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 see, both numerator and denominator are divisible by 2. Dividing numerator and denominator by 2 gives 17/50. Is that the simplest form? 17 is a prime number, so unless 50 is divisible by 17, which it isn't, so yes, 17/50 is the simplified fraction.

Wait, let me just 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 right.

Alternatively, maybe I can approach this problem using another method to cross-verify. Let me think. Another way is to find numbers divisible by 2 and not by 3. So, numbers divisible by 2 are 2, 4, 6, 8, ..., 100. Now, from these numbers, we need to exclude those divisible by 3. So, numbers divisible by both 2 and 3 are numbers divisible by 6, as before. So, the count is same as before.

Alternatively, maybe I can think in terms of inclusion-exclusion. But I think the method I used is correct.

Alternatively, maybe I can calculate the numbers not divisible by 3 and then see how many of them are divisible by 2? Wait, but that might complicate. Let me try.

Total numbers not divisible by 3: total numbers minus numbers divisible by 3. Numbers divisible by 3 between 1 and 100: floor(100/3) = 33 (since 3*33=99). Therefore, numbers not divisible by 3: 100 - 33 = 67. But then, how many of these 67 numbers are divisible by 2? Hmm, that might not be straightforward. Because numbers not divisible by 3 can be even or odd. So, perhaps this approach is more complicated. Let me see.

Alternatively, maybe I can use the principle of inclusion-exclusion. The number 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. Since the question is about numbers divisible by 2 and not by 3, which is numbers divisible by 2 minus numbers divisible by both 2 and 3, which is exactly what I did before. So, that seems consistent.

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 to think is that numbers divisible by 2 but not by 3 are numbers divisible by 2 and co-prime with 3? Wait, not exactly. Because numbers divisible by 2 and not by 3 are numbers divisible by 2, but not divisible by 3. So, they can be divisible by other numbers, but not by 3. So, maybe that's not helpful.

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, perhaps using the formula:

Number of elements in A but not in B = |A| - |A ∩ B|, where A is the set of numbers divisible by 2, and B is the set of numbers divisible by 3. So, that's exactly what I did earlier. So, that gives me 50 - 16 = 34. So, that seems correct.

Alternatively, maybe I can list some numbers to check. Let me take a smaller set, say numbers from 1 to 6. Let's see:

Numbers divisible by 2: 2, 4, 6. Numbers divisible by 3: 3, 6. Numbers divisible by both 2 and 3: 6. So, numbers divisible by 2 but not by 3: 2, 4. So, 2 numbers. Total numbers: 6. Probability is 2/6 = 1/3. Let me check with my formula. Numbers divisible by 2: 3. Numbers divisible by 6: 1. So, 3 - 1 = 2. 2/6 = 1/3. Correct. So, that works.

Another example: numbers from 1 to 12. Numbers divisible by 2: 6 numbers (2,4,6,8,10,12). Numbers divisible by 6: 2 numbers (6, 12). So, numbers divisible by 2 but not by 3: 6 - 2 = 4 numbers: 2,4,8,10. Let me check: 2 is not divisible by 3, 4 is not, 6 is divisible by 3, 8 is not, 10 is not, 12 is divisible by 3. So, yes, 4 numbers. So, 4/12 = 1/3. Wait, same as before? Wait, but in this case, numbers divisible by 2 but not by 3 is 4, which is 1/3 of 12. Wait, but in the previous case, 1/3. So, maybe in general, the probability is 1/3? Wait, but in our original problem, the answer was 17/50, which is approximately 0.34, which is close to 1/3 (approx 0.333...). So, maybe that's a coincidence?

Wait, but let me check with numbers from 1 to 6, probability is 1/3. From 1 to 12, same. Let me check with numbers from 1 to 3. Numbers divisible by 2: 1 number (2). Numbers divisible by 6: 0. So, numbers divisible by 2 but not by 3: 1. Probability 1/3. Hmm, same. Wait, so maybe in general, the probability is 1/3? But in our original problem, the answer is 17/50, which is 0.34, which is slightly higher than 1/3. So, why is that?

Wait, maybe because 100 isn't a multiple of 6? Let me check. Let me think. If the total numbers are a multiple of 6, then the number of numbers divisible by 2 but not by 3 would be (total / 2) - (total / 6) = (total / 2) - (total / 6) = total / 3. So, in that case, the probability would be 1/3. But if the total numbers aren't a multiple of 6, then it's slightly different.

So, in our case, total numbers is 100. Let me check. Let me compute 100 / 2 - 100 / 6. But since we can't have fractions, we need to take floor(100 / 6) = 16. So, 50 - 16 = 34. So, 34 / 100 = 17 / 50. Which is 0.34. Whereas if the total numbers were 6, 12, etc., it would be exactly 1/3. So, in this case, since 100 isn't a multiple of 6, the answer is slightly different.

So, that seems okay. So, my calculation seems correct. Let me check with another example. Let's take numbers from 1 to 7. Numbers divisible by 2: 3 numbers (2,4,6). Numbers divisible by 6: 1 number (6). So, numbers divisible by 2 but not by 3: 2 numbers (2,4). So, probability is 2/7 ≈ 0.2857. If I use the formula: floor(7 / 2) = 3, floor(7 / 6) = 1, so 3 - 1 = 2. 2/7. Which is correct. So, that works. So, the formula seems to hold.

Therefore, going back to the original problem, my calculation of 34 numbers is correct, leading to 17/50. Therefore, the probability is 17/50.

But just to be thorough, let me check the count of numbers divisible by 6 between 1 and 100 again. The numbers divisible by 6 are 6, 12, 18, ..., up to the largest number less than or equal to 100. Let me compute how many terms are in this sequence. The first term is 6, the last term is 96 (since 6*16=96, and 6*17=102 which is over 100). Therefore, there are 16 numbers. So, that's correct. Therefore, numbers divisible by 6: 16. Therefore, numbers divisible by 2 but not by 3: 50 - 16 = 34. So, 34/100 = 17/50. So, that's correct.

Alternatively, maybe I can think of it as follows: For numbers from 1 to 100, how many are divisible by 2 but not by 3. So, numbers divisible by 2 are 2, 4, 6, 8, ..., 100. Now, among these numbers, which are divisible by 3? Those numbers which are multiples of 6. So, as before. So, the count is 16. Therefore, subtracting gives 34. So, same result.

Alternatively, maybe I can think of the numbers divisible by 2 but not by 3 as numbers congruent to 2 mod 6 or 4 mod 6. Because numbers divisible by 2 are either 0 mod 6 or 2 mod 6 or 4 mod 6. Wait, numbers divisible by 2 are even numbers. So, in mod 6, even numbers can be 0, 2, 4 mod 6. But numbers divisible by 3 are 0 mod 3. So, numbers divisible by 2 but not by 3 are numbers that are 2 mod 6 or 4 mod 6. Because 0 mod 6 is divisible by both 2 and 3. Therefore, numbers that are 2 mod 6 or 4 mod 6. So, how many numbers between 1 and 100 are congruent to 2 or 4 mod 6?

