Explanation:
Okay, so I need to find the probability that a randomly selected integer from the set {1, 2, 3, ..., 100} is divisible by 2 and not divisible by 3. Hmm, let me think about how to approach this.

First, probability is generally the number of favorable outcomes divided by the total number of possible outcomes. In this case, the total number of possible outcomes is 100 since there are 100 numbers in the set. So, the denominator of my probability will be 100.

Now, the numerator is the number of integers between 1 and 100 that are divisible by 2 but not by 3. Let me break this down. I need to find numbers that are divisible by 2, which means they are even numbers. But then, from those even numbers, I have to exclude the ones that are also divisible by 3. So, essentially, I need to find the count of numbers divisible by 2, subtract the count of numbers divisible by both 2 and 3.

Wait, let me make sure. If I take numbers divisible by 2, and then remove those that are also divisible by 3, that should give me numbers divisible by 2 but not by 3. Yes, that makes sense. So, how do I calculate that?

First, let me find how many numbers between 1 and 100 are divisible by 2. Well, numbers divisible by 2 are the even numbers. The number of even numbers from 1 to 100 is 100 divided by 2, which is 50. So, there are 50 numbers divisible by 2.

Now, from these 50 numbers, I need to subtract those that are also divisible by 3. So, how many numbers are divisible by both 2 and 3? Well, numbers divisible by both 2 and 3 are numbers divisible by the least common multiple (LCM) of 2 and 3. The LCM of 2 and 3 is 6. Therefore, numbers divisible by 6.

So, how many numbers between 1 and 100 are divisible by 6? That would be 100 divided by 6, but since we can't have a fraction of a number, we need to take the floor of that value. Let me calculate 100 divided by 6. 6 times 16 is 96, and 6 times 17 is 102, which is over 100. So, there are 16 numbers divisible by 6 between 1 and 100.

Wait, let me check that again. If I divide 100 by 6, it's approximately 16.666... So, the integer division would give 16. So, yes, 16 numbers. Therefore, there are 16 numbers divisible by both 2 and 3.

Therefore, the number of numbers divisible by 2 but not by 3 is 50 (divisible by 2) minus 16 (divisible by both 2 and 3), which is 50 - 16 = 34. Therefore, there are 34 numbers that satisfy the condition.

Therefore, the probability is 34 divided by 100. But the question says to express the answer as a common fraction. So, 34/100 can be simplified. Let me check if 34 and 100 have a common factor. Both are divisible by 2. Dividing numerator and denominator by 2 gives 17/50. Is 17 a prime number? Yes, 17 is a prime number, so 17 and 50 have no common factors besides 1. Therefore, the simplified fraction is 17/50.

Wait, let me verify my steps again to make sure I didn't make a mistake. First, total numbers: 100. Numbers divisible by 2: 50. Numbers divisible by 6: floor(100/6) = 16. So, numbers divisible by 2 but not 3: 50 - 16 = 34. Then 34/100 reduces to 17/50. That seems correct.

Alternatively, maybe I can approach this problem using another method to cross-verify. Let me think. The numbers divisible by 2 but not by 3 can be calculated as numbers divisible by 2 minus numbers divisible by both 2 and 3, which is exactly what I did. Alternatively, maybe using inclusion-exclusion principle?

Alternatively, think of numbers divisible by 2 but not by 3. So, numbers that are divisible by 2 but not by 3. So, another way is to consider numbers divisible by 2, which are 50 numbers. Now, among these 50 numbers, how many are divisible by 3? Well, since numbers divisible by both 2 and 3 are numbers divisible by 6, which we already found as 16. Therefore, same result. So, 50 - 16 = 34. So, same answer.

Alternatively, maybe I can think of the numbers not divisible by 3. Wait, but that might complicate. Let me try. The total numbers not divisible by 3: total numbers minus numbers divisible by 3. Total numbers is 100. Numbers divisible by 3: floor(100/3) = 33 (since 3*33=99). Therefore, numbers not divisible by 3: 100 - 33 = 67. But how many of these are divisible by 2? Wait, that's a different approach. So, if I want numbers divisible by 2 and not by 3, maybe I can find numbers divisible by 2 and not by 3 by considering numbers not divisible by 3 and then taking those that are divisible by 2? But that might not be straightforward. Wait, but maybe using some formula?

Alternatively, maybe using the principle of inclusion-exclusion. Let me think. The count of numbers divisible by 2 or 3 is equal to numbers divisible by 2 plus numbers divisible by 3 minus numbers divisible by both 2 and 3. But I don't know if that helps here. Wait, but maybe not. Let me see.

Alternatively, maybe think of the numbers divisible by 2 but not by 3 as numbers divisible by 2 and not divisible by 3. So, another way is to find numbers divisible by 2 and co-prime with 3? Wait, not exactly. Because numbers divisible by 2 but not by 3 are numbers divisible by 2 and not divisible by 3. So, perhaps another way is to think of numbers divisible by 2, and then subtract those divisible by 3. Which is again the same as before.

Alternatively, maybe think of the numbers divisible by 2 but not by 3 as numbers divisible by 2 and not divisible by 3. So, if I can find the count of numbers divisible by 2 and not by 3, which is equal to the count of numbers divisible by 2 minus the count of numbers divisible by both 2 and 3. Which is again the same as before.

So, seems like my original approach is correct. Therefore, 34 numbers. Therefore, probability is 34/100 = 17/50. Let me check with another approach.

Let me think of the numbers from 1 to 100. Let me think of the numbers divisible by 2: 2, 4, 6, 8, ..., 100. So, that's 50 numbers. Now, among these numbers, which are divisible by 3? Well, numbers divisible by both 2 and 3 are numbers divisible by 6. So, numbers divisible by 6 in the list of even numbers. So, numbers divisible by 6 are 6, 12, 18, ..., up to 96. Let me count how many terms are there in this sequence. The first term is 6, the last term is 96. The common difference is 6. So, the number of terms is ((last term - first term)/common difference) + 1. So, ((96 - 6)/6) + 1 = (90/6) + 1 = 15 + 1 = 16. So, 16 numbers. Therefore, same as before. Therefore, subtracting 16 from 50 gives 34. Therefore, 34 numbers. So, probability is 34/100 = 17/50. So, that seems correct.

Alternatively, maybe I can think of the numbers divisible by 2 but not by 3 as numbers divisible by 2 and not divisible by 3. So, another way is to think of numbers divisible by 2 and not divisible by 3. So, for each number divisible by 2, check if it's not divisible by 3. So, for example, take the number 2: 2 is not divisible by 3. 4: same. 6: divisible by 3. 8: not. 10: not. 12: divisible by 3. So, every third even number is divisible by 3. Wait, since every multiple of 6 is divisible by both 2 and 3. So, in the list of even numbers, every third number is divisible by 3. Let me check. Let's take the even numbers: 2, 4, 6, 8, 10, 12, 14, 16, 18,... So, starting from 2, every third number is 6, 12, 18,... So, yes, every third even number is divisible by 3. Therefore, the number of even numbers not divisible by 3 would be total even numbers minus every third even number. So, total even numbers is 50. How many are divisible by 3? Well, since every third even number is divisible by 3, then the number is 50 divided by 3, rounded down? Wait, but earlier we found that it's 16. Let me check. If there are 50 even numbers, and every third one is divisible by 3, then the number of such numbers would be floor(50/3). Let me calculate 50 divided by 3. 3*16=48, so floor(50/3)=16. So, that gives 16 numbers. Therefore, 50 - 16 = 34. So, same result. Therefore, that seems to confirm again that the number is 34. Therefore, probability is 34/100 = 17/50.

Alternatively, maybe think of the probability as follows. The probability that a number is divisible by 2 is 1/2. The probability that a number is not divisible by 3 is 2/3 (since 1/3 of numbers are divisible by 3). But wait, are these events independent? Wait, if they are independent, then the probability that a number is divisible by 2 and not divisible by 3 would be (1/2)*(2/3) = 1/3. But wait, that's different from 17/50. Wait, 1/3 is approximately 0.333..., and 17/50 is 0.34. So, they are close but not the same. Therefore, this suggests that the events are not independent. Therefore, my previous approach is more accurate. So, why is there a discrepancy?

Because the divisibility by 2 and divisibility by 3 are not independent events. So, the probability that a number is divisible by 2 and not divisible by 3 is not simply the product of the probabilities. Therefore, my initial approach of counting is more accurate. Therefore, 17/50 is the correct answer.

But just to check, let me compute 17/50. 17 divided by 50 is 0.34. And 1/3 is approximately 0.333..., so they are close but different. Therefore, the answer is 17/50.

Alternatively, let me think of the total numbers. Let me think of numbers from 1 to 100. Let me think of how many numbers are divisible by 2 but not by 3. Let me think of the numbers divisible by 2: 50 numbers. Now, numbers divisible by 2 and 3 are numbers divisible by 6, which is 16 numbers. Therefore, numbers divisible by 2 but not by 3 is 50 - 16 = 34. Therefore, 34/100 = 17/50. So, that's the answer.

Alternatively, maybe I can think of the numbers not divisible by 3. There are 100 - 33 = 67 numbers not divisible by 3. Now, how many of these are divisible by 2? Well, among the numbers not divisible by 3, half of them are even? Wait, not necessarily. Because numbers not divisible by 3 can be even or odd. So, maybe the count is different. Let me think. Let me try to calculate it.

Total numbers not divisible by 3: 67. How many of these are even? Let me think. The numbers not divisible by 3 can be divided into even and odd numbers. Let me think of the numbers from 1 to 100. Let me think of numbers not divisible by 3. For each number not divisible by 3, it can be either even or odd. Let me think of how many even numbers are there in the numbers not divisible by 3.

