not on the boundary of the rectangle. So, the coordinates must satisfy -5 < x < 5 and -4 < y < 4. Wait, but since the rectangle's vertices are at those coordinates, the sides are at x = -5, x = 5, y = -4, and y = 4. So, strictly inside would mean x is between -5 and 5, not including the endpoints, and same for y.

But hold on, the problem says "strictly inside the rectangular region." So, the coordinates must be such that they are not on the edges. Therefore, for x, the integer values must be from -4 to 4 inclusive? Wait, let me think. If the rectangle spans from x = -5 to x = 5, then the integer coordinates strictly inside would have x-values that are greater than -5 and less than 5. Since x has to be an integer, the possible x-values are -4, -3, -2, -1, 0, 1, 2, 3, 4. Similarly for y-values, since the rectangle spans from y = -4 to y = 4, the strictly inside y-values would be -3, -2, -1, 0, 1, 2, 3.

Wait, let me check that again. If the rectangle includes the points where x is exactly -5 or 5, then strictly inside would exclude those. So, for x, the integer coordinates inside would be from -4 up to 4, inclusive. Similarly for y, from -3 up to 3 inclusive.

So, how many integer x-values are there? Let's calculate. From -4 to 4 inclusive. Let me count them. Starting at -4, -3, -2, -1, 0, 1, 2, 3, 4. That's 9 values. Similarly for y, from -3 to 3 inclusive: -3, -2, -1, 0, 1, 2, 3. That's 7 values.

Therefore, the total number of integer coordinates strictly inside the rectangle would be 9 * 7 = 63.

Wait, but let me make sure I didn't make a mistake here. Let me think again. The rectangle's width is from x=-5 to x=5. So, the length along the x-axis is 10 units. But the number of integer points strictly inside would be the number of integers between -5 and 5, not including the endpoints. So, how many integers are between -5 and 5 exclusive?

For integers between a and b exclusive, the count is b - a - 1. Wait, but here a is -5 and b is 5. So, 5 - (-5) - 1 = 5 + 5 -1 = 9. Which matches the previous count. Similarly for y, between -4 and 4 exclusive. So, 4 - (-4) -1 = 4 + 4 -1 = 7. So, 9*7=63.

But let me check with another approach. For example, if I consider the entire grid of integer coordinates inside the rectangle, including the boundary, then the number of points would be (number of x-values from -5 to 5 inclusive) times (number of y-values from -4 to 4 inclusive).

Number of x-values from -5 to 5 inclusive is 5 - (-5) +1 = 11. Similarly, number of y-values from -4 to 4 inclusive is 4 - (-4) +1 = 9. So total points including boundary is 11*9=99.

But then, if we want strictly inside, we need to subtract the boundary points. However, maybe that's a more complicated way. But let me try.

Alternatively, the number of interior points is equal to (number of interior x-values) * (number of interior y-values). Which is what I did before.

But let me check if that's correct. For example, if I have a rectangle from x=-a to x=a and y=-b to y=b, then the number of interior points would be (2a -1) * (2b -1). Wait, in this case, a=5 and b=4? Wait, no. Wait, if the rectangle spans from -5 to 5 on x, then the length is 10, so the number of interior x-values is 10 - 2 = 8? Wait, no. Wait, confusion here.

Wait, maybe I need to think in terms of grid points. For example, in the x-direction, from -5 to 5 inclusive, there are 11 points. If we want strictly inside, we remove the two endpoints, so 11 - 2 = 9. Similarly, for y-direction, from -4 to 4 inclusive, 9 points. Remove the two endpoints, so 9 - 2 = 7. Therefore, 9*7=63.

Alternatively, if I think of the rectangle as having width 10 and height 8, then the number of interior points would be (width - 2)*(height - 2). Wait, but that would be (10 - 2)*(8 - 2) = 8*6=48? Wait, that contradicts the previous answer. Wait, so which is correct?

Wait, this is confusing. Let me think again.

The formula (width - 2)*(height - 2) is for when the rectangle is aligned with the grid and the sides are integer coordinates. Wait, but actually, the number of interior points in a grid is (number of points along x - 2) * (number of points along y - 2). Wait, but how does that relate to the actual dimensions?