Well, for each residue class mod 6, how many numbers are there?

Total numbers: 100. Let me think. The number of numbers congruent to 2 mod 6 is floor((100 - 2)/6) + 1. Similarly for 4 mod 6. Let me compute that.

For numbers congruent to 2 mod 6: The first number is 2, then 8, 14,..., up to the largest number less than or equal to 100. Let me find the last term. Let me solve 2 + 6k ≤ 100. So, 6k ≤ 98 => k ≤ 98/6 ≈ 16.333. So, k can be 0 to 16, which gives 17 numbers. Wait, but let me check. If k=0: 2, k=1: 8,..., k=16: 2 + 6*16 = 2 + 96 = 98. Then the next term would be 104, which is over 100. So, there are 17 numbers congruent to 2 mod 6.

Similarly, numbers congruent to 4 mod 6: first number 4, then 10, 16,..., up to the largest number ≤ 100. Let me solve 4 + 6k ≤ 100. 6k ≤ 96 => k ≤ 16. So, k=0 to 16, which is 17 numbers. The last term is 4 + 6*16 = 4 + 96 = 100. So, that's exactly 100. Therefore, numbers congruent to 4 mod 6: 17 numbers.

Therefore, total numbers congruent to 2 or 4 mod 6: 17 + 17 = 34. Which matches the previous count. Therefore, 34 numbers. Therefore, probability is 34/100 = 17/50. So, that's another way to get the same answer, which confirms that my answer is correct.

Therefore, I think I can be confident that 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.

---

### 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:

$$
\frac{34}{100}
$$

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 see, both numerator and denominator are divisible by 2. Dividing numerator and denominator by 2 gives 17/50. Is that the simplest form? 17 is a prime number, so unless 50 is divisible by 17, which it isn't, so yes, 17/50 is the simplified fraction.

Wait, let me just 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 right.

Alternatively, maybe I can approach this problem using another method to cross-verify. Let me think. Another way is to find numbers divisible by 2 and not by 3. So, numbers divisible by 2 are 2, 4, 6, 8, ..., 100. Now, from these numbers, we need to exclude those divisible by 3. So, numbers divisible by both 2 and 3 are numbers divisible by 6, as before. So, the count is same as before.

Alternatively, maybe I can think in terms of inclusion-exclusion. But I think the method I used is correct.

Alternatively, maybe I can calculate the numbers not divisible by 3 and then see how many of them are divisible by 2? Wait, but that might complicate. Let me try.

Total numbers not divisible by 3: total numbers minus numbers divisible by 3. Numbers divisible by 3 between 1 and 100: floor(100/3) = 33 (since 3*33=99). Therefore, numbers not divisible by 3: 100 - 33 = 67. But then, how many of these 67 numbers are divisible by 2? Hmm, that might not be straightforward. Because numbers not divisible by 3 can be even or odd. So, perhaps this approach is more complicated. Let me see.

Alternatively, maybe I can use the principle of inclusion-exclusion. The number 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. Since the question is about numbers divisible by 2 and not by 3, which is numbers divisible by 2 minus numbers divisible by both 2 and 3, which is exactly what I did before. So, that seems consistent.

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 to think is that numbers divisible by 2 but not by 3 are numbers divisible by 2 and co-prime with 3? Wait, not exactly. Because numbers divisible by 2 and not by 3 are numbers divisible by 2, but not divisible by 3. So, they can be divisible by other numbers, but not by 3. So, maybe that's not helpful.

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, perhaps using the formula:

Number of elements in A but not in B = |A| - |A ∩ B|, where A is the set of numbers divisible by 2, and B is the set of numbers divisible by 3. So, that's exactly what I did earlier. So, that gives me 50 - 16 = 34. So, that seems correct.

Alternatively, maybe I can list some numbers to check. Let me take a smaller set, say numbers from 1 to 6. Let's see:

Numbers divisible by 2: 2, 4, 6. Numbers divisible by 3: 3, 6. Numbers divisible by both 2 and 3: 6. So, numbers divisible by 2 but not by 3: 2, 4. So, 2 numbers. Total numbers: 6. Probability is 2/6 = 1/3. Let me check with my formula. Numbers divisible by 2: 3. Numbers divisible by 6: 1. So, 3 - 1 = 2. 2/6 = 1/3. Correct. So, that works.

Another example: numbers from 1 to 12. Numbers divisible by 2: 6 numbers (2,4,6,8,10,12). Numbers divisible by 6: 2 numbers (6, 12). So, numbers divisible by 2 but not by 3: 6 - 2 = 4 numbers: 2,4,8,10. Let me check: 2 is not divisible by 3, 4 is not, 6 is divisible by 3, 8 is not, 10 is not, 12 is divisible by 3. So, yes, 4 numbers. So, 4/12 = 1/3. Wait, same as before? Wait, but in this case, numbers divisible by 2 but not by 3 is 4, which is 1/3 of 12. Wait, but in the previous case, 1/3. So, maybe in general, the probability is 1/3? Wait, but in our original problem, the answer was 17/50, which is approximately 0.34, which is close to 1/3 (approx 0.333...). So, maybe that's a coincidence?

Wait, but let me check with numbers from 1 to 6, probability is 1/3. From 1 to 12, same. Let me check with numbers from 1 to 3. Numbers divisible by 2: 1 number (2). Numbers divisible by 6: 0. So, numbers divisible by 2 but not by 3: 1. Probability 1/3. Hmm, same. Wait, so maybe in general, the probability is 1/3? But in our original problem, the answer is 17/50, which is 0.34, which is slightly higher than 1/3. So, why is that?

Wait, maybe because 100 isn't a multiple of 6? Let me check. Let me think. If the total numbers are a multiple of 6, then the number of numbers divisible by 2 but not by 3 would be (total / 2) - (total / 6) = (total / 2) - (total / 6) = total / 3. So, in that case, the probability would be 1/3. But if the total numbers aren't a multiple of 6, then it's slightly different.

So, in our case, total numbers is 100. Let me check. Let me compute 100 / 2 - 100 / 6. But since we can't have fractions, we need to take floor(100 / 6) = 16. So, 50 - 16 = 34. So, 34 / 100 = 17 / 50. Which is 0.34. Whereas if the total numbers were 6, 12, etc., it would be exactly 1/3. So, in this case, since 100 isn't a multiple of 6, the answer is slightly different.

So, that seems okay. So, my calculation seems correct. Let me check with another example. Let's take numbers from 1 to 7. Numbers divisible by 2: 3 numbers (2,4,6). Numbers divisible by 6: 1 number (6). So, numbers divisible by 2 but not by 3: 2 numbers (2,4). So, probability is 2/7 ≈ 0.2857. If I use the formula: floor(7 / 2) = 3, floor(7 / 6) = 1, so 3 - 1 = 2. 2/7. Which is correct. So, that works. So, the formula seems to hold.