Alternatively, maybe use inclusion-exclusion again. The number of numbers divisible by 2 and not by 3 is equal to total numbers divisible by 2 minus numbers divisible by both 2 and 3, which is 50 - 16 = 34. So, same as before. Alternatively, if I want to compute it as numbers not divisible by 3 and divisible by 2, which is same as numbers divisible by 2 and not by 3. So, maybe another way is to compute numbers divisible by 2 and not by 3 as numbers divisible by 2 minus numbers divisible by 6. Which is again same as before.

Alternatively, maybe think of the numbers not divisible by 3. There are 67 numbers. How many of them are even? Let me think. Let me think of numbers not divisible by 3. Let me think of numbers from 1 to 100. Let me think of numbers not divisible by 3. For each residue class modulo 6. Wait, maybe that's overcomplicating. Alternatively, think that numbers not divisible by 3 are numbers congruent to 1, 2, 4, 5 mod 6. So, in each block of 6 numbers, there are 4 numbers not divisible by 3. Out of these, how many are even? Let me check. In numbers 1-6: numbers not divisible by 3 are 1, 2, 4, 5. Among these, even numbers are 2 and 4. So, 2 out of 4. Therefore, in each block of 6 numbers, there are 2 even numbers not divisible by 3. Therefore, in 100 numbers, how many blocks of 6 are there? 100 divided by 6 is 16 with a remainder of 4. So, 16 full blocks and 4 extra numbers.

In each full block, there are 2 even numbers not divisible by 3. Therefore, 16 blocks contribute 16*2 = 32 numbers. Then, the remaining 4 numbers are 97, 98, 99, 100. Let me check which of these are not divisible by 3. 97: 9+7=16, not divisible by 3. 98: 9+8=17, not divisible by 3. 99: 9+9=18, divisible by 3. 100: 1+0+0=1, not divisible by 3. Therefore, among the remaining 4 numbers, the numbers not divisible by 3 are 97, 98, 100. So, three numbers. Out of these, how many are even? 97 is odd, 98 is even, 100 is even. Therefore, 2 even numbers. Therefore, total even numbers not divisible by 3 is 32 + 2 = 34. Therefore, same result. Therefore, 34 numbers. Therefore, probability is 34/100 = 17/50. So, that's another way to get the same answer. Therefore, this confirms that the answer is indeed 17/50.

Therefore, I think my answer is correct. So, the probability is 17/50.

**Final Answer**
The probability is \boxed{\dfrac{17}{50}}.

To determine the probability that a randomly selected integer from the set $\{1, 2, 3, \ldots, 100\}$ is **divisible by 2 and not divisible by 3**, we proceed step by step using principles of counting and probability.

---

### Step 1: Total Number of Outcomes

The total number of integers in the set is:

$$
100
$$

This will be the denominator of our probability.

---

### Step 2: Count of Numbers Divisible by 2

Numbers divisible by 2 are the even numbers in the set. Since every second number is even, the count is:

$$
\left\lfloor \frac{100}{2} \right\rfloor = 50
$$

So, there are **50** numbers divisible by 2.

---

### Step 3: Count of Numbers Divisible by Both 2 and 3

A number divisible by both 2 and 3 is divisible by their **least common multiple (LCM)**, which is:

$$
\text{LCM}(2, 3) = 6
$$

So, we count how many numbers between 1 and 100 are divisible by 6:

$$
\left\lfloor \frac{100}{6} \right\rfloor = 16
$$

Thus, there are **16** numbers divisible by both 2 and 3.

---

### Step 4: Count of Numbers Divisible by 2 but Not by 3

To find the count of numbers divisible by 2 **but not** by 3, we subtract the count of numbers divisible by both 2 and 3 from the count of numbers divisible by 2:

$$
50 - 16 = 34
$$

So, there are **34** numbers that are divisible by 2 but not by 3.

---

### Step 5: Compute the Probability

The probability is the ratio of favorable outcomes (numbers divisible by 2 but not by 3) to the total number of outcomes:

$$
\text{Probability} = \frac{34}{100}
$$

Simplify the fraction:

$$
\frac{34}{100} = \frac{17}{50}
$$

---

### Final Answer

$$
\boxed{\dfrac{17}{50}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that a randomly selected integer from the set {1, 2, 3, ..., 100} is divisible by 2 and not divisible by 3. Hmm, let me think about how to approach this.

First, probability is generally the number of favorable outcomes divided by the total number of possible outcomes. In this case, the total number of possible outcomes is 100 since there are 100 numbers in the set. So, the denominator of my probability will be 100.

Now, the numerator is the number of integers between 1 and 100 that are divisible by 2 but not by 3. Let me break this down. I need to find numbers that are divisible by 2, which means they are even numbers. But then, from those even numbers, I have to exclude the ones that are also divisible by 3. So, essentially, I need to find the count of numbers divisible by 2, subtract the count of numbers divisible by both 2 and 3.

Wait, let me make sure. If I take numbers divisible by 2, and then remove those that are also divisible by 3, that should give me numbers divisible by 2 but not by 3. Yes, that makes sense. So, how do I calculate that?

First, let me find how many numbers between 1 and 100 are divisible by 2. Well, numbers divisible by 2 are the even numbers. The number of even numbers from 1 to 100 is 100 divided by 2, which is 50. So, there are 50 numbers divisible by 2.

Now, from these 50 numbers, I need to subtract those that are also divisible by 3. So, how many numbers are divisible by both 2 and 3? Well, numbers divisible by both 2 and 3 are numbers divisible by the least common multiple (LCM) of 2 and 3. The LCM of 2 and 3 is 6. Therefore, numbers divisible by 6.

So, how many numbers between 1 and 100 are divisible by 6? That would be 100 divided by 6, but since we can't have a fraction of a number, we need to take the floor of that value. Let me calculate 100 divided by 6. 6 times 16 is 96, and 6 times 17 is 102, which is over 100. So, there are 16 numbers divisible by 6 between 1 and 100.

Wait, let me check that again. If I divide 100 by 6, it's approximately 16.666... So, the integer division would give 16. So, yes, 16 numbers. Therefore, there are 16 numbers divisible by both 2 and 3.

Therefore, the number of numbers divisible by 2 but not by 3 is 50 (divisible by 2) minus 16 (divisible by both 2 and 3), which is 50 - 16 = 34. Therefore, there are 34 numbers that satisfy the condition.

Therefore, the probability is 34 divided by 100. But the question says to express the answer as a common fraction. So, 34/100 can be simplified. Let me check if 34 and 100 have a common factor. Both are divisible by 2. Dividing numerator and denominator by 2 gives 17/50. Is 17 a prime number? Yes, 17 is a prime number, so 17 and 50 have no common factors besides 1. Therefore, the simplified fraction is 17/50.

Wait, let me verify my steps again to make sure I didn't make a mistake. First, total numbers: 100. Numbers divisible by 2: 50. Numbers divisible by 6: floor(100/6) = 16. So, numbers divisible by 2 but not 3: 50 - 16 = 34. Then 34/100 reduces to 17/50. That seems correct.

Alternatively, maybe I can approach this problem using another method to cross-verify. Let me think. The numbers divisible by 2 but not by 3 can be calculated as numbers divisible by 2 minus numbers divisible by both 2 and 3, which is exactly what I did. Alternatively, maybe using inclusion-exclusion principle?

Alternatively, think of numbers divisible by 2 but not by 3. So, numbers that are divisible by 2 but not by 3. So, another way is to consider numbers divisible by 2, which are 50 numbers. Now, among these 50 numbers, how many are divisible by 3? Well, since numbers divisible by both 2 and 3 are numbers divisible by 6, which we already found as 16. Therefore, same result. So, 50 - 16 = 34. So, same answer.

Alternatively, maybe I can think of the numbers not divisible by 3. Wait, but that might complicate. Let me try. The total numbers not divisible by 3: total numbers minus numbers divisible by 3. Total numbers is 100. Numbers divisible by 3: floor(100/3) = 33 (since 3*33=99). Therefore, numbers not divisible by 3: 100 - 33 = 67. But how many of these are divisible by 2? Wait, that's a different approach. So, if I want numbers divisible by 2 and not by 3, maybe I can find numbers divisible by 2 and not by 3 by considering numbers not divisible by 3 and then taking those that are divisible by 2? But that might not be straightforward. Wait, but maybe using some formula?

Alternatively, maybe using the principle of inclusion-exclusion. Let me think. The count of numbers divisible by 2 or 3 is equal to numbers divisible by 2 plus numbers divisible by 3 minus numbers divisible by both 2 and 3. But I don't know if that helps here. Wait, but maybe not. Let me see.

Alternatively, maybe think of the numbers divisible by 2 but not by 3 as numbers divisible by 2 and not divisible by 3. So, another way is to find numbers divisible by 2 and co-prime with 3? Wait, not exactly. Because numbers divisible by 2 but not by 3 are numbers divisible by 2 and not divisible by 3. So, perhaps another way is to think of numbers divisible by 2, and then subtract those divisible by 3. Which is again the same as before.

Alternatively, maybe think of the numbers divisible by 2 but not by 3 as numbers divisible by 2 and not divisible by 3. So, if I can find the count of numbers divisible by 2 and not by 3, which is equal to the count of numbers divisible by 2 minus the count of numbers divisible by both 2 and 3. Which is again the same as before.

So, seems like my original approach is correct. Therefore, 34 numbers. Therefore, probability is 34/100 = 17/50. Let me check with another approach.

Let me think of the numbers from 1 to 100. Let me think of the numbers divisible by 2: 2, 4, 6, 8, ..., 100. So, that's 50 numbers. Now, among these numbers, which are divisible by 3? Well, numbers divisible by both 2 and 3 are numbers divisible by 6. So, numbers divisible by 6 in the list of even numbers. So, numbers divisible by 6 are 6, 12, 18, ..., up to 96. Let me count how many terms are there in this sequence. The first term is 6, the last term is 96. The common difference is 6. So, the number of terms is ((last term - first term)/common difference) + 1. So, ((96 - 6)/6) + 1 = (90/6) + 1 = 15 + 1 = 16. So, 16 numbers. Therefore, same as before. Therefore, subtracting 16 from 50 gives 34. Therefore, 34 numbers. So, probability is 34/100 = 17/50. So, that seems correct.