Wait, for example, if you have a rectangle that spans from x=0 to x=1, and y=0 to y=1, then the number of interior points would be 0, because the only integer points are (0,0), (0,1), (1,0), (1,1). But if you consider strictly inside, there are no points. However, if you use the formula (width - 2)*(height - 2), where width is 1 and height is 1, you get (1 - 2)*(1 - 2) = (-1)*(-1)=1, which is wrong. So that formula is not correct.

Therefore, my previous approach was better. Let me think again.

The number of integer points strictly inside the rectangle is equal to the number of integer points in the interior. For that, we need to find all integer coordinates (x, y) such that -5 < x < 5 and -4 < y < 4.

Since x must be an integer, the possible x-values are integers from -4 to 4 inclusive. Similarly, y-values are integers from -3 to 3 inclusive.

So, how many integers are there from -4 to 4 inclusive? Let's calculate. From -4 to 4, that's 4 - (-4) +1 = 9 numbers. Similarly, from -3 to 3 inclusive, that's 3 - (-3) +1 = 7 numbers. Therefore, total number of points is 9*7=63.

But let me check with another example. Suppose we have a rectangle from x=-1 to x=1, y=-1 to y=1. Then strictly inside would be x from -0 to 0? Wait, no. Wait, strictly inside would be x between -1 and 1, not including the endpoints. So x can be 0. Similarly y can be 0. So only one point (0,0).

Using the formula: number of x-values from -1 to 1 exclusive is 1 - (-1) -1 = 1 +1 -1=1. Similarly for y. So 1*1=1. Correct.

Alternatively, if we have rectangle from x=-2 to x=2, y=-2 to y=2. Then strictly inside would be x from -1, 0, 1 (3 values) and y from -1, 0, 1 (3 values). So total 9 points. Using formula: (2 - (-2) -1) * (2 - (-2) -1) = (4 -1)*(4 -1)= 3*3=9. Correct.

So in general, for a rectangle from x=-a to x=a and y=-b to y=b, the number of interior points is (2a -1)*(2b -1). Wait, in the first example, a=1, b=1: (2*1 -1)*(2*1 -1)=1*1=1. Correct. In the second example, a=2, b=2: (4-1)*(4-1)=9. Correct.

But in our problem, the rectangle is from x=-5 to x=5 and y=-4 to y=4. So, a=5, b=4. Therefore, number of interior points would be (2*5 -1)*(2*4 -1)= (10 -1)*(8 -1)=9*7=63. Which matches our previous result.

Therefore, the answer is 63.

But let me check again with another approach. Let me think of the entire grid.

The rectangle includes all points where x is between -5 and 5 inclusive, and y between -4 and 4 inclusive. So, total number of points is (5 - (-5) +1) * (4 - (-4) +1) = 11 * 9 = 99.

Now, the number of points on the boundary. How many points are on the boundary?

The boundary consists of four sides: left, right, top, bottom. However, we have to be careful not to double count the corners.

Number of points on the left side (x=-5): y ranges from -4 to 4 inclusive. That's 9 points. Similarly, right side (x=5): 9 points.

Top side (y=4): x ranges from -5 to 5 inclusive. That's 11 points. Similarly, bottom side (y=-4): 11 points.

But the four corners are counted twice here. Each corner is on two sides. So total boundary points: 9 + 9 + 11 + 11 - 4 = 9+9=18, 11+11=22, total 18+22=40, minus 4 corners counted twice: 40 - 4 = 36.

Therefore, total boundary points are 36.

Therefore, the number of interior points is total points minus boundary points: 99 - 36 = 63.

So that's another way to get 63.

Therefore, this confirms the answer is 63.

But let me check again the boundary points calculation.

Left side: x=-5, y from -4 to 4 inclusive. That's 9 points. Similarly right side: 9 points.

Top side: y=4, x from -5 to 5 inclusive. That's 11 points. Similarly bottom side: 11 points.

But the four corners: (-5,4), (-5,-4), (5,4), (5,-4). Each corner is counted in two sides. So total overcount is 4 points.

Therefore, total boundary points: 9 + 9 + 11 + 11 - 4 = 36.

Total points: 99.

Interior points: 99 - 36 = 63.

Yes, that seems correct.