Therefore, going back to the original problem, my calculation of 34 numbers is correct, leading to 17/50. Therefore, the probability is 17/50.

But just to be thorough, let me check the count of numbers divisible by 6 between 1 and 100 again. The numbers divisible by 6 are 6, 12, 18, ..., up to the largest number less than or equal to 100. Let me compute how many terms are in this sequence. The first term is 6, the last term is 96 (since 6*16=96, and 6*17=102 which is over 100). Therefore, there are 16 numbers. So, that's correct. Therefore, numbers divisible by 6: 16. Therefore, numbers divisible by 2 but not by 3: 50 - 16 = 34. So, 34/100 = 17/50. So, that's correct.

Alternatively, maybe I can think of it as follows: For numbers from 1 to 100, how many are divisible by 2 but not by 3. So, numbers divisible by 2 are 2, 4, 6, 8, ..., 100. Now, among these numbers, which are divisible by 3? Those numbers which are multiples of 6. So, as before. So, the count is 16. Therefore, subtracting gives 34. So, same result.

Alternatively, maybe I can think of the numbers divisible by 2 but not by 3 as numbers congruent to 2 mod 6 or 4 mod 6. Because numbers divisible by 2 are either 0 mod 6 or 2 mod 6 or 4 mod 6. Wait, numbers divisible by 2 are even numbers. So, in mod 6, even numbers can be 0, 2, 4 mod 6. But numbers divisible by 3 are 0 mod 3. So, numbers divisible by 2 but not by 3 are numbers that are 2 mod 6 or 4 mod 6. Because 0 mod 6 is divisible by both 2 and 3. Therefore, numbers that are 2 mod 6 or 4 mod 6. So, how many numbers between 1 and 100 are congruent to 2 or 4 mod 6?

Well, for each residue class mod 6, how many numbers are there?

Total numbers: 100. Let me think. The number of numbers congruent to 2 mod 6 is floor((100 - 2)/6) + 1. Similarly for 4 mod 6. Let me compute that.

For numbers congruent to 2 mod 6: The first number is 2, then 8, 14,..., up to the largest number less than or equal to 100. Let me find the last term. Let me solve 2 + 6k ≤ 100. So, 6k ≤ 98 => k ≤ 98/6 ≈ 16.333. So, k can be 0 to 16, which gives 17 numbers. Wait, but let me check. If k=0: 2, k=1: 8,..., k=16: 2 + 6*16 = 2 + 96 = 98. Then the next term would be 104, which is over 100. So, there are 17 numbers congruent to 2 mod 6.

Similarly, numbers congruent to 4 mod 6: first number 4, then 10, 16,..., up to the largest number ≤ 100. Let me solve 4 + 6k ≤ 100. 6k ≤ 96 => k ≤ 16. So, k=0 to 16, which is 17 numbers. The last term is 4 + 6*16 = 4 + 96 = 100. So, that's exactly 100. Therefore, numbers congruent to 4 mod 6: 17 numbers.

Therefore, total numbers congruent to 2 or 4 mod 6: 17 + 17 = 34. Which matches the previous count. Therefore, 34 numbers. Therefore, probability is 34/100 = 17/50. So, that's another way to get the same answer, which confirms that my answer is correct.

Therefore, I think I can be confident that 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.

---

### 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:

$$
\frac{34}{100}
$$

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, we need to find the minimum value of $a^2$. Given that $a$ is between -6 and -2, inclusive. Let me think about $a^2$. Since $a$ is negative, squaring it will make it positive. The function $f(a) = a^2$ is a parabola opening upwards, so its minimum occurs at the vertex. However, since $a$ is restricted to the interval $[-6, -2]$, which is entirely on the negative side, the minimum of $a^2$ will occur at the point closest to zero. Since $a$ is between -6 and -2, the closest to zero is -2. Therefore, the minimum value of $a^2$ is $(-2)^2 = 4$.

Wait, but hold on. Let me check that again. If $a$ is between -6 and -2, then $a^2$ will be between $(-6)^2 = 36$ and $(-2)^2 = 4$. So yes, the minimum of $a^2$ is 4, and the maximum is 36. Therefore, the minimum value of $a^2$ is 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 negative. But maybe I need to check if there's a combination where $\frac{1}{b^2}$ is larger and $a^2$ is smaller? Wait, but we already took the maximum of $\frac{1}{b^2}$ and the minimum of $a^2$. So that should give the maximum of the entire expression. But let me check if that's correct.

Alternatively, maybe there's a case where even though $\frac{1}{b^2}$ is smaller, but $a^2$ is much smaller, leading to a larger overall value? Wait, but since $\frac{1}{b^2}$ is positive and $a^2$ is positive, so the expression is $\frac{1}{b^2} - a^2$. So if I want to maximize this, I need to maximize the first term and minimize the second term. So that should be correct.

But let me compute $\frac{1}{9} - 4 = \frac{1}{9} - \frac{36}{9} = -\frac{35}{9}$. But is this the maximum? Wait, maybe there's a case where $\frac{1}{b^2}$ is smaller but $a^2$ is even smaller? But $a^2$ can't be smaller than 4. So that's the minimum. So maybe that is the maximum? But that seems like a negative number. Is there a possibility that the expression can be positive?

Wait, let me check with some numbers. Let me 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 term: $-2 + 1/3 = -5/3$

Second term: $1/3 + 2 = 7/3$

Multiply them: $(-5/3)(7/3) = -35/9$, which is the same as before. So that's negative. But maybe if I take other values?

Wait, let me try $a = -2$ and $b = 5$. Then the expression is $\left(-2 + 1/5\right)\left(1/5 - (-2)\right)$. Let me compute:

First term: $-2 + 1/5 = -9/5$

Second term: $1/5 + 2 = 11/5$

Multiply: $(-9/5)(11/5) = -99/25 ≈ -3.96$, which is more negative than -35/9 ≈ -3.888... So worse.

What if I take $a = -6$ and $b = 3$? Then:

First term: $-6 + 1/3 = -17/3$

Second term: $1/3 - (-6) = 1/3 + 6 = 19/3$

Multiply: (-17/3)(19/3) = -323/9 ≈ -35.888... which is even worse.

Alternatively, if I take $a = -2$ and $b = 3$, which gives -35/9 ≈ -3.888...

Wait, but maybe if I take $a$ closer to zero? Wait, but $a$ is between -6 and -2. So the closest to zero is -2. So maybe that's the minimum of $a^2$. But then, if I take $a = -2$ and $b = 3$, which gives the maximum of $\frac{1}{b^2}$ and minimum of $a^2$, but the result is negative. Is there a way to get a positive value?

Wait, let me think again. The original expression is $\frac{1}{b^2} - a^2$. If $\frac{1}{b^2} > a^2$, then the expression is positive. So maybe there are values where $\frac{1}{b^2} > a^2$?