Alternatively, maybe I can think of the numbers divisible by 2 but not by 3 as numbers divisible by 2 and not divisible by 3. So, another way is to think of numbers divisible by 2 and not divisible by 3. So, for each number divisible by 2, check if it's not divisible by 3. So, for example, take the number 2: 2 is not divisible by 3. 4: same. 6: divisible by 3. 8: not. 10: not. 12: divisible by 3. So, every third even number is divisible by 3. Wait, since every multiple of 6 is divisible by both 2 and 3. So, in the list of even numbers, every third number is divisible by 3. Let me check. Let's take the even numbers: 2, 4, 6, 8, 10, 12, 14, 16, 18,... So, starting from 2, every third number is 6, 12, 18,... So, yes, every third even number is divisible by 3. Therefore, the number of even numbers not divisible by 3 would be total even numbers minus every third even number. So, total even numbers is 50. How many are divisible by 3? Well, since every third even number is divisible by 3, then the number is 50 divided by 3, rounded down? Wait, but earlier we found that it's 16. Let me check. If there are 50 even numbers, and every third one is divisible by 3, then the number of such numbers would be floor(50/3). Let me calculate 50 divided by 3. 3*16=48, so floor(50/3)=16. So, that gives 16 numbers. Therefore, 50 - 16 = 34. So, same result. Therefore, that seems to confirm again that the number is 34. Therefore, probability is 34/100 = 17/50.

Alternatively, maybe think of the probability as follows. The probability that a number is divisible by 2 is 1/2. The probability that a number is not divisible by 3 is 2/3 (since 1/3 of numbers are divisible by 3). But wait, are these events independent? Wait, if they are independent, then the probability that a number is divisible by 2 and not divisible by 3 would be (1/2)*(2/3) = 1/3. But wait, that's different from 17/50. Wait, 1/3 is approximately 0.333..., and 17/50 is 0.34. So, they are close but not the same. Therefore, this suggests that the events are not independent. Therefore, my previous approach is more accurate. So, why is there a discrepancy?

Because the divisibility by 2 and divisibility by 3 are not independent events. So, the probability that a number is divisible by 2 and not divisible by 3 is not simply the product of the probabilities. Therefore, my initial approach of counting is more accurate. Therefore, 17/50 is the correct answer.

But just to check, let me compute 17/50. 17 divided by 50 is 0.34. And 1/3 is approximately 0.333..., so they are close but different. Therefore, the answer is 17/50.

Alternatively, let me think of the total numbers. Let me think of numbers from 1 to 100. Let me think of how many numbers are divisible by 2 but not by 3. Let me think of the numbers divisible by 2: 50 numbers. Now, numbers divisible by 2 and 3 are numbers divisible by 6, which is 16 numbers. Therefore, numbers divisible by 2 but not by 3 is 50 - 16 = 34. Therefore, 34/100 = 17/50. So, that's the answer.

Alternatively, maybe I can think of the numbers not divisible by 3. There are 100 - 33 = 67 numbers not divisible by 3. Now, how many of these are divisible by 2? Well, among the numbers not divisible by 3, half of them are even? Wait, not necessarily. Because numbers not divisible by 3 can be even or odd. So, maybe the count is different. Let me think. Let me try to calculate it.

Total numbers not divisible by 3: 67. How many of these are even? Let me think. The numbers not divisible by 3 can be divided into even and odd numbers. Let me think of the numbers from 1 to 100. Let me think of numbers not divisible by 3. For each number not divisible by 3, it can be either even or odd. Let me think of how many even numbers are there in the numbers not divisible by 3.

Alternatively, maybe use inclusion-exclusion again. The number of numbers divisible by 2 and not by 3 is equal to total numbers divisible by 2 minus numbers divisible by both 2 and 3, which is 50 - 16 = 34. So, same as before. Alternatively, if I want to compute it as numbers not divisible by 3 and divisible by 2, which is same as numbers divisible by 2 and not by 3. So, maybe another way is to compute numbers divisible by 2 and not by 3 as numbers divisible by 2 minus numbers divisible by 6. Which is again same as before.

Alternatively, maybe think of the numbers not divisible by 3. There are 67 numbers. How many of them are even? Let me think. Let me think of numbers not divisible by 3. Let me think of numbers from 1 to 100. Let me think of numbers not divisible by 3. For each residue class modulo 6. Wait, maybe that's overcomplicating. Alternatively, think that numbers not divisible by 3 are numbers congruent to 1, 2, 4, 5 mod 6. So, in each block of 6 numbers, there are 4 numbers not divisible by 3. Out of these, how many are even? Let me check. In numbers 1-6: numbers not divisible by 3 are 1, 2, 4, 5. Among these, even numbers are 2 and 4. So, 2 out of 4. Therefore, in each block of 6 numbers, there are 2 even numbers not divisible by 3. Therefore, in 100 numbers, how many blocks of 6 are there? 100 divided by 6 is 16 with a remainder of 4. So, 16 full blocks and 4 extra numbers.

In each full block, there are 2 even numbers not divisible by 3. Therefore, 16 blocks contribute 16*2 = 32 numbers. Then, the remaining 4 numbers are 97, 98, 99, 100. Let me check which of these are not divisible by 3. 97: 9+7=16, not divisible by 3. 98: 9+8=17, not divisible by 3. 99: 9+9=18, divisible by 3. 100: 1+0+0=1, not divisible by 3. Therefore, among the remaining 4 numbers, the numbers not divisible by 3 are 97, 98, 100. So, three numbers. Out of these, how many are even? 97 is odd, 98 is even, 100 is even. Therefore, 2 even numbers. Therefore, total even numbers not divisible by 3 is 32 + 2 = 34. Therefore, same result. Therefore, 34 numbers. Therefore, probability is 34/100 = 17/50. So, that's another way to get the same answer. Therefore, this confirms that the answer is indeed 17/50.

Therefore, I think my answer is correct. So, the probability is 17/50.

**Final Answer**
The probability is \boxed{\dfrac{17}{50}}.

To determine the probability that a randomly selected integer from the set $\{1, 2, 3, \ldots, 100\}$ is **divisible by 2 and not divisible by 3**, we proceed step by step using principles of counting and probability.

---

### Step 1: Total Number of Outcomes

The total number of integers in the set is:

$$
100
$$

This will be the denominator of our probability.

---

### Step 2: Count of Numbers Divisible by 2

Numbers divisible by 2 are the even numbers in the set. Since every second number is even, the count is:

$$
\left\lfloor \frac{100}{2} \right\rfloor = 50
$$

So, there are **50** numbers divisible by 2.

---

### Step 3: Count of Numbers Divisible by Both 2 and 3

A number divisible by both 2 and 3 is divisible by their **least common multiple (LCM)**, which is:

$$
\text{LCM}(2, 3) = 6
$$

So, we count how many numbers between 1 and 100 are divisible by 6:

$$
\left\lfloor \frac{100}{6} \right\rfloor = 16
$$

Thus, there are **16** numbers divisible by both 2 and 3.

---

### Step 4: Count of Numbers Divisible by 2 but Not by 3

To find the count of numbers divisible by 2 **but not** by 3, we subtract the count of numbers divisible by both 2 and 3 from the count of numbers divisible by 2:

$$
50 - 16 = 34
$$

So, there are **34** numbers that are divisible by 2 but not by 3.

---

### Step 5: Compute the Probability

The probability is the ratio of favorable outcomes (numbers divisible by 2 but not by 3) to the total number of outcomes:

$$
\text{Probability} = \frac{34}{100}
$$

Simplify the fraction:

$$
\frac{34}{100} = \frac{17}{50}
$$

---

### Final Answer

$$
\boxed{\dfrac{17}{50}}
$$

Solution: Since $100 = 50\cdot 2$, there are 50 integers in the set that are divisible by 2.  The numbers among these that are also divisible by 3 are the multiples of 6 in the set.  Dividing 100 by 6 gives $16\frac23$, so there are 16 multiples of 6 in the set, which leaves $50-16 = 34$ multiples of 2 that are not also multiples of 3.  There are 100 numbers in the set so the desired probability is $\dfrac{34}{100} = \boxed{\dfrac{17}{50}}$.


Explanation:
Okay, so I need to find the greatest possible value of the expression $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$ given that $-6 \leq a \leq -2$ and $3 \leq b \leq 5$. Hmm, let me think about how to approach this.

First, maybe I should simplify the expression. Let me expand the product to see if it becomes easier to handle. Let me write it out:

$$
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)
$$

This looks like a difference of squares. Wait, the formula is $(x + y)(x - y) = x^2 - y^2$. Let me check if that applies here. Let me set $x = \frac{1}{b}$ and $y = a$. Then the expression becomes:

$$
\left(\frac{1}{b} + a\right)\left(\frac{1}{b} - a\right) = \left(\frac{1}{b}\right)^2 - a^2
$$

Oh! So the original expression simplifies to $\frac{1}{b^2} - a^2$. That's much simpler. So instead of dealing with the product, I can just work with $\frac{1}{b^2} - a^2$. Now, I need to find the maximum value of this expression given the constraints on $a$ and $b$.

Since the expression is $\frac{1}{b^2} - a^2$, to maximize this, I need to maximize $\frac{1}{b^2}$ and minimize $a^2$. Because $\frac{1}{b^2}$ is being subtracted by $a^2$, so maximizing the first term and minimizing the second term will give the maximum overall value.

Let me confirm that logic. If I have $X - Y$, then to maximize it, I need to maximize $X$ and minimize $Y$. Yes, that makes sense. So, since $X = \frac{1}{b^2}$ and $Y = a^2$, I need to find the maximum of $X$ and the minimum of $Y$.

