Explanation:
Okay, so I need to find the value of this function f(n) at three different points: π/3, √45, and 8^(2/3). Then add them all together. Let me start by recalling what the function f(n) does.

The function f(n) is defined piecewise. If n is greater than or equal to 4, then f(n) is the floor of n. If n is less than 4, then f(n) is the ceiling of n. So first, I need to figure out for each of these three numbers whether they are greater than or equal to 4 or less than 4. Then apply the appropriate function: floor or ceiling.

Let me take them one by one.

First, let's compute f(π/3). Let me recall that π is approximately 3.1416, so π divided by 3 would be... Let me calculate that. π/3 ≈ 3.1416 / 3 ≈ 1.0472. So π/3 is approximately 1.0472. Since this is less than 4, according to the function definition, I need to take the ceiling of π/3. The ceiling function rounds a number up to the nearest integer. So ceiling of 1.0472 is 2, because it's the smallest integer greater than or equal to 1.0472. So f(π/3) is 2. Let me note that down.

Next, let's compute f(√45). Let me figure out what √45 is. Well, 45 is between 36 (which is 6²) and 49 (which is 7²). So √45 is between 6 and 7. Let me calculate it more precisely. √45 can be simplified as √(9*5) = 3√5. Since √5 is approximately 2.236, so 3*2.236 ≈ 6.708. Therefore, √45 ≈ 6.708. Since this is greater than 4, we need to apply the floor function. The floor function gives the greatest integer less than or equal to the number. So floor(6.708) is 6. Therefore, f(√45) is 6. Got that.

Third, let's compute f(8^(2/3)). Let me think about 8 raised to the power of 2/3. Remember that exponents can be broken down. 8^(2/3) is the same as (8^(1/3))^2. The cube root of 8 is 2, since 2³ = 8. Then squaring that gives 2² = 4. Therefore, 8^(2/3) is exactly 4. Now, since 4 is equal to 4, according to the function definition, we use the floor function. So floor(4) is 4. Therefore, f(8^(2/3)) is 4.

Now, adding them all together: f(π/3) + f(√45) + f(8^(2/3)) = 2 + 6 + 4. Let me compute that. 2 + 6 is 8, and 8 + 4 is 12. Therefore, the total should be 12.

Wait, let me double-check each step to make sure I didn't make any mistakes.

Starting with π/3: π is approximately 3.1416, divided by 3 is approximately 1.047. That's definitely less than 4, so ceiling(1.047) is 2. Correct.

For √45: √45 is approximately 6.708, which is greater than 4, so floor(6.708) is 6. Correct.

For 8^(2/3): Let me verify that again. 8^(1/3) is 2, then squared is 4. So that's exactly 4. So floor(4) is 4. Correct.

Adding them up: 2 + 6 + 4. 2 + 6 is 8, plus 4 is 12. Seems straightforward.

Wait, but let me check if there's any possibility that 8^(2/3) is not exactly 4? Let me think. 8 is 2³, so 8^(2/3) is (2³)^(2/3) = 2^(3*(2/3)) = 2² = 4. Yes, that's exact. So no issues there.

Another thing to check: for the ceiling function. For π/3 ≈ 1.047, ceiling is 2. If it was exactly an integer, like 1.0, ceiling would still be 1. But since it's 1.047, which is not an integer, ceiling is 2. Correct.

For √45, since it's irrational, approximately 6.708, floor is 6. Correct.

So all steps seem correct. Therefore, the answer should be 12.

But just to be thorough, let me check if there's any chance that the function definition is different. The function f(n) is floor(n) if n ≥ 4, else ceiling(n). So for numbers exactly equal to 4, floor(4) is 4, which is correct.

Another thing: sometimes floor and ceiling can be confusing with integers. For example, if n is an integer, then floor(n) = ceiling(n) = n. So if n is exactly 4, then floor(4) = 4, which is correct.

So, all steps are correct. Therefore, the answer is 12.

**Final Answer**
\boxed{12}

To evaluate the expression
$$
f\left(\frac{\pi}{3}\right) + f(\sqrt{45}) + f(8^{2/3}),
$$
we begin by analyzing the function $ f(n) $, which is defined as:

$$
f(n) =
\begin{cases}
\lfloor n \rfloor & \text{if } n \geq 4, \\
\lceil n \rceil & \text{if } n < 4.
\end{cases}
$$

This means that for any input $ n $, we apply the **floor function** if $ n \geq 4 $, and the **ceiling function** if $ n < 4 $.