But given that $a$ is between -6 and -2, so $a^2$ is between 4 and 36. And $\frac{1}{b^2}$ is between $\frac{1}{25}$ and $\frac{1}{9}$. So $\frac{1}{b^2}$ is at most $\frac{1}{9} ≈ 0.111$, while $a^2$ is at least 4. Therefore, $\frac{1}{b^2} - a^2$ is always negative. Therefore, the maximum value is the least negative one, 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 $\frac{1}{9} - 4 = -\frac{35}{9}$. But the problem says "greatest possible value", so maybe that is the answer? But let me check again.

Wait, but 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 expand it step by step:

Multiply $a$ by $\frac{1}{b}$: $a \cdot \frac{1}{b} = \frac{a}{b}$

Multiply $a$ by $-a$: $-a^2$

Multiply $\frac{1}{b}$ by $\frac{1}{b}$: $\frac{1}{b^2}$

Multiply $\frac{1}{b}$ by $-a$: $-\frac{a}{b}$

So adding all these terms together:

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

The $\frac{a}{b}$ and $-\frac{a}{b}$ cancel out, leaving $-a^2 + \frac{1}{b^2}$, which is the same as $\frac{1}{b^2} - a^2$. So that's correct. Therefore, the expression simplifies to $\frac{1}{b^2} - a^2$, which is always negative, as we saw. Therefore, the maximum value is indeed when $\frac{1}{b^2}$ is maximum and $a^2$ is minimum, which is $\frac{1}{9} - 4 = -\frac{35}{9}$. But let me check if there's another combination where maybe $\frac{1}{b^2} - a^2$ is larger (i.e., less negative). For example, if I take $a$ slightly larger (closer to zero) and $b$ slightly larger? Wait, but $a$ can't be larger than -2. So the minimum $a^2$ is 4. If I take $a = -2$, then $a^2 = 4$. If I take $a$ closer to -2, but since -2 is the upper bound, there's no closer. So that's the minimum. Similarly, for $b$, the maximum of $\frac{1}{b^2}$ is at $b = 3$. Therefore, that combination gives the maximum value of the expression.

But let me check another possibility. Suppose I take $a = -2$ and $b = 3$, which gives $\frac{1}{9} - 4 = -35/9$. What if I take $a = -2$ and $b = 5$? Then $\frac{1}{25} - 4 = -\frac{99}{25} ≈ -3.96$, which is more negative. If I take $a = -6$ and $b = 3$, then $\frac{1}{9} - 36 = -\frac{323}{9} ≈ -35.888...$. If I take $a = -2$ and $b = 3$, that's the least negative. So that seems to be the maximum.

But wait, maybe there's another way to approach this problem? Let me think. Maybe instead of simplifying first, I can consider the original expression and analyze it differently. Let me try that.

Original expression: $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me denote $x = a$ and $y = \frac{1}{b}$. Then the expression becomes $(x + y)(y - x) = y^2 - x^2$, which is the same as before. So that's the same result. Therefore, regardless of substitution, we get to the same expression. Therefore, the conclusion is the same.

Alternatively, maybe I can think of this as a function of two variables and try to find its maximum over the given intervals. But since we already simplified it to $\frac{1}{b^2} - a^2$, and since $a$ and $b$ are independent variables (the constraints on $a$ and $b$ are separate), then to maximize $\frac{1}{b^2} - a^2$, we need to maximize $\frac{1}{b^2}$ and minimize $a^2$. Therefore, the maximum occurs at $b = 3$ and $a = -2$, giving $\frac{1}{9} - 4 = -\frac{35}{9}$. Therefore, the answer should be $-\frac{35}{9}$. But let me check once more if there's a possibility that the maximum occurs at some other point where maybe $\frac{1}{b^2} - a^2$ is larger.

Wait, for example, suppose $a$ is not at its minimum $a^2$, but maybe there's a combination where even though $a^2$ is larger, $\frac{1}{b^2}$ is significantly larger? But since $\frac{1}{b^2}$ is at most $\frac{1}{9}$, and $a^2$ is at least 4, so even if $a^2$ is larger, say 36, then $\frac{1}{b^2} - a^2$ would be $\frac{1}{9} - 36 = -\frac{323}{9}$, which is worse. So no, that's not helpful.

Alternatively, if I take $b$ larger, making $\frac{1}{b^2}$ smaller, but maybe $a$ is such that $a^2$ is smaller? But $a$ can't be smaller in magnitude than -2. So $a^2$ can't be smaller than 4. Therefore, even if I take $b$ larger, making $\frac{1}{b^2}$ smaller, the expression becomes more negative. Therefore, the maximum is indeed at $b = 3$ and $a = -2$.

But let me check another angle. Suppose I consider the original expression $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Maybe there's a way to write this as something else? For example, maybe think of it as $\left(\frac{1}{b} + a\right)\left(\frac{1}{b} - a\right)$. Which is the same as $\left(\frac{1}{b}\right)^2 - a^2$, as before. So no difference.

Alternatively, maybe think of it as a quadratic in terms of $a$ or $b$. Let me try treating it as a quadratic in $a$. Let me write the expression:

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

So it's a quadratic in $a$ with coefficient of $a^2$ being -1, which is a downward opening parabola. Therefore, for fixed $b$, the maximum of this expression with respect to $a$ occurs at the vertex. But since the coefficient of $a^2$ is negative, the maximum is at the vertex. However, since $a$ is constrained between -6 and -2, the maximum for each fixed $b$ would be either at the vertex or at the endpoints of the interval for $a$. Wait, but the vertex of the quadratic $-a^2 + c$ (where $c$ is a constant) is at $a = 0$, but our $a$ is constrained to be between -6 and -2, which is entirely on the left of the vertex. Therefore, since the parabola opens downward, the maximum on the interval would be at the point closest to the vertex, which is $a = -2$. Therefore, for each fixed $b$, the maximum value of the expression is achieved when $a = -2$. Therefore, to maximize the entire expression, we need to maximize $\frac{1}{b^2}$, which again occurs at $b = 3$. Therefore, the maximum value is indeed at $a = -2$, $b = 3$, giving $-35/9$.

Alternatively, if I consider the expression as a function of $b$, for fixed $a$, then $\frac{1}{b^2}$ is a decreasing function of $b$, so for fixed $a$, the maximum occurs at the minimum $b$, which is 3. Therefore, again, the maximum occurs at $b = 3$ and $a = -2$.

Therefore, all approaches lead to the same conclusion. Therefore, the greatest possible value is $-35/9$. However, the problem says "greatest possible value", and since all values are negative, the "greatest" is the one closest to zero, which is indeed $-35/9$. Therefore, I think that is the answer.

But just to be thorough, let me check another point. Suppose $a = -2$ and $b = 3$, which gives us $-35/9$. What if I take $a = -2$ and $b = 3.5$? Then $\frac{1}{(3.5)^2} - (-2)^2 = \frac{1}{12.25} - 4 ≈ 0.0816 - 4 ≈ -3.9184$, which is more negative than -35/9 ≈ -3.888... So worse. Similarly, if I take $a = -2$ and $b = 3$, which is the maximum for $\frac{1}{b^2}$, so that's better.