So first, let's find the maximum value of $\frac{1}{b^2}$. Since $b$ is between 3 and 5, inclusive, let's see how $\frac{1}{b^2}$ behaves. The function $f(b) = \frac{1}{b^2}$ is a decreasing function for $b > 0$. That means as $b$ increases, $f(b)$ decreases. Therefore, the maximum value of $\frac{1}{b^2}$ occurs at the smallest value of $b$, which is 3. So the maximum of $\frac{1}{b^2}$ is $\frac{1}{3^2} = \frac{1}{9}$.

Next, let's find the minimum value of $a^2$. Given that $a$ is between -6 and -2, inclusive. Since $a$ is negative, squaring it will make it positive. The square of a number is minimized when the absolute value is minimized. So, since $a$ is between -6 and -2, the absolute value of $a$ is between 2 and 6. Therefore, the minimum value of $a^2$ occurs when $|a|$ is smallest, which is when $a$ is closest to zero. Since $a$ is between -6 and -2, the closest to zero is -2. Therefore, the minimum value of $a^2$ is $(-2)^2 = 4$.

Therefore, substituting back into the expression $\frac{1}{b^2} - a^2$, the maximum value would be $\frac{1}{9} - 4$. Wait, but that would be $\frac{1}{9} - 4 = -\frac{35}{9}$. But that seems negative. However, maybe I made a mistake here? Because if I take the maximum of $\frac{1}{b^2}$ and the minimum of $a^2$, then subtract them, but maybe there's a combination where $\frac{1}{b^2}$ is not maximum and $a^2$ is not minimum, but their combination gives a higher value?

Wait, let me think again. The expression is $\frac{1}{b^2} - a^2$. So if I want to maximize this, I need to maximize $\frac{1}{b^2}$ and minimize $a^2$. But if I take the maximum of $\frac{1}{b^2}$ and the minimum of $a^2$, then that would be the maximum of the entire expression. But let me check with numbers.

Suppose I take $b = 3$ (maximizing $\frac{1}{b^2}$) and $a = -2$ (minimizing $a^2$). Then the expression is $\frac{1}{9} - 4 = -\frac{35}{9}$. But maybe if I take $b$ larger, so that $\frac{1}{b^2}$ is smaller, but $a$ is more negative, so $a^2$ is larger. Wait, but that would make the expression even smaller. Alternatively, if I take $b$ smaller (but $b$ can't be smaller than 3), so that's already the minimum. Alternatively, if I take $a$ less negative, so closer to zero, but that would make $a^2$ smaller. Wait, but $a$ is between -6 and -2. So the minimum $a^2$ is at $a = -2$, as I thought. So maybe that is indeed the case.

But wait, maybe there's a different combination where $\frac{1}{b^2} - a^2$ is larger? Let me check with some numbers. Let me try $b = 3$ and $a = -2$: $\frac{1}{9} - 4 = -\frac{35}{9} \approx -3.888...$. What if I take $b = 5$ and $a = -2$: $\frac{1}{25} - 4 = -\frac{99}{25} = -3.96$, which is worse. If I take $b = 3$ and $a = -6$: $\frac{1}{9} - 36 = -\frac{323}{9} \approx -35.888...$, which is worse. If I take $b = 3$ and $a = -3$: $\frac{1}{9} - 9 = -\frac{80}{9} \approx -8.888...$, worse. If I take $b = 5$ and $a = -6$: $\frac{1}{25} - 36 = -\frac{899}{25} = -35.96$, worse. So seems like the maximum is indeed when $b$ is as small as possible and $a$ is as close to zero as possible. But that gives me a negative number. Is there a way to get a positive value?

Wait, maybe I made a mistake in simplifying the original expression? Let me check again. The original expression is $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me expand it again:

First, multiply $a$ by $\frac{1}{b}$: $a \cdot \frac{1}{b}$

Then $a \cdot (-a)$: $-a^2$

Then $\frac{1}{b} \cdot \frac{1}{b}$: $\frac{1}{b^2}$

Then $\frac{1}{b} \cdot (-a)$: $-\frac{a}{b}$

So altogether:

$a \cdot \frac{1}{b} - a^2 + \frac{1}{b^2} - \frac{a}{b}$

Wait, the first term and the last term are $a/b - a/b$, which cancel each other. So we are left with $-a^2 + \frac{1}{b^2}$. So yes, that's correct. So the expression simplifies to $\frac{1}{b^2} - a^2$. So that's correct.

Therefore, the expression is $\frac{1}{b^2} - a^2$. So if I want to maximize this, I need to maximize $\frac{1}{b^2}$ and minimize $a^2$. But since $\frac{1}{b^2}$ is maximized when $b$ is minimized (i.e., $b=3$) and $a^2$ is minimized when $a$ is closest to zero (i.e., $a=-2$), then the maximum value is $\frac{1}{9} - 4 = -\frac{35}{9}$. But that's negative. However, maybe there's a different combination where $\frac{1}{b^2} - a^2$ is positive?

Wait, let me check. Suppose $a$ is negative and $b$ is positive. Then $\frac{1}{b^2}$ is positive, and $a^2$ is positive. So the expression is positive minus positive. Depending on the values, it could be positive or negative. For example, if $\frac{1}{b^2} > a^2$, then the expression is positive. So maybe there are values where this is positive?

So perhaps my previous reasoning was wrong. Maybe I need to consider when $\frac{1}{b^2} - a^2$ is positive. Let me think. So if $\frac{1}{b^2} > a^2$, then the expression is positive. So maybe there are some combinations where this is true, and then we need to find the maximum among those?

But how do I know if there are such combinations? Let me check with some numbers. Let me take $a = -2$ and $b = 3$. Then $\frac{1}{9} - 4$ is negative. If I take $a = -1$, but wait, $a$ is between -6 and -2, so $a$ can't be -1. How about $a = -2$ and $b = 5$: $\frac{1}{25} - 4$ is still negative. If I take $a = -2$ and $b = 3$, same thing. How about if I take $a$ closer to zero? Wait, $a$ can't be closer to zero than -2. So the minimum $a^2$ is 4. The maximum $\frac{1}{b^2}$ is 1/9. So 1/9 is about 0.111, and 4 is 4. So 0.111 - 4 is negative. Therefore, the entire expression is negative. Therefore, the maximum value is the least negative value, which would be when $\frac{1}{b^2}$ is as large as possible and $a^2$ is as small as possible. Therefore, the maximum value is indeed $\frac{1}{9} - 4 = -\frac{35}{9}$. But the problem says "the greatest possible value", so maybe that's the answer? But maybe I need to check if there are other combinations where the expression is larger?

Wait, but if the expression is always negative, then the greatest possible value is the one closest to zero. So maybe I need to check if there are other combinations where $\frac{1}{b^2} - a^2$ is larger (i.e., less negative) than $-35/9$?

Wait, let me think. Let me consider that maybe if I take $a$ more negative, but then $a^2$ becomes larger, which would make the expression smaller. If I take $a$ less negative (closer to zero), then $a^2$ is smaller, so the expression becomes larger (less negative). Therefore, the maximum occurs when $a$ is as close to zero as possible (i.e., $a = -2$) and $b$ as small as possible (i.e., $b = 3$). Therefore, the maximum value is $\frac{1}{9} - 4 = -\frac{35}{9}$. But maybe I need to check if there's a case where $\frac{1}{b^2} - a^2$ is positive?

Wait, let me check with $a = -1/2$, but $a$ is between -6 and -2, so that's not allowed. How about $a = -2$ and $b = 1$? But $b$ is between 3 and 5, so $b$ can't be 1. So even if $b$ is as small as 3, $\frac{1}{b^2}$ is 1/9, and $a^2$ is 4. So 1/9 - 4 is negative. Therefore, the expression is always negative. Therefore, the maximum value is indeed $-35/9$. But the problem says "the greatest possible value", so maybe that is the answer? But let me check again.

Alternatively, maybe I made a mistake in simplifying the original expression. Let me check again:

Original expression: $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me compute this again step by step.

Multiply the two binomials:

First term: $a \times \frac{1}{b} = \frac{a}{b}$

Second term: $a \times (-a) = -a^2$

Third term: $\frac{1}{b} \times \frac{1}{b} = \frac{1}{b^2}$

Fourth term: $\frac{1}{b} \times (-a) = -\frac{a}{b}$

So adding all terms: $\frac{a}{b} - a^2 + \frac{1}{b^2} - \frac{a}{b}$. The $\frac{a}{b}$ and $-\frac{a}{b}$ cancel out, leaving $-a^2 + \frac{1}{b^2}$. So that's correct. Therefore, the expression is indeed $\frac{1}{b^2} - a^2$. Therefore, the conclusion is that the maximum value is when $\frac{1}{b^2}$ is maximum and $a^2$ is minimum, which gives $-35/9$. But maybe I need to check if there's another way to approach this problem?

Alternatively, maybe I can think of the expression as $\frac{1}{b^2} - a^2$ and consider it as a function of $a$ and $b$, and then find its maximum over the given intervals. Since $a$ and $b$ are independent variables, maybe I can analyze them separately?

For the term $\frac{1}{b^2}$, as $b$ increases, $\frac{1}{b^2}$ decreases, so maximum at $b=3$, minimum at $b=5$. For the term $a^2$, since $a$ is negative, $a^2$ is minimized when $a$ is closest to zero, which is $a=-2$, and maximized when $a=-6$. Therefore, since the expression is $\frac{1}{b^2} - a^2$, to maximize it, we need to maximize $\frac{1}{b^2}$ and minimize $a^2$, which gives us $b=3$ and $a=-2$, leading to $\frac{1}{9} - 4 = -\frac{35}{9}$. Therefore, that seems to be the answer.

But let me check if there's a case where $\frac{1}{b^2} - a^2$ is larger. For example, suppose $a$ is more negative, but $b$ is smaller? Wait, $b$ can't be smaller than 3. If $a$ is more negative, $a^2$ increases, so $\frac{1}{b^2} - a^2$ decreases. If $a$ is less negative (closer to zero), $a^2$ decreases, so $\frac{1}{b^2} - a^2$ increases. Therefore, the maximum occurs at the minimal $a^2$ and maximal $\frac{1}{b^2}$, which is $a=-2$ and $b=3$. Therefore, the answer is $-35/9$. But the problem says "the greatest possible value", so maybe that's correct? But maybe I need to check if there's a different combination where the expression is larger?