Alternatively, if I think of the interior points as those not on the boundary. So, for x, the interior points are those not equal to -5 or 5. Similarly for y.

So, for x: total x-values from -5 to 5 inclusive is 11. Subtract the two boundary points (-5 and 5), so 11 - 2 = 9.

For y: total y-values from -4 to 4 inclusive is 9. Subtract the two boundary points (-4 and 4), so 9 - 2 = 7.

Therefore, interior points: 9 * 7 = 63.

Same result.

Therefore, I think 63 is the correct answer.

But just to be thorough, let me think of a smaller rectangle where I can count manually.

Take a rectangle with vertices at (1,1), (1,-1), (-1,1), (-1,-1). So, this is a square from -1 to 1 on both axes.

Total points including boundary: (1 - (-1) +1) * (1 - (-1) +1) = 3*3=9.

Boundary points: left side (x=-1): y from -1 to 1: 3 points. Right side (x=1): 3 points. Top side (y=1): x from -1 to 1: 3 points. Bottom side (y=-1): 3 points. Total: 3+3+3+3=12. Subtract 4 corners counted twice: 12 - 4 = 8.

Interior points: total points 9 - boundary points 8 = 1. Which is correct, since the only interior point is (0,0).

Alternatively, using the formula: (2*1 -1)*(2*1 -1)=1*1=1. Correct.

Another example: rectangle from (-2,-2) to (2,2). Total points: (2 - (-2) +1)*(2 - (-2)+1)=5*5=25.

Boundary points: left side (x=-2): y from -2 to 2: 5 points. Right side (x=2): 5 points. Top side (y=2): x from -2 to 2: 5 points. Bottom side (y=-2): 5 points. Total: 5*4=20. Subtract 4 corners: 20 - 4=16.

Interior points: 25 - 16=9. Which is (2*2 -1)*(2*2 -1)=3*3=9. Correct.

So, the formula works.

Therefore, returning to the original problem, the answer is 63.

But just to make sure, let me think of another way.

For the x-coordinate: from -5 to 5, exclusive. So, integers from -4 to 4. How many numbers is that? Let's count them.

Starting from -4: -4, -3, -2, -1, 0, 1, 2, 3, 4. That's 9 numbers.

Similarly for y: from -3 to 3. That's -3, -2, -1, 0, 1, 2, 3. 7 numbers.

So, 9*7=63.

Yes.

Alternatively, if I think of the number of integer points in the rectangle, it's (number of x-values) * (number of y-values).

But since we want strictly inside, we need to exclude the boundaries.

Therefore, the answer is 63.

I think that's solid.

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

To determine how many **integer coordinates** lie **strictly inside** the rectangle with vertices at $(5,4)$, $(5,-4)$, $(-5,4)$, and $(-5,-4)$, we begin by analyzing the region defined by these points.

---

### Step 1: Understand the Rectangle's Boundaries

The rectangle is centered at the origin and spans:

- **Horizontally** from $x = -5$ to $x = 5$
- **Vertically** from $y = -4$ to $y = 4$

This means the rectangle includes all points $(x, y)$ such that:
$$
-5 \leq x \leq 5 \quad \text{and} \quad -4 \leq y \leq 4
$$

However, the question asks for **strictly inside** the rectangle, which means we must **exclude** the boundary points. That is, we want:
$$
-5 < x < 5 \quad \text{and} \quad -4 < y < 4
$$

---

### Step 2: Count the Valid Integer Coordinates

We now count the integer values of $x$ and $y$ that satisfy the above inequalities.

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

- **For $y$:** 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

Each valid $x$-value can be paired with each valid $y$-value to form a unique point inside the rectangle. Therefore, the total number of such points is:
$$
9 \times 7 = 63
$$

---

### Final Answer

$$
\boxed{63}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many integer coordinates are strictly inside this rectangle with vertices at (5,4), (5,-4), (-5,4), and (-5,-4). Let me start by visualizing the rectangle. The rectangle is centered at the origin, right? Because the x-coordinates go from -5 to 5 and the y-coordinates go from -4 to 4. So, it's a rectangle that's 10 units wide (from -5 to 5 on the x-axis) and 8 units tall (from -4 to 4 on the y-axis).