---

### Step 1: Evaluate $ f\left(\frac{\pi}{3}\right) $

We know that $ \pi \approx 3.1416 $, so:

$$
\frac{\pi}{3} \approx \frac{3.1416}{3} \approx 1.0472
$$

Since $ 1.0472 < 4 $, we apply the **ceiling function**:

$$
f\left(\frac{\pi}{3}\right) = \lceil 1.0472 \rceil = 2
$$

---

### Step 2: Evaluate $ f(\sqrt{45}) $

We simplify $ \sqrt{45} $:

$$
\sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5} \approx 3 \cdot 2.236 = 6.708
$$

Since $ 6.708 \geq 4 $, we apply the **floor function**:

$$
f(\sqrt{45}) = \lfloor 6.708 \rfloor = 6
$$

---

### Step 3: Evaluate $ f(8^{2/3}) $

We simplify $ 8^{2/3} $:

$$
8^{2/3} = (8^{1/3})^2 = (2)^2 = 4
$$

Since $ 4 \geq 4 $, we apply the **floor function**:

$$
f(8^{2/3}) = \lfloor 4 \rfloor = 4
$$

---

### Final Step: Add the Results

$$
f\left(\frac{\pi}{3}\right) + f(\sqrt{45}) + f(8^{2/3}) = 2 + 6 + 4 = \boxed{12}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of this function f(n) at three different points: π/3, √45, and 8^(2/3). Then add them all together. Let me start by recalling what the function f(n) does.

The function f(n) is defined piecewise. If n is greater than or equal to 4, then f(n) is the floor of n. If n is less than 4, then f(n) is the ceiling of n. So first, I need to figure out for each of these three numbers whether they are greater than or equal to 4 or less than 4. Then apply the appropriate function: floor or ceiling.

Let me take them one by one.

First, let's compute f(π/3). Let me recall that π is approximately 3.1416, so π divided by 3 would be... Let me calculate that. π/3 ≈ 3.1416 / 3 ≈ 1.0472. So π/3 is approximately 1.0472. Since this is less than 4, according to the function definition, I need to take the ceiling of π/3. The ceiling function rounds a number up to the nearest integer. So ceiling of 1.0472 is 2, because it's the smallest integer greater than or equal to 1.0472. So f(π/3) is 2. Let me note that down.

Next, let's compute f(√45). Let me figure out what √45 is. Well, 45 is between 36 (which is 6²) and 49 (which is 7²). So √45 is between 6 and 7. Let me calculate it more precisely. √45 can be simplified as √(9*5) = 3√5. Since √5 is approximately 2.236, so 3*2.236 ≈ 6.708. Therefore, √45 ≈ 6.708. Since this is greater than 4, we need to apply the floor function. The floor function gives the greatest integer less than or equal to the number. So floor(6.708) is 6. Therefore, f(√45) is 6. Got that.

Third, let's compute f(8^(2/3)). Let me think about 8 raised to the power of 2/3. Remember that exponents can be broken down. 8^(2/3) is the same as (8^(1/3))^2. The cube root of 8 is 2, since 2³ = 8. Then squaring that gives 2² = 4. Therefore, 8^(2/3) is exactly 4. Now, since 4 is equal to 4, according to the function definition, we use the floor function. So floor(4) is 4. Therefore, f(8^(2/3)) is 4.

Now, adding them all together: f(π/3) + f(√45) + f(8^(2/3)) = 2 + 6 + 4. Let me compute that. 2 + 6 is 8, and 8 + 4 is 12. Therefore, the total should be 12.

Wait, let me double-check each step to make sure I didn't make any mistakes.

Starting with π/3: π is approximately 3.1416, divided by 3 is approximately 1.047. That's definitely less than 4, so ceiling(1.047) is 2. Correct.

For √45: √45 is approximately 6.708, which is greater than 4, so floor(6.708) is 6. Correct.

For 8^(2/3): Let me verify that again. 8^(1/3) is 2, then squared is 4. So that's exactly 4. So floor(4) is 4. Correct.

Adding them up: 2 + 6 + 4. 2 + 6 is 8, plus 4 is 12. Seems straightforward.

Wait, but let me check if there's any possibility that 8^(2/3) is not exactly 4? Let me think. 8 is 2³, so 8^(2/3) is (2³)^(2/3) = 2^(3*(2/3)) = 2² = 4. Yes, that's exact. So no issues there.