Wait, let me think again. Suppose I take $a = -2$ and $b = 3$, then the original expression is $\left(-2 + \frac{1}{3}\right)\left(\frac{1}{3} - (-2)\right)$. Let me compute that:

First, compute $-2 + 1/3 = -5/3$.

Then compute $1/3 - (-2) = 1/3 + 2 = 7/3$.

Multiply them: $-5/3 * 7/3 = -35/9$. So that's correct.

Alternatively, if I take $a = -2$ and $b = 5$, then the original expression is $\left(-2 + 1/5\right)\left(1/5 - (-2)\right) = (-9/5)(11/5) = -99/25 ≈ -3.96$, which is less than -35/9 ≈ -3.888... So indeed, -35/9 is larger (i.e., less negative).

Alternatively, if I take $a = -3$ and $b = 3$, then the original expression is $\left(-3 + 1/3\right)\left(1/3 - (-3)\right) = (-8/3)(10/3) = -80/9 ≈ -8.888...$, which is worse.

Alternatively, if I take $a = -2$ and $b = 3$, which gives -35/9. What if I take $a = -2$ and $b = 3.5$? Let me compute that. $b = 3.5 = 7/2$. Then $\frac{1}{b^2} = 4/49 ≈ 0.0816$. Then $a^2 = 4$. So the expression is 4/49 - 4 ≈ -3.918, which is worse than -35/9 ≈ -3.888. So worse.

Alternatively, if I take $a = -2$ and $b = 3$, which is the maximum of $\frac{1}{b^2}$ and minimum of $a^2$, so that's the best.

Alternatively, what if I take $a$ slightly more negative but $b$ slightly smaller? But $b$ can't be smaller than 3. So no.

Alternatively, maybe there's a case where $\frac{1}{b^2} - a^2$ is positive? Let me check with $a = -1/2$, but $a$ is between -6 and -2, so that's not allowed. How about $a = -2$ and $b = 1$? But $b$ is between 3 and 5. So even if $b$ is 3, $\frac{1}{b^2}$ is 1/9 ≈ 0.111, and $a^2 = 4$. So 0.111 - 4 is negative. Therefore, the expression is always negative. Therefore, the maximum value is indeed -35/9. Therefore, the answer is \boxed{-\dfrac{35}{9}}.

Wait, but the problem says "the greatest possible value". But maybe I made a mistake in the sign? Let me check again. The original expression is $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me plug in $a = -2$ and $b = 3$:

First term: $a + 1/b = -2 + 1/3 = -5/3$

Second term: $1/b - a = 1/3 - (-2) = 1/3 + 2 = 7/3$

Multiply them: (-5/3)(7/3) = -35/9. So that's correct. So the value is negative. But maybe there's a different combination where the product is positive?

Wait, suppose $a$ is negative and $1/b - a$ is positive? Let me think. Since $a$ is negative, $-a$ is positive. So $1/b - a = 1/b + |a|$. Since both $1/b$ and $|a|$ are positive, this term is positive. The first term is $a + 1/b$. Since $a$ is negative and $1/b$ is positive, depending on the values, this could be positive or negative. For example, if $a$ is close to zero, then $a + 1/b$ is positive. If $a$ is more negative, then $a + 1/b$ could be negative.

Therefore, the product $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$ could be positive or negative. For example, if $a + 1/b$ is positive and $1/b - a$ is positive, then the product is positive. If $a + 1/b$ is negative and $1/b - a$ is positive, then the product is negative. Therefore, maybe there are cases where the product is positive, and we need to find the maximum among those?

So perhaps my previous reasoning was incomplete. Let me think again. So the expression is $\frac{1}{b^2} - a^2$, which can be positive or negative. So to find the maximum value, we need to consider both possibilities.

So first, let's find when $\frac{1}{b^2} - a^2$ is positive. That is, when $\frac{1}{b^2} > a^2$. Which implies that $|a| < \frac{1}{b}$. Since $a$ is negative, $|a| = -a$. Therefore, the inequality becomes $-a < \frac{1}{b}$, which implies $a > -\frac{1}{b}$. But since $a$ is between -6 and -2, and $b$ is between 3 and 5, let's see for which values of $a$ and $b$ this holds.

So $a > -\frac{1}{b}$. Since $b$ is between 3 and 5, $-\frac{1}{b}$ is between $-1/5$ and $-1/3$. But $a$ is between -6 and -2, which is all less than $-1/3$. Therefore, $a > -\frac{1}{b}$ would mean that $a$ is greater than a number between $-1/5$ and $-1/3$, but since $a$ is at most -2, which is less than $-1/3$, there is no overlap. Therefore, there are no values of $a$ and $b$ in the given ranges where $|a| < \frac{1}{b}$. Therefore, $\frac{1}{b^2} - a^2$ is always negative. Therefore, the expression is always negative, and the maximum value is the one closest to zero, which is when $\frac{1}{b^2}$ is as large as possible and $a^2$ as small as possible. Therefore, the maximum value is indeed $-35/9$.

But let me check with specific numbers. Let me take $a = -2$ and $b = 3$. Then $|a| = 2$, and $\frac{1}{b} = 1/3 ≈ 0.333$. So $|a| = 2 > 1/3$, so $\frac{1}{b^2} - a^2$ is negative. If I take $a = -1.5$, but $a$ is between -6 and -2, so $a = -1.5$ is not allowed. If I take $a = -2$ and $b = 3$, then $|a| = 2$, which is greater than $1/b = 1/3$. Therefore, the expression is negative. Therefore, there are no cases where the expression is positive. Therefore, the maximum value is indeed $-35/9$.

But let me check another case. Suppose $a = -2$ and $b = 3$. Then the original expression is $\left(-2 + 1/3\right)\left(1/3 - (-2)\right) = (-5/3)(7/3) = -35/9$. If I take $a = -2$ and $b = 3.5$, then the expression is $\left(-2 + 1/3.5\right)\left(1/3.5 - (-2)\right)$. Let me compute that:

First, $1/3.5 = 2/7 ≈ 0.2857$. So $-2 + 2/7 = -12/7 ≈ -1.714$. Then $1/3.5 - (-2) = 2/7 + 2 = 16/7 ≈ 2.2857$. Multiply them: (-12/7)(16/7) = -192/49 ≈ -3.918, which is less than -35/9 ≈ -3.888. So worse.

If I take $a = -2$ and $b = 3$, which gives -35/9 ≈ -3.888. If I take $a = -2$ and $b = 3$, that's the maximum. If I take $a = -2$ and $b = 3$, that's the best. If I take $a = -2$ and $b = 3$, that's the maximum. Therefore, the answer is -35/9.

But the problem says "the greatest possible value". But maybe I need to check if there's another combination where the expression is larger? For example, if I take $a$ slightly more negative but $b$ slightly larger? Wait, but $b$ larger would make $\frac{1}{b^2}$ smaller, and $a$ more negative would make $a^2$ larger. So both terms would make the expression smaller. Therefore, no.

Alternatively, if I take $a$ slightly less negative (closer to zero) but $b$ slightly larger? Wait, $a$ can't be closer to zero than -2. So if I take $a = -2$ and $b = 3$, that's the closest to zero for $a$ and the smallest $b$. Therefore, that's the best.

Alternatively, maybe there's a case where $a + 1/b$ is positive and $1/b - a$ is positive, leading to a positive product? But as we saw earlier, since $a$ is between -6 and -2, and $b$ is between 3 and 5, $a + 1/b$ is $a +$ something positive. Since $a$ is negative, but how big is $1/b$? For $b = 3$, $1/b = 1/3 ≈ 0.333$. So $a + 1/b$ would be $-2 + 0.333 ≈ -1.666$, which is negative. For $a = -2$ and $b = 3$, it's negative. If $a$ is closer to zero, say $a = -2$, but that's the closest. If $a$ is -2, then $a + 1/b$ is negative. If $a$ is more negative, like -3, then $a + 1/b$ is even more negative. Therefore, $a + 1/b$ is always negative. Therefore, the first term is negative, the second term is positive, so the product is negative. Therefore, the expression is always negative, and the maximum value is -35/9.

Therefore, I think that is the answer. However, the problem says "the greatest possible value", and maybe I need to check if there's a different approach where the answer is positive? Let me think again.

Wait, maybe I made a mistake in the sign when simplifying? Let me check again:

Original expression: $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me think of it as $(\frac{1}{b} + a)(\frac{1}{b} - a)$. Which is $\left(\frac{1}{b}\right)^2 - a^2$. So that's correct. Therefore, the expression is $\frac{1}{b^2} - a^2$. Therefore, if I want to maximize this, I need to maximize $\frac{1}{b^2}$ and minimize $a^2$. Therefore, the answer is $\frac{1}{9} - 4 = -\frac{35}{9}$. Therefore, I think that is correct.

But maybe the problem expects a positive answer? Let me check once more with different values. Suppose $a = -2$ and $b = 3$, then the original expression is $\left(-2 + 1/3\right)\left(1/3 - (-2)\right) = (-5/3)(7/3) = -35/9$. If I take $a = -2$ and $b = 3$, that's the maximum. If I take $a = -2$ and $b = 3$, that's the maximum. Therefore, I think that is correct.

Alternatively, maybe I need to consider the expression as a function of $a$ and $b$ and find its maximum. Let me think of it as $f(a, b) = \frac{1}{b^2} - a^2$. Since $a$ and $b$ are independent variables, the maximum occurs at the maximum of $\frac{1}{b^2}$ and the minimum of $a^2$. Therefore, the maximum of $f(a, b)$ is $\frac{1}{3^2} - (-2)^2 = \frac{1}{9} - 4 = -\frac{35}{9}$. Therefore, that's the answer.