But the question is about integer coordinates strictly inside the rectangle. That means I need to count all the points (x, y) where x and y are integers, and the point is not on the boundary of the rectangle. So, the coordinates must satisfy -5 < x < 5 and -4 < y < 4. Wait, but since the rectangle's vertices are at those coordinates, the sides are at x = -5, x = 5, y = -4, and y = 4. So, strictly inside would mean x is between -5 and 5, not including the endpoints, and same for y.

But hold on, the problem says "strictly inside the rectangular region." So, the coordinates must be such that they are not on the edges. Therefore, for x, the integer values must be from -4 to 4 inclusive? Wait, let me think. If the rectangle spans from x = -5 to x = 5, then the integer coordinates strictly inside would have x-values that are greater than -5 and less than 5. Since x has to be an integer, the possible x-values are -4, -3, -2, -1, 0, 1, 2, 3, 4. Similarly for y-values, since the rectangle spans from y = -4 to y = 4, the strictly inside y-values would be -3, -2, -1, 0, 1, 2, 3.

Wait, let me check that again. If the rectangle includes the points where x is exactly -5 or 5, then strictly inside would exclude those. So, for x, the integer coordinates inside would be from -4 up to 4, inclusive. Similarly for y, from -3 up to 3 inclusive.

So, how many integer x-values are there? Let's calculate. From -4 to 4 inclusive. Let me count them. Starting at -4, -3, -2, -1, 0, 1, 2, 3, 4. That's 9 values. Similarly for y, from -3 to 3 inclusive: -3, -2, -1, 0, 1, 2, 3. That's 7 values.

Therefore, the total number of integer coordinates strictly inside the rectangle would be 9 * 7 = 63.

Wait, but let me make sure I didn't make a mistake here. Let me think again. The rectangle's width is from x=-5 to x=5. So, the length along the x-axis is 10 units. But the number of integer points strictly inside would be the number of integers between -5 and 5, not including the endpoints. So, how many integers are between -5 and 5 exclusive?

For integers between a and b exclusive, the count is b - a - 1. Wait, but here a is -5 and b is 5. So, 5 - (-5) - 1 = 5 + 5 -1 = 9. Which matches the previous count. Similarly for y, between -4 and 4 exclusive. So, 4 - (-4) -1 = 4 + 4 -1 = 7. So, 9*7=63.

But let me check with another approach. For example, if I consider the entire grid of integer coordinates inside the rectangle, including the boundary, then the number of points would be (number of x-values from -5 to 5 inclusive) times (number of y-values from -4 to 4 inclusive).

Number of x-values from -5 to 5 inclusive is 5 - (-5) +1 = 11. Similarly, number of y-values from -4 to 4 inclusive is 4 - (-4) +1 = 9. So total points including boundary is 11*9=99.

But then, if we want strictly inside, we need to subtract the boundary points. However, maybe that's a more complicated way. But let me try.

Alternatively, the number of interior points is equal to (number of interior x-values) * (number of interior y-values). Which is what I did before.

But let me check if that's correct. For example, if I have a rectangle from x=-a to x=a and y=-b to y=b, then the number of interior points would be (2a -1) * (2b -1). Wait, in this case, a=5 and b=4? Wait, no. Wait, if the rectangle spans from -5 to 5 on x, then the length is 10, so the number of interior x-values is 10 - 2 = 8? Wait, no. Wait, confusion here.

Wait, maybe I need to think in terms of grid points. For example, in the x-direction, from -5 to 5 inclusive, there are 11 points. If we want strictly inside, we remove the two endpoints, so 11 - 2 = 9. Similarly, for y-direction, from -4 to 4 inclusive, 9 points. Remove the two endpoints, so 9 - 2 = 7. Therefore, 9*7=63.

Alternatively, if I think of the rectangle as having width 10 and height 8, then the number of interior points would be (width - 2)*(height - 2). Wait, but that would be (10 - 2)*(8 - 2) = 8*6=48? Wait, that contradicts the previous answer. Wait, so which is correct?

Wait, this is confusing. Let me think again.

The formula (width - 2)*(height - 2) is for when the rectangle is aligned with the grid and the sides are integer coordinates. Wait, but actually, the number of interior points in a grid is (number of points along x - 2) * (number of points along y - 2). Wait, but how does that relate to the actual dimensions?