Alternatively, if I take $a = -3$ and $b = 3$. Then $a^2 = 9$, so $\frac{1}{9} - 9 = -80/9 ≈ -8.888...$, which is worse. If I take $a = -2.5$ and $b = 3$, then $a^2 = 6.25$, so $\frac{1}{9} - 6.25 ≈ 0.111 - 6.25 ≈ -6.139$, which is worse than -35/9 ≈ -3.888. So indeed, the maximum occurs at $a = -2$, $b = 3$.

Wait, but let me check if there's a case where $a$ is not at its minimum $a^2$, but maybe $b$ is such that $\frac{1}{b^2}$ is larger? But $b$ can't be smaller than 3, so $\frac{1}{b^2}$ can't be larger than 1/9. Therefore, no.

Alternatively, maybe there's a combination where $\frac{1}{b^2} - a^2$ is larger? For example, if $a$ is negative and $b$ is such that $\frac{1}{b^2}$ is larger? But again, $b$ can't be smaller than 3. So no.

Alternatively, maybe I made a mistake in assuming that maximizing $\frac{1}{b^2}$ and minimizing $a^2$ gives the maximum of the expression. Let me think again. Suppose that $\frac{1}{b^2} - a^2$ is the expression. Since both terms are independent (because $a$ and $b$ are independent variables), then to maximize the expression, we need to maximize $\frac{1}{b^2}$ and minimize $a^2$. Therefore, yes, that is correct.

Alternatively, if there was a relationship between $a$ and $b$, maybe we could have a different approach, but since they are independent, we can treat them separately.

Therefore, I think my conclusion is correct. The maximum value is $-35/9$. However, the problem says "greatest possible value", and since all values are negative, the greatest is the one closest to zero, which is indeed $-35/9$. Therefore, the answer is $\boxed{-\dfrac{35}{9}}$.

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

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

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

Multiply them: (-5/3)(7/3) = -35/9. Correct. So that's negative. But maybe there's a case where the expression is positive? Let me think. Suppose $a$ is negative and $1/b$ is positive. Then $a + 1/b$ could be positive or negative depending on the values. Similarly, $1/b - a$ is positive because $1/b$ is positive and $-a$ is positive (since $a$ is negative). Therefore, the second term is positive. The first term, $a + 1/b$, could be positive or negative. If $a + 1/b$ is positive, then the product is positive. If it's negative, the product is negative. Therefore, maybe there are values where the expression is positive?

Wait, so maybe I need to check if there are values where $a + 1/b$ is positive. Let me see. For $a$ between -6 and -2, and $b$ between 3 and 5. Let me find if there exists $a$ and $b$ such that $a + 1/b > 0$. That would mean $a > -1/b$. Since $a$ is at least -6 and $b$ is at most 5, so $-1/b$ is at least -1/5 = -0.2. Therefore, $a$ is between -6 and -2, which is less than -0.2. Therefore, $a + 1/b$ is always less than -0.2 + 1/3 ≈ -0.2 + 0.333 ≈ 0.133? Wait, no. Wait, if $a$ is between -6 and -2, and $b$ is between 3 and 5, then $1/b$ is between 1/5 and 1/3. So $a + 1/b$ is between (-6 + 1/5) and (-2 + 1/3). Let me compute:

Lower bound: -6 + 1/5 = -5.8

Upper bound: -2 + 1/3 ≈ -1.666...

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 entire expression is always negative. Therefore, the maximum value is indeed the least negative one, which is -35/9. Therefore, my answer is correct.

But let me check with another example. Let me take $a = -2$ and $b = 3$. Then, as before, the expression is -35/9 ≈ -3.888...

If I take $a = -2$ and $b = 3.5$, then:

$a + 1/b = -2 + 2/7 ≈ -2 + 0.2857 ≈ -1.7143$

$1/b - a = 2/7 - (-2) = 2/7 + 2 ≈ 2.2857$

Multiply them: (-1.7143)(2.2857) ≈ -3.928, which is more negative than -3.888... So worse.

If I take $a = -2$ and $b = 3.1$:

$a + 1/b ≈ -2 + 1/3.1 ≈ -2 + 0.3226 ≈ -1.6774$

$1/b - a ≈ 0.3226 + 2 = 2.3226$

Multiply: (-1.6774)(2.3226) ≈ -3.896, which is still more negative than -3.888...

Wait, so maybe there's a point where it's less negative? Let me try with $a = -2$ and $b = 3$. Let me compute the exact value:

$a + 1/b = -2 + 1/3 = -5/3 ≈ -1.6667$

$1/b - a = 1/3 - (-2) = 1/3 + 2 = 7/3 ≈ 2.3333$

Multiply: (-5/3)(7/3) = -35/9 ≈ -3.888...

If I take $a = -2$ and $b = 3. Let me try $b = 3. Let me try $b = 3. Let me try $b = 3. Let me try $b = 3. Let me try $b = 3. Let me try $b = 3. Let me try $b = 3. Wait, I think I need to check if there's a value of $b$ where $\frac{1}{b^2} - a^2$ is larger. Wait, but since $a$ is fixed at -2, and $b$ is varying, then $\frac{1}{b^2}$ is maximized at $b = 3$, so that's the maximum. Therefore, even if I take $a = -2$ and $b = 3$, that's the maximum for that $a$. Therefore, the conclusion is correct.

Alternatively, suppose I take $a$ slightly larger than -2, say $a = -2 + \epsilon$, where $\epsilon$ is a small positive number. Then $a^2$ would be slightly less than 4. However, since $a$ is constrained to be at least -6 and at most -2, so $a$ can't be greater than -2. Therefore, the minimum $a^2$ is indeed 4 at $a = -2$. Therefore, even if I take $a$ slightly larger (closer to -2), but since $a$ can't be larger than -2, that's not allowed. Therefore, the minimum $a^2$ is 4.

Therefore, I think my answer is correct. Therefore, the greatest possible value is $-35/9$.

But just to be 100% sure, let me check if there's a case where $a$ is not at -2 and $b$ is not at 3, but the combination gives a higher value. For example, suppose $a = -3$ and $b = 3$. Then the expression is $\frac{1}{9} - 9 = -80/9 ≈ -8.888...$, which is worse. If $a = -2.5$ and $b = 3$, then $\frac{1}{9} - 6.25 ≈ -6.138...$, worse. If $a = -2$ and $b = 3.5$, as before, it's worse. If $a = -2$ and $b = 3$, it's -35/9. If I take $a = -2$ and $b = 3$, that's the best. Therefore, I think that is the answer.

Therefore, I think 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
$$
\frac{1}{b^2} - a^2
$$
under the given constraints.

- The term $\frac{1}{b^2}$ is **maximized** when $b$ is **as small as possible**, since $\frac{1}{b^2}$ is a **decreasing function** of $b$ for $b > 0$.
- The term $a^2$ is **minimized** when $a$ is **as close to zero as possible**, since $a^2$ is a **parabola** that increases as $|a|$ increases.