Alternatively, maybe there's a mistake in the problem statement? Or maybe I misread the problem? Let me check again.

The problem says: If $-6 \leq a \leq -2$ and $3 \leq b \leq 5$, what is the greatest possible value of $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$? Express your answer as a common fraction.

No, I think I read it correctly. So the answer is -35/9. But maybe the problem expects a positive value? Let me check once again with different values. Suppose $a = -1/2$ and $b = 1$, but those values are outside the given ranges. If $a = -2$ and $b = 3$, then the value is -35/9. If I take $a = -2$ and $b = 3$, that's the maximum. Therefore, I think that is correct.

Alternatively, maybe I need to consider that the original expression can be rewritten as $\frac{1}{b^2} - a^2$, but maybe there's a different way to write it? For example, maybe factor it differently? Let me see:

Alternatively, $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right) = \left(\frac{1}{b} + a\right)\left(\frac{1}{b} - a\right) = \left(\frac{1}{b}\right)^2 - a^2$. So that's the same as before. Therefore, no other way.

Alternatively, maybe I can think of this as a quadratic in terms of $a$ or $b$? For example, if I fix $b$, then the expression is $-a^2 + \frac{1}{b^2}$, which is a downward opening parabola in terms of $a$, so its maximum occurs at the vertex. But since $a$ is constrained between -6 and -2, the maximum of $-a^2 + \frac{1}{b^2}$ would occur at the minimum of $a^2$, which is at $a = -2$. Similarly, if I fix $a$, then the expression is $\frac{1}{b^2} - a^2$, which is a decreasing function in $b$, so maximum at $b = 3$. Therefore, again, the maximum occurs at $a = -2$ and $b = 3$, giving $-35/9$.

Therefore, I think that is the correct answer. Therefore, the greatest possible value is $-35/9$, which is a common fraction. Therefore, the answer is $\boxed{-\dfrac{35}{9}}$.

**Final Answer**
\boxed{-\dfrac{35}{9}}

To find the **greatest possible value** of the expression
$$
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)
$$
given the constraints
$$
-6 \leq a \leq -2 \quad \text{and} \quad 3 \leq b \leq 5,
$$
we begin by simplifying the expression.

---

### Step 1: Simplify the Expression

We recognize the expression as a **difference of squares**:

$$
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right) = \left(\frac{1}{b}\right)^2 - a^2 = \frac{1}{b^2} - a^2
$$

So, the original expression simplifies to:

$$
\frac{1}{b^2} - a^2
$$

---

### Step 2: Analyze the Expression

We now want to **maximize** the expression:

$$
\frac{1}{b^2} - a^2
$$

This is a function of two variables, $a$ and $b$, with the following constraints:

- $a \in [-6, -2]$
- $b \in [3, 5]$

To **maximize** the expression, we need to:

- **Maximize** $\frac{1}{b^2}$, which is a **decreasing** function of $b$. So, it is maximized when $b$ is **smallest**, i.e., $b = 3$.
- **Minimize** $a^2$, which is a **non-negative** function. Since $a$ is negative, $a^2$ is minimized when $a$ is **closest to zero**, i.e., $a = -2$.

---

### Step 3: Plug in the Optimal Values

Using the values that maximize $\frac{1}{b^2}$ and minimize $a^2$:

- $b = 3 \Rightarrow \frac{1}{b^2} = \frac{1}{9}$
- $a = -2 \Rightarrow a^2 = 4$

So the expression becomes:

$$
\frac{1}{9} - 4 = -\frac{35}{9}
$$

---

### Step 4: Confirm the Expression is Always Negative

Since $a$ is negative and $b$ is positive, the original expression is a product of a **negative** and a **positive** term, so the result is **always negative**. Therefore, the **greatest possible value** is the one **closest to zero**, which occurs when $\frac{1}{b^2}$ is as large as possible and $a^2$ is as small as possible.

---

### Final Answer

$$
\boxed{-\dfrac{35}{9}}
$$

[Assistant]
Explanation:
Okay, so I need to find the greatest possible value of the expression $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$ given that $-6 \leq a \leq -2$ and $3 \leq b \leq 5$. Hmm, let me think about how to approach this.

First, maybe I should simplify the expression. Let me expand the product to see if it becomes easier to handle. Let me write it out:

$$
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)
$$

This looks like a difference of squares. Wait, the formula is $(x + y)(x - y) = x^2 - y^2$. Let me check if that applies here. Let me set $x = \frac{1}{b}$ and $y = a$. Then the expression becomes:

$$
\left(\frac{1}{b} + a\right)\left(\frac{1}{b} - a\right) = \left(\frac{1}{b}\right)^2 - a^2
$$

Oh! So the original expression simplifies to $\frac{1}{b^2} - a^2$. That's much simpler. So instead of dealing with the product, I can just work with $\frac{1}{b^2} - a^2$. Now, I need to find the maximum value of this expression given the constraints on $a$ and $b$.

Since the expression is $\frac{1}{b^2} - a^2$, to maximize this, I need to maximize $\frac{1}{b^2}$ and minimize $a^2$. Because $\frac{1}{b^2}$ is being subtracted by $a^2$, so maximizing the first term and minimizing the second term will give the maximum overall value.

Let me confirm that logic. If I have $X - Y$, then to maximize it, I need to maximize $X$ and minimize $Y$. Yes, that makes sense. So, since $X = \frac{1}{b^2}$ and $Y = a^2$, I need to find the maximum of $X$ and the minimum of $Y$.

So first, let's find the maximum value of $\frac{1}{b^2}$. Since $b$ is between 3 and 5, inclusive, let's see how $\frac{1}{b^2}$ behaves. The function $f(b) = \frac{1}{b^2}$ is a decreasing function for $b > 0$. That means as $b$ increases, $f(b)$ decreases. Therefore, the maximum value of $\frac{1}{b^2}$ occurs at the smallest value of $b$, which is 3. So the maximum of $\frac{1}{b^2}$ is $\frac{1}{3^2} = \frac{1}{9}$.

Next, let's find the minimum value of $a^2$. Given that $a$ is between -6 and -2, inclusive. Since $a$ is negative, squaring it will make it positive. The square of a number is minimized when the absolute value is minimized. So, since $a$ is between -6 and -2, the absolute value of $a$ is between 2 and 6. Therefore, the minimum value of $a^2$ occurs when $|a|$ is smallest, which is when $a$ is closest to zero. Since $a$ is between -6 and -2, the closest to zero is -2. Therefore, the minimum value of $a^2$ is $(-2)^2 = 4$.

Therefore, substituting back into the expression $\frac{1}{b^2} - a^2$, the maximum value would be $\frac{1}{9} - 4$. Wait, but that would be $\frac{1}{9} - 4 = -\frac{35}{9}$. But that seems negative. However, maybe I made a mistake here? Because if I take the maximum of $\frac{1}{b^2}$ and the minimum of $a^2$, then subtract them, but maybe there's a combination where $\frac{1}{b^2}$ is not maximum and $a^2$ is not minimum, but their combination gives a higher value?

Wait, let me think again. The expression is $\frac{1}{b^2} - a^2$. So if I want to maximize this, I need to maximize $\frac{1}{b^2}$ and minimize $a^2$. But if I take the maximum of $\frac{1}{b^2}$ and the minimum of $a^2$, then that would be the maximum of the entire expression. But let me check with numbers.

Suppose I take $b = 3$ (maximizing $\frac{1}{b^2}$) and $a = -2$ (minimizing $a^2$). Then the expression is $\frac{1}{9} - 4 = -\frac{35}{9}$. But maybe if I take $b$ larger, so that $\frac{1}{b^2}$ is smaller, but $a$ is more negative, so $a^2$ is larger. Wait, but that would make the expression even smaller. Alternatively, if I take $b$ smaller (but $b$ can't be smaller than 3), so that's already the minimum. Alternatively, if I take $a$ less negative, so closer to zero, but that would make $a^2$ smaller. Wait, but $a$ is between -6 and -2. So the minimum $a^2$ is at $a = -2$, as I thought. So maybe that is indeed the case.

But wait, maybe there's a different combination where $\frac{1}{b^2} - a^2$ is larger? Let me check with some numbers. Let me try $b = 3$ and $a = -2$: $\frac{1}{9} - 4 = -\frac{35}{9} \approx -3.888...$. What if I take $b = 5$ and $a = -2$: $\frac{1}{25} - 4 = -\frac{99}{25} = -3.96$, which is worse. If I take $b = 3$ and $a = -6$: $\frac{1}{9} - 36 = -\frac{323}{9} \approx -35.888...$, which is worse. If I take $b = 3$ and $a = -3$: $\frac{1}{9} - 9 = -\frac{80}{9} \approx -8.888...$, worse. If I take $b = 5$ and $a = -6$: $\frac{1}{25} - 36 = -\frac{899}{25} = -35.96$, worse. So seems like the maximum is indeed when $b$ is as small as possible and $a$ is as close to zero as possible. But that gives me a negative number. Is there a way to get a positive value?

Wait, maybe I made a mistake in simplifying the original expression? Let me check again. The original expression is $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me expand it again:

First, multiply $a$ by $\frac{1}{b}$: $a \cdot \frac{1}{b}$

Then $a \cdot (-a)$: $-a^2$

Then $\frac{1}{b} \cdot \frac{1}{b}$: $\frac{1}{b^2}$

Then $\frac{1}{b} \cdot (-a)$: $-\frac{a}{b}$

So altogether:

$a \cdot \frac{1}{b} - a^2 + \frac{1}{b^2} - \frac{a}{b}$

Wait, the first term and the last term are $a/b - a/b$, which cancel each other. So we are left with $-a^2 + \frac{1}{b^2}$. So yes, that's correct. So the expression simplifies to $\frac{1}{b^2} - a^2$. So that's correct.