Wait, for example, if you have a rectangle that spans from x=0 to x=1, and y=0 to y=1, then the number of interior points would be 0, because the only integer points are (0,0), (0,1), (1,0), (1,1). But if you consider strictly inside, there are no points. However, if you use the formula (width - 2)*(height - 2), where width is 1 and height is 1, you get (1 - 2)*(1 - 2) = (-1)*(-1)=1, which is wrong. So that formula is not correct.

Therefore, my previous approach was better. Let me think again.

The number of integer points strictly inside the rectangle is equal to the number of integer points in the interior. For that, we need to find all integer coordinates (x, y) such that -5 < x < 5 and -4 < y < 4.

Since x must be an integer, the possible x-values are integers from -4 to 4 inclusive. Similarly, y-values are integers from -3 to 3 inclusive.

So, how many integers are there from -4 to 4 inclusive? Let's calculate. From -4 to 4, that's 4 - (-4) +1 = 9 numbers. Similarly, from -3 to 3 inclusive, that's 3 - (-3) +1 = 7 numbers. Therefore, total number of points is 9*7=63.

But let me check with another example. Suppose we have a rectangle from x=-1 to x=1, y=-1 to y=1. Then strictly inside would be x from -0 to 0? Wait, no. Wait, strictly inside would be x between -1 and 1, not including the endpoints. So x can be 0. Similarly y can be 0. So only one point (0,0).

Using the formula: number of x-values from -1 to 1 exclusive is 1 - (-1) -1 = 1 +1 -1=1. Similarly for y. So 1*1=1. Correct.

Alternatively, if we have rectangle from x=-2 to x=2, y=-2 to y=2. Then strictly inside would be x from -1, 0, 1 (3 values) and y from -1, 0, 1 (3 values). So total 9 points. Using formula: (2 - (-2) -1) * (2 - (-2) -1) = (4 -1)*(4 -1)= 3*3=9. Correct.

So in general, for a rectangle from x=-a to x=a and y=-b to y=b, the number of interior points is (2a -1)*(2b -1). Wait, in the first example, a=1, b=1: (2*1 -1)*(2*1 -1)=1*1=1. Correct. In the second example, a=2, b=2: (4-1)*(4-1)=9. Correct.

But in our problem, the rectangle is from x=-5 to x=5 and y=-4 to y=4. So, a=5, b=4. Therefore, number of interior points would be (2*5 -1)*(2*4 -1)= (10 -1)*(8 -1)=9*7=63. Which matches our previous result.

Therefore, the answer is 63.

But let me check again with another approach. Let me think of the entire grid.

The rectangle includes all points where x is between -5 and 5 inclusive, and y between -4 and 4 inclusive. So, total number of points is (5 - (-5) +1) * (4 - (-4) +1) = 11 * 9 = 99.

Now, the number of points on the boundary. How many points are on the boundary?

The boundary consists of four sides: left, right, top, bottom. However, we have to be careful not to double count the corners.

Number of points on the left side (x=-5): y ranges from -4 to 4 inclusive. That's 9 points. Similarly, right side (x=5): 9 points.

Top side (y=4): x ranges from -5 to 5 inclusive. That's 11 points. Similarly, bottom side (y=-4): 11 points.

But the four corners are counted twice here. Each corner is on two sides. So total boundary points: 9 + 9 + 11 + 11 - 4 = 9+9=18, 11+11=22, total 18+22=40, minus 4 corners counted twice: 40 - 4 = 36.

Therefore, total boundary points are 36.

Therefore, the number of interior points is total points minus boundary points: 99 - 36 = 63.

So that's another way to get 63.

Therefore, this confirms the answer is 63.

But let me check again the boundary points calculation.

Left side: x=-5, y from -4 to 4 inclusive. That's 9 points. Similarly right side: 9 points.

Top side: y=4, x from -5 to 5 inclusive. That's 11 points. Similarly bottom side: 11 points.

But the four corners: (-5,4), (-5,-4), (5,4), (5,-4). Each corner is counted in two sides. So total overcount is 4 points.

Therefore, total boundary points: 9 + 9 + 11 + 11 - 4 = 36.

Total points: 99.