Another thing to check: for the ceiling function. For π/3 ≈ 1.047, ceiling is 2. If it was exactly an integer, like 1.0, ceiling would still be 1. But since it's 1.047, which is not an integer, ceiling is 2. Correct.

For √45, since it's irrational, approximately 6.708, floor is 6. Correct.

So all steps seem correct. Therefore, the answer should be 12.

But just to be thorough, let me check if there's any chance that the function definition is different. The function f(n) is floor(n) if n ≥ 4, else ceiling(n). So for numbers exactly equal to 4, floor(4) is 4, which is correct.

Another thing: sometimes floor and ceiling can be confusing with integers. For example, if n is an integer, then floor(n) = ceiling(n) = n. So if n is exactly 4, then floor(4) = 4, which is correct.

So, all steps are correct. Therefore, the answer is 12.

**Final Answer**
\boxed{12}

To evaluate the expression
$$
f\left(\frac{\pi}{3}\right) + f(\sqrt{45}) + f(8^{2/3}),
$$
we begin by analyzing the function $ f(n) $, which is defined as:

$$
f(n) =
\begin{cases}
\lfloor n \rfloor & \text{if } n \geq 4, \\
\lceil n \rceil & \text{if } n < 4.
\end{cases}
$$

This means that for any input $ n $, we apply the **floor function** if $ n \geq 4 $, and the **ceiling function** if $ n < 4 $.

---

### Step 1: Evaluate $ f\left(\frac{\pi}{3}\right) $

We know that $ \pi \approx 3.1416 $, so:

$$
\frac{\pi}{3} \approx \frac{3.1416}{3} \approx 1.0472
$$

Since $ 1.0472 < 4 $, we apply the **ceiling function**:

$$
f\left(\frac{\pi}{3}\right) = \lceil 1.0472 \rceil = 2
$$

---

### Step 2: Evaluate $ f(\sqrt{45}) $

We simplify $ \sqrt{45} $:

$$
\sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5} \approx 3 \cdot 2.236 = 6.708
$$

Since $ 6.708 \geq 4 $, we apply the **floor function**:

$$
f(\sqrt{45}) = \lfloor 6.708 \rfloor = 6
$$

---

### Step 3: Evaluate $ f(8^{2/3}) $

We simplify $ 8^{2/3} $:

$$
8^{2/3} = (8^{1/3})^2 = (2)^2 = 4
$$

Since $ 4 \geq 4 $, we apply the **floor function**:

$$
f(8^{2/3}) = \lfloor 4 \rfloor = 4
$$

---

### Final Step: Add the Results

$$
f\left(\frac{\pi}{3}\right) + f(\sqrt{45}) + f(8^{2/3}) = 2 + 6 + 4 = \boxed{12}
$$

Solution: We begin by finding $f\left(\frac{\pi}{3}\right)$. Since we know that $\pi \approx 3.14$, $\frac{\pi}{3}$ must be slightly greater than $1$, so $f\left( \frac{\pi}{3} \right)= \left\lceil \frac{\pi}{3} \right\rceil = 2$. To find $f(\sqrt{45})$, we realize that $\sqrt{36} < \sqrt{45} < \sqrt{49}$, so $6 < \sqrt{45} < 7$. Therefore, since $\sqrt{45} \geq 4$, we have that $f(\sqrt{45}) = \lfloor \sqrt{45} \rfloor = 6$. Finally, we consider $f(8^{2/3})$. We know that $8^{2/3}= (\sqrt[3]{8})^2 = 2^2 = 4$, so $f(8^{2/3})= \lfloor 8^{2/3} \rfloor = 4$.  			 Thus, we have that $f\left(\frac{\pi}{3}\right) + f(\sqrt{45}) + f(8^{2/3}) = 2 + 6 + 4 = \boxed{12}$.


Explanation:
Okay, so I need to figure out the fifth number in the list of positive integers that have exactly three positive divisors. Let me start by recalling some number theory concepts.

First, I remember that the number of divisors a number has is related to its prime factorization. If a number n can be expressed as $ n = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k} $, where the $ p_i $ are prime numbers, then the number of positive divisors of n is $ (a_1 + 1)(a_2 + 1)\dots(a_k + 1) $.

So, if a number has exactly three positive divisors, then the product $ (a_1 + 1)(a_2 + 1)\dots(a_k + 1) = 3 $. Since 3 is a prime number, the only way to express it as a product of integers greater than 1 is 3 itself. That means there must be only one term in the product, so $ a_1 + 1 = 3 $, which implies $ a_1 = 2 $. Therefore, the number must be the square of a prime number.