Given the constraints:
- $b \in [3, 5]$, so the **maximum** of $\frac{1}{b^2}$ occurs at $b = 3$:
  $$
  \frac{1}{b^2} = \frac{1}{3^2} = \frac{1}{9}
  $$
- $a \in [-6, -2]$, so the **minimum** of $a^2$ occurs at $a = -2$:
  $$
  a^2 = (-2)^2 = 4
  $$

---

### Step 3: Compute the Maximum Value

Substituting these values into the simplified expression:

$$
\frac{1}{b^2} - a^2 = \frac{1}{9} - 4 = \frac{1}{9} - \frac{36}{9} = -\frac{35}{9}
$$

---

### Step 4: Confirm the Result

We can verify that this is the **maximum** value by considering other combinations of $a$ and $b$ within the given ranges. For example:

- If $a = -2$ and $b = 5$, then $\frac{1}{b^2} = \frac{1}{25}$, and $a^2 = 4$, so the expression becomes:
  $$
  \frac{1}{25} - 4 = -\frac{99}{25} \approx -3.96
  $$
  which is **more negative** than $-\frac{35}{9} \approx -3.888$.

- If $a = -6$ and $b = 3$, then $\frac{1}{b^2} = \frac{1}{9}$, and $a^2 = 36$, so the expression becomes:
  $$
  \frac{1}{9} - 36 = -\frac{323}{9} \approx -35.888
  $$
  which is **much more negative**.

Thus, the **maximum** value of the expression occurs when $a = -2$ and $b = 3$, and the value is:

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

[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, we need to find the minimum value of $a^2$. Given that $a$ is between -6 and -2, inclusive. Let me think about $a^2$. Since $a$ is negative, squaring it will make it positive. The function $f(a) = a^2$ is a parabola opening upwards, so its minimum occurs at the vertex. However, since $a$ is restricted to the interval $[-6, -2]$, which is entirely on the negative side, the minimum of $a^2$ will occur at the point closest to zero. Since $a$ is between -6 and -2, the closest to zero is -2. Therefore, the minimum value of $a^2$ is $(-2)^2 = 4$.

Wait, but hold on. Let me check that again. If $a$ is between -6 and -2, then $a^2$ will be between $(-6)^2 = 36$ and $(-2)^2 = 4$. So yes, the minimum of $a^2$ is 4, and the maximum is 36. Therefore, the minimum value of $a^2$ is 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 negative. But maybe I need to check if there's a combination where $\frac{1}{b^2}$ is larger and $a^2$ is smaller? Wait, but we already took the maximum of $\frac{1}{b^2}$ and the minimum of $a^2$. So that should give the maximum of the entire expression. But let me check if that's correct.

Alternatively, maybe there's a case where even though $\frac{1}{b^2}$ is smaller, but $a^2$ is much smaller, leading to a larger overall value? Wait, but since $\frac{1}{b^2}$ is positive and $a^2$ is positive, so the expression is $\frac{1}{b^2} - a^2$. So if I want to maximize this, I need to maximize the first term and minimize the second term. So that should be correct.

But let me compute $\frac{1}{9} - 4 = \frac{1}{9} - \frac{36}{9} = -\frac{35}{9}$. But is this the maximum? Wait, maybe there's a case where $\frac{1}{b^2}$ is smaller but $a^2$ is even smaller? But $a^2$ can't be smaller than 4. So that's the minimum. So maybe that is the maximum? But that seems like a negative number. Is there a possibility that the expression can be positive?

Wait, let me check with some numbers. Let me 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 term: $-2 + 1/3 = -5/3$

Second term: $1/3 + 2 = 7/3$

Multiply them: $(-5/3)(7/3) = -35/9$, which is the same as before. So that's negative. But maybe if I take other values?

Wait, let me try $a = -2$ and $b = 5$. Then the expression is $\left(-2 + 1/5\right)\left(1/5 - (-2)\right)$. Let me compute:

First term: $-2 + 1/5 = -9/5$

Second term: $1/5 + 2 = 11/5$

Multiply: $(-9/5)(11/5) = -99/25 ≈ -3.96$, which is more negative than -35/9 ≈ -3.888... So worse.

What if I take $a = -6$ and $b = 3$? Then:

First term: $-6 + 1/3 = -17/3$

Second term: $1/3 - (-6) = 1/3 + 6 = 19/3$

Multiply: (-17/3)(19/3) = -323/9 ≈ -35.888... which is even worse.

Alternatively, if I take $a = -2$ and $b = 3$, which gives -35/9 ≈ -3.888...

Wait, but maybe if I take $a$ closer to zero? Wait, but $a$ is between -6 and -2. So the closest to zero is -2. So maybe that's the minimum of $a^2$. But then, if I take $a = -2$ and $b = 3$, which gives the maximum of $\frac{1}{b^2}$ and minimum of $a^2$, but the result is negative. Is there a way to get a positive value?

Wait, let me think again. The original expression is $\frac{1}{b^2} - a^2$. If $\frac{1}{b^2} > a^2$, then the expression is positive. So maybe there are values where $\frac{1}{b^2} > a^2$?

But given that $a$ is between -6 and -2, so $a^2$ is between 4 and 36. And $\frac{1}{b^2}$ is between $\frac{1}{25}$ and $\frac{1}{9}$. So $\frac{1}{b^2}$ is at most $\frac{1}{9} ≈ 0.111$, while $a^2$ is at least 4. Therefore, $\frac{1}{b^2} - a^2$ is always negative. Therefore, the maximum value is the least negative one, 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 $\frac{1}{9} - 4 = -\frac{35}{9}$. But the problem says "greatest possible value", so maybe that is the answer? But let me check again.

Wait, but 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 expand it step by step:

Multiply $a$ by $\frac{1}{b}$: $a \cdot \frac{1}{b} = \frac{a}{b}$

Multiply $a$ by $-a$: $-a^2$

Multiply $\frac{1}{b}$ by $\frac{1}{b}$: $\frac{1}{b^2}$

Multiply $\frac{1}{b}$ by $-a$: $-\frac{a}{b}$

So adding all these terms together:

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

The $\frac{a}{b}$ and $-\frac{a}{b}$ cancel out, leaving $-a^2 + \frac{1}{b^2}$, which is the same as $\frac{1}{b^2} - a^2$. So that's correct. Therefore, the expression simplifies to $\frac{1}{b^2} - a^2$, which is always negative, as we saw. Therefore, the maximum value is indeed when $\frac{1}{b^2}$ is maximum and $a^2$ is minimum, which is $\frac{1}{9} - 4 = -\frac{35}{9}$. But let me check if there's another combination where maybe $\frac{1}{b^2} - a^2$ is larger (i.e., less negative). For example, if I take $a$ slightly larger (closer to zero) and $b$ slightly larger? Wait, but $a$ can't be larger than -2. So the minimum $a^2$ is 4. If I take $a = -2$, then $a^2 = 4$. If I take $a$ closer to -2, but since -2 is the upper bound, there's no closer. So that's the minimum. Similarly, for $b$, the maximum of $\frac{1}{b^2}$ is at $b = 3$. Therefore, that combination gives the maximum value of the expression.