Interior points: 99 - 36 = 63.

Yes, that seems correct.

Alternatively, if I think of the interior points as those not on the boundary. So, for x, the interior points are those not equal to -5 or 5. Similarly for y.

So, for x: total x-values from -5 to 5 inclusive is 11. Subtract the two boundary points (-5 and 5), so 11 - 2 = 9.

For y: total y-values from -4 to 4 inclusive is 9. Subtract the two boundary points (-4 and 4), so 9 - 2 = 7.

Therefore, interior points: 9 * 7 = 63.

Same result.

Therefore, I think 63 is the correct answer.

But just to be thorough, let me think of a smaller rectangle where I can count manually.

Take a rectangle with vertices at (1,1), (1,-1), (-1,1), (-1,-1). So, this is a square from -1 to 1 on both axes.

Total points including boundary: (1 - (-1) +1) * (1 - (-1) +1) = 3*3=9.

Boundary points: left side (x=-1): y from -1 to 1: 3 points. Right side (x=1): 3 points. Top side (y=1): x from -1 to 1: 3 points. Bottom side (y=-1): 3 points. Total: 3+3+3+3=12. Subtract 4 corners counted twice: 12 - 4 = 8.

Interior points: total points 9 - boundary points 8 = 1. Which is correct, since the only interior point is (0,0).

Alternatively, using the formula: (2*1 -1)*(2*1 -1)=1*1=1. Correct.

Another example: rectangle from (-2,-2) to (2,2). Total points: (2 - (-2) +1)*(2 - (-2)+1)=5*5=25.

Boundary points: left side (x=-2): y from -2 to 2: 5 points. Right side (x=2): 5 points. Top side (y=2): x from -2 to 2: 5 points. Bottom side (y=-2): 5 points. Total: 5*4=20. Subtract 4 corners: 20 - 4=16.

Interior points: 25 - 16=9. Which is (2*2 -1)*(2*2 -1)=3*3=9. Correct.

So, the formula works.

Therefore, returning to the original problem, the answer is 63.

But just to make sure, let me think of another way.

For the x-coordinate: from -5 to 5, exclusive. So, integers from -4 to 4. How many numbers is that? Let's count them.

Starting from -4: -4, -3, -2, -1, 0, 1, 2, 3, 4. That's 9 numbers.

Similarly for y: from -3 to 3. That's -3, -2, -1, 0, 1, 2, 3. 7 numbers.

So, 9*7=63.

Yes.

Alternatively, if I think of the number of integer points in the rectangle, it's (number of x-values) * (number of y-values).

But since we want strictly inside, we need to exclude the boundaries.

Therefore, the answer is 63.

I think that's solid.

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

To determine how many **integer coordinates** lie **strictly inside** the rectangle with vertices at $(5,4)$, $(5,-4)$, $(-5,4)$, and $(-5,-4)$, we begin by analyzing the region defined by these points.

---

### Step 1: Understand the Rectangle's Boundaries

The rectangle is centered at the origin and spans:

- **Horizontally** from $x = -5$ to $x = 5$
- **Vertically** from $y = -4$ to $y = 4$

This means the rectangle includes all points $(x, y)$ such that:
$$
-5 \leq x \leq 5 \quad \text{and} \quad -4 \leq y \leq 4
$$

However, the question asks for **strictly inside** the rectangle, which means we must **exclude** the boundary points. That is, we want:
$$
-5 < x < 5 \quad \text{and} \quad -4 < y < 4
$$

---

### Step 2: Count the Valid Integer Coordinates

We now count the integer values of $x$ and $y$ that satisfy the above inequalities.

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

- **For $y$:** 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

Each valid $x$-value can be paired with each valid $y$-value to form a unique point inside the rectangle. Therefore, the total number of such points is:
$$
9 \times 7 = 63
$$

---

### Final Answer

$$
\boxed{63}
$$

Solution: Points with integer coordinates are called lattice points. The length of the rectangle is $5 - (-5) = 10$ units. There will be 9 lattice positions between the two vertical sides of the rectangle. The height of the rectangle is $4 - (-4) = 8$ units. There will be 7 lattice positions between the top and bottom of the rectangle. That's a total of $9 \times 7 = \boxed{63}$ lattice points.