Therefore, the expression is $\frac{1}{b^2} - a^2$. So if I want to maximize this, I need to maximize $\frac{1}{b^2}$ and minimize $a^2$. But since $\frac{1}{b^2}$ is maximized when $b$ is minimized (i.e., $b=3$) and $a^2$ is minimized when $a$ is closest to zero (i.e., $a=-2$), then the maximum value is $\frac{1}{9} - 4 = -\frac{35}{9}$. But that's negative. However, maybe there's a different combination where $\frac{1}{b^2} - a^2$ is positive?

Wait, let me check. Suppose $a$ is negative and $b$ is positive. Then $\frac{1}{b^2}$ is positive, and $a^2$ is positive. So the expression is positive minus positive. Depending on the values, it could be positive or negative. For example, if $\frac{1}{b^2} > a^2$, then the expression is positive. So maybe there are values where this is positive?

So perhaps my previous reasoning was wrong. Maybe I need to consider when $\frac{1}{b^2} - a^2$ is positive. Let me think. So if $\frac{1}{b^2} > a^2$, then the expression is positive. So maybe there are some combinations where this is true, and then we need to find the maximum among those?

But how do I know if there are such combinations? Let me check with some numbers. Let me take $a = -2$ and $b = 3$. Then $\frac{1}{9} - 4$ is negative. If I take $a = -1$, but wait, $a$ is between -6 and -2, so $a$ can't be -1. How about $a = -2$ and $b = 5$: $\frac{1}{25} - 4$ is still negative. If I take $a = -2$ and $b = 3$, same thing. How about if I take $a$ closer to zero? Wait, $a$ can't be closer to zero than -2. So the minimum $a^2$ is 4. The maximum $\frac{1}{b^2}$ is 1/9. So 1/9 is about 0.111, and 4 is 4. So 0.111 - 4 is negative. Therefore, the entire expression is negative. Therefore, the maximum value is the least negative value, which would be when $\frac{1}{b^2}$ is as large as possible and $a^2$ is as small as possible. Therefore, the maximum value is indeed $\frac{1}{9} - 4 = -\frac{35}{9}$. But the problem says "the greatest possible value", so maybe that's the answer? But maybe I need to check if there are other combinations where the expression is larger?

Wait, but if the expression is always negative, then the greatest possible value is the one closest to zero. So maybe I need to check if there are other combinations where $\frac{1}{b^2} - a^2$ is larger (i.e., less negative) than $-35/9$?

Wait, let me think. Let me consider that maybe if I take $a$ more negative, but then $a^2$ becomes larger, which would make the expression smaller. If I take $a$ less negative (closer to zero), then $a^2$ is smaller, so the expression becomes larger (less negative). Therefore, the maximum occurs when $a$ is as close to zero as possible (i.e., $a = -2$) and $b$ as small as possible (i.e., $b = 3$). Therefore, the maximum value is $\frac{1}{9} - 4 = -\frac{35}{9}$. But maybe I need to check if there's a case where $\frac{1}{b^2} - a^2$ is positive?

Wait, let me check with $a = -1/2$, but $a$ is between -6 and -2, so that's not allowed. How about $a = -2$ and $b = 1$? But $b$ is between 3 and 5, so $b$ can't be 1. So even if $b$ is as small as 3, $\frac{1}{b^2}$ is 1/9, and $a^2$ is 4. So 1/9 - 4 is negative. Therefore, the expression is always negative. Therefore, the maximum value is indeed $-35/9$. But the problem says "the greatest possible value", so maybe that is the answer? But let me check again.

Alternatively, maybe I made a mistake in simplifying the original expression. Let me check again:

Original expression: $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me compute this again step by step.

Multiply the two binomials:

First term: $a \times \frac{1}{b} = \frac{a}{b}$

Second term: $a \times (-a) = -a^2$

Third term: $\frac{1}{b} \times \frac{1}{b} = \frac{1}{b^2}$

Fourth term: $\frac{1}{b} \times (-a) = -\frac{a}{b}$

So adding all terms: $\frac{a}{b} - a^2 + \frac{1}{b^2} - \frac{a}{b}$. The $\frac{a}{b}$ and $-\frac{a}{b}$ cancel out, leaving $-a^2 + \frac{1}{b^2}$. So that's correct. Therefore, the expression is indeed $\frac{1}{b^2} - a^2$. Therefore, the conclusion is that the maximum value is when $\frac{1}{b^2}$ is maximum and $a^2$ is minimum, which gives $-35/9$. But maybe I need to check if there's another way to approach this problem?

Alternatively, maybe I can think of the expression as $\frac{1}{b^2} - a^2$ and consider it as a function of $a$ and $b$, and then find its maximum over the given intervals. Since $a$ and $b$ are independent variables, maybe I can analyze them separately?

For the term $\frac{1}{b^2}$, as $b$ increases, $\frac{1}{b^2}$ decreases, so maximum at $b=3$, minimum at $b=5$. For the term $a^2$, since $a$ is negative, $a^2$ is minimized when $a$ is closest to zero, which is $a=-2$, and maximized when $a=-6$. Therefore, since the expression is $\frac{1}{b^2} - a^2$, to maximize it, we need to maximize $\frac{1}{b^2}$ and minimize $a^2$, which gives us $b=3$ and $a=-2$, leading to $\frac{1}{9} - 4 = -\frac{35}{9}$. Therefore, that seems to be the answer.

But let me check if there's a case where $\frac{1}{b^2} - a^2$ is larger. For example, suppose $a$ is more negative, but $b$ is smaller? Wait, $b$ can't be smaller than 3. If $a$ is more negative, $a^2$ increases, so $\frac{1}{b^2} - a^2$ decreases. If $a$ is less negative (closer to zero), $a^2$ decreases, so $\frac{1}{b^2} - a^2$ increases. Therefore, the maximum occurs at the minimal $a^2$ and maximal $\frac{1}{b^2}$, which is $a=-2$ and $b=3$. Therefore, the answer is $-35/9$. But the problem says "the greatest possible value", so maybe that's correct? But maybe I need to check if there's a different combination where the expression is larger?

Wait, let me think again. Suppose I take $a = -2$ and $b = 3$, then the original expression is $\left(-2 + \frac{1}{3}\right)\left(\frac{1}{3} - (-2)\right)$. Let me compute that:

First, compute $-2 + 1/3 = -5/3$.

Then compute $1/3 - (-2) = 1/3 + 2 = 7/3$.

Multiply them: $-5/3 * 7/3 = -35/9$. So that's correct.

Alternatively, if I take $a = -2$ and $b = 5$, then the original expression is $\left(-2 + 1/5\right)\left(1/5 - (-2)\right) = (-9/5)(11/5) = -99/25 ≈ -3.96$, which is less than -35/9 ≈ -3.888... So indeed, -35/9 is larger (i.e., less negative).

Alternatively, if I take $a = -3$ and $b = 3$, then the original expression is $\left(-3 + 1/3\right)\left(1/3 - (-3)\right) = (-8/3)(10/3) = -80/9 ≈ -8.888...$, which is worse.

Alternatively, if I take $a = -2$ and $b = 3$, which gives -35/9. What if I take $a = -2$ and $b = 3.5$? Let me compute that. $b = 3.5 = 7/2$. Then $\frac{1}{b^2} = 4/49 ≈ 0.0816$. Then $a^2 = 4$. So the expression is 4/49 - 4 ≈ -3.918, which is worse than -35/9 ≈ -3.888. So worse.

Alternatively, if I take $a = -2$ and $b = 3$, which is the maximum of $\frac{1}{b^2}$ and minimum of $a^2$, so that's the best.

Alternatively, what if I take $a$ slightly more negative but $b$ slightly smaller? But $b$ can't be smaller than 3. So no.

Alternatively, maybe there's a case where $\frac{1}{b^2} - a^2$ is positive? Let me check with $a = -1/2$, but $a$ is between -6 and -2, so that's not allowed. How about $a = -2$ and $b = 1$? But $b$ is between 3 and 5. So even if $b$ is 3, $\frac{1}{b^2}$ is 1/9 ≈ 0.111, and $a^2 = 4$. So 0.111 - 4 is negative. Therefore, the expression is always negative. Therefore, the maximum value is indeed -35/9. Therefore, the answer is \boxed{-\dfrac{35}{9}}.

Wait, but the problem says "the greatest possible value". But maybe I made a mistake in the sign? Let me check again. The original expression is $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me plug in $a = -2$ and $b = 3$:

First term: $a + 1/b = -2 + 1/3 = -5/3$

Second term: $1/b - a = 1/3 - (-2) = 1/3 + 2 = 7/3$

Multiply them: (-5/3)(7/3) = -35/9. So that's correct. So the value is negative. But maybe there's a different combination where the product is positive?

Wait, suppose $a$ is negative and $1/b - a$ is positive? Let me think. Since $a$ is negative, $-a$ is positive. So $1/b - a = 1/b + |a|$. Since both $1/b$ and $|a|$ are positive, this term is positive. The first term is $a + 1/b$. Since $a$ is negative and $1/b$ is positive, depending on the values, this could be positive or negative. For example, if $a$ is close to zero, then $a + 1/b$ is positive. If $a$ is more negative, then $a + 1/b$ could be negative.

Therefore, the product $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$ could be positive or negative. For example, if $a + 1/b$ is positive and $1/b - a$ is positive, then the product is positive. If $a + 1/b$ is negative and $1/b - a$ is positive, then the product is negative. Therefore, maybe there are cases where the product is positive, and we need to find the maximum among those?

So perhaps my previous reasoning was incomplete. Let me think again. So the expression is $\frac{1}{b^2} - a^2$, which can be positive or negative. So to find the maximum value, we need to consider both possibilities.

So first, let's find when $\frac{1}{b^2} - a^2$ is positive. That is, when $\frac{1}{b^2} > a^2$. Which implies that $|a| < \frac{1}{b}$. Since $a$ is negative, $|a| = -a$. Therefore, the inequality becomes $-a < \frac{1}{b}$, which implies $a > -\frac{1}{b}$. But since $a$ is between -6 and -2, and $b$ is between 3 and 5, let's see for which values of $a$ and $b$ this holds.