But let me check another possibility. Suppose I take $a = -2$ and $b = 3$, which gives $\frac{1}{9} - 4 = -35/9$. What if I take $a = -2$ and $b = 5$? Then $\frac{1}{25} - 4 = -\frac{99}{25} ≈ -3.96$, which is more negative. If I take $a = -6$ and $b = 3$, then $\frac{1}{9} - 36 = -\frac{323}{9} ≈ -35.888...$. If I take $a = -2$ and $b = 3$, that's the least negative. So that seems to be the maximum.

But wait, maybe there's another way to approach this problem? Let me think. Maybe instead of simplifying first, I can consider the original expression and analyze it differently. Let me try that.

Original expression: $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Let me denote $x = a$ and $y = \frac{1}{b}$. Then the expression becomes $(x + y)(y - x) = y^2 - x^2$, which is the same as before. So that's the same result. Therefore, regardless of substitution, we get to the same expression. Therefore, the conclusion is the same.

Alternatively, maybe I can think of this as a function of two variables and try to find its maximum over the given intervals. But since we already simplified it to $\frac{1}{b^2} - a^2$, and since $a$ and $b$ are independent variables (the constraints on $a$ and $b$ are separate), then to maximize $\frac{1}{b^2} - a^2$, we need to maximize $\frac{1}{b^2}$ and minimize $a^2$. Therefore, the maximum occurs at $b = 3$ and $a = -2$, giving $\frac{1}{9} - 4 = -\frac{35}{9}$. Therefore, the answer should be $-\frac{35}{9}$. But let me check once more if there's a possibility that the maximum occurs at some other point where maybe $\frac{1}{b^2} - a^2$ is larger.

Wait, for example, suppose $a$ is not at its minimum $a^2$, but maybe there's a combination where even though $a^2$ is larger, $\frac{1}{b^2}$ is significantly larger? But since $\frac{1}{b^2}$ is at most $\frac{1}{9}$, and $a^2$ is at least 4, so even if $a^2$ is larger, say 36, then $\frac{1}{b^2} - a^2$ would be $\frac{1}{9} - 36 = -\frac{323}{9}$, which is worse. So no, that's not helpful.

Alternatively, if I take $b$ larger, making $\frac{1}{b^2}$ smaller, but maybe $a$ is such that $a^2$ is smaller? But $a$ can't be smaller in magnitude than -2. So $a^2$ can't be smaller than 4. Therefore, even if I take $b$ larger, making $\frac{1}{b^2}$ smaller, the expression becomes more negative. Therefore, the maximum is indeed at $b = 3$ and $a = -2$.

But let me check another angle. Suppose I consider the original expression $\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)$. Maybe there's a way to write this as something else? For example, maybe think of it as $\left(\frac{1}{b} + a\right)\left(\frac{1}{b} - a\right)$. Which is the same as $\left(\frac{1}{b}\right)^2 - a^2$, as before. So no difference.

Alternatively, maybe think of it as a quadratic in terms of $a$ or $b$. Let me try treating it as a quadratic in $a$. Let me write the expression:

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

So it's a quadratic in $a$ with coefficient of $a^2$ being -1, which is a downward opening parabola. Therefore, for fixed $b$, the maximum of this expression with respect to $a$ occurs at the vertex. But since the coefficient of $a^2$ is negative, the maximum is at the vertex. However, since $a$ is constrained between -6 and -2, the maximum for each fixed $b$ would be either at the vertex or at the endpoints of the interval for $a$. Wait, but the vertex of the quadratic $-a^2 + c$ (where $c$ is a constant) is at $a = 0$, but our $a$ is constrained to be between -6 and -2, which is entirely on the left of the vertex. Therefore, since the parabola opens downward, the maximum on the interval would be at the point closest to the vertex, which is $a = -2$. Therefore, for each fixed $b$, the maximum value of the expression is achieved when $a = -2$. Therefore, to maximize the entire expression, we need to maximize $\frac{1}{b^2}$, which again occurs at $b = 3$. Therefore, the maximum value is indeed at $a = -2$, $b = 3$, giving $-35/9$.

Alternatively, if I consider the expression as a function of $b$, for fixed $a$, then $\frac{1}{b^2}$ is a decreasing function of $b$, so for fixed $a$, the maximum occurs at the minimum $b$, which is 3. Therefore, again, the maximum occurs at $b = 3$ and $a = -2$.

Therefore, all approaches lead to the same conclusion. Therefore, the greatest possible value is $-35/9$. However, the problem says "greatest possible value", and since all values are negative, the "greatest" is the one closest to zero, which is indeed $-35/9$. Therefore, I think that is the answer.

But just to be thorough, let me check another point. Suppose $a = -2$ and $b = 3$, which gives us $-35/9$. What if I take $a = -2$ and $b = 3.5$? Then $\frac{1}{(3.5)^2} - (-2)^2 = \frac{1}{12.25} - 4 ≈ 0.0816 - 4 ≈ -3.9184$, which is more negative than -35/9 ≈ -3.888... So worse. Similarly, if I take $a = -2$ and $b = 3$, which is the maximum for $\frac{1}{b^2}$, so that's better.

Alternatively, if I take $a = -3$ and $b = 3$. Then $a^2 = 9$, so $\frac{1}{9} - 9 = -80/9 ≈ -8.888...$, which is worse. If I take $a = -2.5$ and $b = 3$, then $a^2 = 6.25$, so $\frac{1}{9} - 6.25 ≈ 0.111 - 6.25 ≈ -6.139$, which is worse than -35/9 ≈ -3.888. So indeed, the maximum occurs at $a = -2$, $b = 3$.

Wait, but let me check if there's a case where $a$ is not at its minimum $a^2$, but maybe $b$ is such that $\frac{1}{b^2}$ is larger? But $b$ can't be smaller than 3, so $\frac{1}{b^2}$ can't be larger than 1/9. Therefore, no.

Alternatively, maybe there's a combination where $\frac{1}{b^2} - a^2$ is larger? For example, if $a$ is negative and $b$ is such that $\frac{1}{b^2}$ is larger? But again, $b$ can't be smaller than 3. So no.

Alternatively, maybe I made a mistake in assuming that maximizing $\frac{1}{b^2}$ and minimizing $a^2$ gives the maximum of the expression. Let me think again. Suppose that $\frac{1}{b^2} - a^2$ is the expression. Since both terms are independent (because $a$ and $b$ are independent variables), then to maximize the expression, we need to maximize $\frac{1}{b^2}$ and minimize $a^2$. Therefore, yes, that is correct.

Alternatively, if there was a relationship between $a$ and $b$, maybe we could have a different approach, but since they are independent, we can treat them separately.

Therefore, I think my conclusion is correct. The maximum value is $-35/9$. However, the problem says "greatest possible value", and since all values are negative, the greatest is the one closest to zero, which is indeed $-35/9$. Therefore, the answer is $\boxed{-\dfrac{35}{9}}$.

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

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

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

Multiply them: (-5/3)(7/3) = -35/9. Correct. So that's negative. But maybe there's a case where the expression is positive? Let me think. Suppose $a$ is negative and $1/b$ is positive. Then $a + 1/b$ could be positive or negative depending on the values. Similarly, $1/b - a$ is positive because $1/b$ is positive and $-a$ is positive (since $a$ is negative). Therefore, the second term is positive. The first term, $a + 1/b$, could be positive or negative. If $a + 1/b$ is positive, then the product is positive. If it's negative, the product is negative. Therefore, maybe there are values where the expression is positive?

Wait, so maybe I need to check if there are values where $a + 1/b$ is positive. Let me see. For $a$ between -6 and -2, and $b$ between 3 and 5. Let me find if there exists $a$ and $b$ such that $a + 1/b > 0$. That would mean $a > -1/b$. Since $a$ is at least -6 and $b$ is at most 5, so $-1/b$ is at least -1/5 = -0.2. Therefore, $a$ is between -6 and -2, which is less than -0.2. Therefore, $a + 1/b$ is always less than -0.2 + 1/3 ≈ -0.2 + 0.333 ≈ 0.133? Wait, no. Wait, if $a$ is between -6 and -2, and $b$ is between 3 and 5, then $1/b$ is between 1/5 and 1/3. So $a + 1/b$ is between (-6 + 1/5) and (-2 + 1/3). Let me compute:

Lower bound: -6 + 1/5 = -5.8

Upper bound: -2 + 1/3 ≈ -1.666...

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 entire expression is always negative. Therefore, the maximum value is indeed the least negative one, which is -35/9. Therefore, my answer is correct.

But let me check with another example. Let me take $a = -2$ and $b = 3$. Then, as before, the expression is -35/9 ≈ -3.888...

If I take $a = -2$ and $b = 3.5$, then:

$a + 1/b = -2 + 2/7 ≈ -2 + 0.2857 ≈ -1.7143$

$1/b - a = 2/7 - (-2) = 2/7 + 2 ≈ 2.2857$

Multiply them: (-1.7143)(2.2857) ≈ -3.928, which is more negative than -3.888... So worse.

If I take $a = -2$ and $b = 3.1$:

$a + 1/b ≈ -2 + 1/3.1 ≈ -2 + 0.3226 ≈ -1.6774$

$1/b - a ≈ 0.3226 + 2 = 2.3226$

Multiply: (-1.6774)(2.3226) ≈ -3.896, which is still more negative than -3.888...

Wait, so maybe there's a point where it's less negative? Let me try with $a = -2$ and $b = 3$. Let me compute the exact value:

$a + 1/b = -2 + 1/3 = -5/3 ≈ -1.6667$

$1/b - a = 1/3 - (-2) = 1/3 + 2 = 7/3 ≈ 2.3333$

Multiply: (-5/3)(7/3) = -35/9 ≈ -3.888...

If I take $a = -2$ and $b = 3. Let me try $b = 3. Let me try $b = 3. Let me try $b = 3. Let me try $b = 3. Let me try $b = 3. Let me try $b = 3. Let me try $b = 3. Wait, I think I need to check if there's a value of $b$ where $\frac{1}{b^2} - a^2$ is larger. Wait, but since $a$ is fixed at -2, and $b$ is varying, then $\frac{1}{b^2}$ is maximized at $b = 3$, so that's the maximum. Therefore, even if I take $a = -2$ and $b = 3$, that's the maximum for that $a$. Therefore, the conclusion is correct.

Alternatively, suppose I take $a$ slightly larger than -2, say $a = -2 + \epsilon$, where $\epsilon$ is a small positive number. Then $a^2$ would be slightly less than 4. However, since $a$ is constrained to be at least -6 and at most -2, so $a$ can't be greater than -2. Therefore, the minimum $a^2$ is indeed 4 at $a = -2$. Therefore, even if I take $a$ slightly larger (closer to -2), but since $a$ can't be larger than -2, that's not allowed. Therefore, the minimum $a^2$ is 4.

Therefore, I think my answer is correct. Therefore, the greatest possible value is $-35/9$.

But just to be 100% sure, let me check if there's a case where $a$ is not at -2 and $b$ is not at 3, but the combination gives a higher value. For example, suppose $a = -3$ and $b = 3$. Then the expression is $\frac{1}{9} - 9 = -80/9 ≈ -8.888...$, which is worse. If $a = -2.5$ and $b = 3$, then $\frac{1}{9} - 6.25 ≈ -6.138...$, worse. If $a = -2$ and $b = 3.5$, as before, it's worse. If $a = -2$ and $b = 3$, it's -35/9. If I take $a = -2$ and $b = 3$, that's the best. Therefore, I think that is the answer.

Therefore, I think 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
$$
\frac{1}{b^2} - a^2
$$
under the given constraints.

- The term $\frac{1}{b^2}$ is **maximized** when $b$ is **as small as possible**, since $\frac{1}{b^2}$ is a **decreasing function** of $b$ for $b > 0$.
- The term $a^2$ is **minimized** when $a$ is **as close to zero as possible**, since $a^2$ is a **parabola** that increases as $|a|$ increases.

Given the constraints:
- $b \in [3, 5]$, so the **maximum** of $\frac{1}{b^2}$ occurs at $b = 3$:
  $$
  \frac{1}{b^2} = \frac{1}{3^2} = \frac{1}{9}
  $$
- $a \in [-6, -2]$, so the **minimum** of $a^2$ occurs at $a = -2$:
  $$
  a^2 = (-2)^2 = 4
  $$

---

### Step 3: Compute the Maximum Value

Substituting these values into the simplified expression:

$$
\frac{1}{b^2} - a^2 = \frac{1}{9} - 4 = \frac{1}{9} - \frac{36}{9} = -\frac{35}{9}
$$

---

### Step 4: Confirm the Result

We can verify that this is the **maximum** value by considering other combinations of $a$ and $b$ within the given ranges. For example:

- If $a = -2$ and $b = 5$, then $\frac{1}{b^2} = \frac{1}{25}$, and $a^2 = 4$, so the expression becomes:
  $$
  \frac{1}{25} - 4 = -\frac{99}{25} \approx -3.96
  $$
  which is **more negative** than $-\frac{35}{9} \approx -3.888$.

- If $a = -6$ and $b = 3$, then $\frac{1}{b^2} = \frac{1}{9}$, and $a^2 = 36$, so the expression becomes:
  $$
  \frac{1}{9} - 36 = -\frac{323}{9} \approx -35.888
  $$
  which is **much more negative**.

Thus, the **maximum** value of the expression occurs when $a = -2$ and $b = 3$, and the value is:

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

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 $\frac{1}{3^2} - (-2)^2 = \boxed{-\frac{35}{9}}$.