So $a > -\frac{1}{b}$. Since $b$ is between 3 and 5, $-\frac{1}{b}$ is between $-1/5$ and $-1/3$. But $a$ is between -6 and -2, which is all less than $-1/3$. Therefore, $a > -\frac{1}{b}$ would mean that $a$ is greater than a number between $-1/5$ and $-1/3$, but since $a$ is at most -2, which is less than $-1/3$, there is no overlap. Therefore, there are no values of $a$ and $b$ in the given ranges where $|a| < \frac{1}{b}$. Therefore, $\frac{1}{b^2} - a^2$ is always negative. Therefore, the expression is always negative, and the maximum value is the one closest to zero, which is when $\frac{1}{b^2}$ is as large as possible and $a^2$ as small as possible. Therefore, the maximum value is indeed $-35/9$.

But let me check with specific numbers. Let me take $a = -2$ and $b = 3$. Then $|a| = 2$, and $\frac{1}{b} = 1/3 ≈ 0.333$. So $|a| = 2 > 1/3$, so $\frac{1}{b^2} - a^2$ is negative. If I take $a = -1.5$, but $a$ is between -6 and -2, so $a = -1.5$ is not allowed. If I take $a = -2$ and $b = 3$, then $|a| = 2$, which is greater than $1/b = 1/3$. Therefore, the expression is negative. Therefore, there are no cases where the expression is positive. Therefore, the maximum value is indeed $-35/9$.

But let me check another case. Suppose $a = -2$ and $b = 3$. Then the original expression is $\left(-2 + 1/3\right)\left(1/3 - (-2)\right) = (-5/3)(7/3) = -35/9$. If I take $a = -2$ and $b = 3.5$, then the expression is $\left(-2 + 1/3.5\right)\left(1/3.5 - (-2)\right)$. Let me compute that:

First, $1/3.5 = 2/7 ≈ 0.2857$. So $-2 + 2/7 = -12/7 ≈ -1.714$. Then $1/3.5 - (-2) = 2/7 + 2 = 16/7 ≈ 2.2857$. Multiply them: (-12/7)(16/7) = -192/49 ≈ -3.918, which is less than -35/9 ≈ -3.888. So worse.

If I take $a = -2$ and $b = 3$, which gives -35/9 ≈ -3.888. If I take $a = -2$ and $b = 3$, that's the maximum. If I take $a = -2$ and $b = 3$, that's the best. If I take $a = -2$ and $b = 3$, that's the maximum. Therefore, the answer is -35/9.

But the problem says "the greatest possible value". But maybe I need to check if there's another combination where the expression is larger? For example, if I take $a$ slightly more negative but $b$ slightly larger? Wait, but $b$ larger would make $\frac{1}{b^2}$ smaller, and $a$ more negative would make $a^2$ larger. So both terms would make the expression smaller. Therefore, no.

Alternatively, if I take $a$ slightly less negative (closer to zero) but $b$ slightly larger? Wait, $a$ can't be closer to zero than -2. So if I take $a = -2$ and $b = 3$, that's the closest to zero for $a$ and the smallest $b$. Therefore, that's the best.

Alternatively, maybe there's a case where $a + 1/b$ is positive and $1/b - a$ is positive, leading to a positive product? But as we saw earlier, since $a$ is between -6 and -2, and $b$ is between 3 and 5, $a + 1/b$ is $a +$ something positive. Since $a$ is negative, but how big is $1/b$? For $b = 3$, $1/b = 1/3 ≈ 0.333$. So $a + 1/b$ would be $-2 + 0.333 ≈ -1.666$, which is negative. For $a = -2$ and $b = 3$, it's negative. If $a$ is closer to zero, say $a = -2$, but that's the closest. If $a$ is -2, then $a + 1/b$ is negative. If $a$ is more negative, like -3, then $a + 1/b$ is even more negative. Therefore, $a + 1/b$ is always negative. Therefore, the first term is negative, the second term is positive, so the product is negative. Therefore, the expression is always negative, and the maximum value is -35/9.

Therefore, I think that is the answer. However, the problem says "the greatest possible value", and maybe I need to check if there's a different approach where the answer is positive? Let me think again.

Wait, maybe I made a mistake in the sign when simplifying? Let me check again:

Original expression: $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me think of it as $(\frac{1}{b} + a)(\frac{1}{b} - a)$. Which is $\left(\frac{1}{b}\right)^2 - a^2$. So that's correct. Therefore, the expression is $\frac{1}{b^2} - a^2$. Therefore, if I want to maximize this, I need to maximize $\frac{1}{b^2}$ and minimize $a^2$. Therefore, the answer is $\frac{1}{9} - 4 = -\frac{35}{9}$. Therefore, I think that is correct.

But maybe the problem expects a positive answer? Let me check once more with different values. Suppose $a = -2$ and $b = 3$, then the original expression is $\left(-2 + 1/3\right)\left(1/3 - (-2)\right) = (-5/3)(7/3) = -35/9$. If I take $a = -2$ and $b = 3$, that's the maximum. If I take $a = -2$ and $b = 3$, that's the maximum. Therefore, I think that is correct.

Alternatively, maybe I need to consider the expression as a function of $a$ and $b$ and find its maximum. Let me think of it as $f(a, b) = \frac{1}{b^2} - a^2$. Since $a$ and $b$ are independent variables, the maximum occurs at the maximum of $\frac{1}{b^2}$ and the minimum of $a^2$. Therefore, the maximum of $f(a, b)$ is $\frac{1}{3^2} - (-2)^2 = \frac{1}{9} - 4 = -\frac{35}{9}$. Therefore, that's the answer.

Alternatively, maybe there's a mistake in the problem statement? Or maybe I misread the problem? Let me check again.

The problem says: If $-6 \leq a \leq -2$ and $3 \leq b \leq 5$, what is the greatest possible value of $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$? Express your answer as a common fraction.

No, I think I read it correctly. So the answer is -35/9. But maybe the problem expects a positive value? Let me check once again with different values. Suppose $a = -1/2$ and $b = 1$, but those values are outside the given ranges. If $a = -2$ and $b = 3$, then the value is -35/9. If I take $a = -2$ and $b = 3$, that's the maximum. Therefore, I think that is correct.

Alternatively, maybe I need to consider that the original expression can be rewritten as $\frac{1}{b^2} - a^2$, but maybe there's a different way to write it? For example, maybe factor it differently? Let me see:

Alternatively, $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right) = \left(\frac{1}{b} + a\right)\left(\frac{1}{b} - a\right) = \left(\frac{1}{b}\right)^2 - a^2$. So that's the same as before. Therefore, no other way.

Alternatively, maybe I can think of this as a quadratic in terms of $a$ or $b$? For example, if I fix $b$, then the expression is $-a^2 + \frac{1}{b^2}$, which is a downward opening parabola in terms of $a$, so its maximum occurs at the vertex. But since $a$ is constrained between -6 and -2, the maximum of $-a^2 + \frac{1}{b^2}$ would occur at the minimum of $a^2$, which is at $a = -2$. Similarly, if I fix $a$, then the expression is $\frac{1}{b^2} - a^2$, which is a decreasing function in $b$, so maximum at $b = 3$. Therefore, again, the maximum occurs at $a = -2$ and $b = 3$, giving $-35/9$.

Therefore, I think that is the correct answer. Therefore, the greatest possible value is $-35/9$, which is a common fraction. Therefore, the answer is $\boxed{-\dfrac{35}{9}}$.

**Final Answer**
\boxed{-\dfrac{35}{9}}

To find the **greatest possible value** of the expression
$$
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)
$$
given the constraints
$$
-6 \leq a \leq -2 \quad \text{and} \quad 3 \leq b \leq 5,
$$
we begin by simplifying the expression.

---

### Step 1: Simplify the Expression

We recognize the expression as a **difference of squares**:

$$
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right) = \left(\frac{1}{b}\right)^2 - a^2 = \frac{1}{b^2} - a^2
$$

So, the original expression simplifies to:

$$
\frac{1}{b^2} - a^2
$$

---

### Step 2: Analyze the Expression

We now want to **maximize** the expression:

$$
\frac{1}{b^2} - a^2
$$

This is a function of two variables, $a$ and $b$, with the following constraints:

- $a \in [-6, -2]$
- $b \in [3, 5]$

To **maximize** the expression, we need to:

- **Maximize** $\frac{1}{b^2}$, which is a **decreasing** function of $b$. So, it is maximized when $b$ is **smallest**, i.e., $b = 3$.
- **Minimize** $a^2$, which is a **non-negative** function. Since $a$ is negative, $a^2$ is minimized when $a$ is **closest to zero**, i.e., $a = -2$.

---

### Step 3: Plug in the Optimal Values

Using the values that maximize $\frac{1}{b^2}$ and minimize $a^2$:

- $b = 3 \Rightarrow \frac{1}{b^2} = \frac{1}{9}$
- $a = -2 \Rightarrow a^2 = 4$

So the expression becomes:

$$
\frac{1}{9} - 4 = -\frac{35}{9}
$$

---

### Step 4: Confirm the Expression is Always Negative

Since $a$ is negative and $b$ is positive, the original expression is a product of a **negative** and a **positive** term, so the result is **always negative**. Therefore, the **greatest possible value** is the one **closest to zero**, which occurs when $\frac{1}{b^2}$ is as large as possible and $a^2$ is as small as possible.

---

### Final Answer

$$
\boxed{-\dfrac{35}{9}}
$$

Solution: The given expression expands to $\frac{1}{b^2} - a^2$.  Thus we want $b$ to have the least possible magnitude and $a$ to also have the least possible magnitude.  Our maximum value is therefore $\frac{1}{3^2} - (-2)^2 = \boxed{-\frac{35}{9}}$.