Wait, let me check that again. If the number has exactly three divisors, then it must be of the form $ p^2 $, where p is a prime. Because if it's a square of a prime, then its divisors are 1, p, and $ p^2 $. That's exactly three divisors.

But hold on, could there be another case? For example, if the number had two different prime factors? Let me think. Suppose a number is $ p \times q $, where p and q are distinct primes. Then the number of divisors would be (1+1)(1+1) = 4. So that's four divisors. So that's not three. If the number was $ p^3 $, then the number of divisors would be 3 + 1 = 4. So that's four again. So, no, the only way to get exactly three divisors is to have a square of a prime.

Therefore, the numbers with exactly three positive divisors are squares of primes. So, to list them in ascending order, I need to list the squares of primes in order.

So, let me list the primes in order and then square them. The primes are 2, 3, 5, 7, 11, 13, etc. So their squares are:

- 2² = 4
- 3² = 9
- 5² = 25
- 7² = 49
- 11² = 121
- 13² = 169
- and so on.

Therefore, the numbers with exactly three divisors in ascending order are 4, 9, 25, 49, 121, 169,... So the fifth number would be 121? Wait, let me count again.

First number: 4 (which is 2²)
Second: 9 (3²)
Third: 25 (5²)
Fourth: 49 (7²)
Fifth: 121 (11²)

Yes, so the fifth number is 121. Therefore, the answer should be 121.

But wait, let me make sure there are no other numbers with exactly three divisors that I might have missed. For example, could there be a composite number that isn't a square of a prime but still has exactly three divisors? Let me think. Suppose there is a number that is a product of two primes, but with exponents... Wait, if it's a product of two distinct primes, then as I said before, it has four divisors. If it's a cube of a prime, then it has four divisors. If it's a product of a prime squared and another prime, then the number of divisors would be (2+1)(1+1) = 6. So that's six divisors.

Alternatively, if a number is 1, but 1 has only one divisor. So no. So, indeed, only squares of primes have exactly three divisors.

Therefore, the list is indeed the squares of primes in order. Therefore, the fifth one is 11 squared, which is 121.

But let me check the first few numbers to make sure. Let me list numbers starting from 1 and check how many divisors they have.

1: 1 divisor (1)
2: divisors 1, 2 (2)
3: 1, 3 (2)
4: 1, 2, 4 (3) → first number
5: 1, 5 (2)
6: 1, 2, 3, 6 (4)
7: 1, 7 (2)
8: 1, 2, 4, 8 (4)
9: 1, 3, 9 (3) → second number
10: 1, 2, 5, 10 (4)
11: 1, 11 (2)
12: 1, 2, 3, 4, 6, 12 (6)
13: 1, 13 (2)
14: 1, 2, 7, 14 (4)
15: 1, 3, 5, 15 (4)
16: 1, 2, 4, 8, 16 (5)
17: 1, 17 (2)
18: 1, 2, 3, 6, 9, 18 (6)
19: 1, 19 (2)
20: 1, 2, 4, 5, 10, 20 (6)
21: 1, 3, 7, 21 (4)
22: 1, 2, 11, 22 (4)
23: 1, 23 (2)
24: 1, 2, 3, 4, 6, 8, 12, 24 (8)
25: 1, 5, 25 (3) → third number
26: 1, 2, 13, 26 (4)
27: 1, 3, 9, 27 (4)
28: 1, 2, 4, 7, 14, 28 (6)
29: 1, 29 (2)
30: 1, 2, 3, 5, 6, 10, 15, 30 (8)
31: 1, 31 (2)
32: 1, 2, 4, 8, 16, 32 (6)
33: 1, 3, 11, 33 (4)
34: 1, 2, 17, 34 (4)
35: 1, 5, 7, 35 (4)
36: 1, 2, 3, 4, 6, 9, 12, 18, 36 (9)
37: 1, 37 (2)
38: 1, 2, 19, 38 (4)
39: 1, 3, 13, 39 (4)
40: 1, 2, 4, 5, 8, 10, 20, 40 (8)
41: 1, 41 (2)
42: 1, 2, 3, 6, 7, 14, 21, 42 (8)
43: 1, 43 (2)
44: 1, 2, 4, 11, 22, 44 (6)
45: 1, 3, 5, 9, 15, 45 (6)
46: 1, 2, 23, 46 (4)
47: 1, 47 (2)
48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 (10)
49: 1, 7, 49 (3) → fourth number
50: 1, 2, 5, 10, 25, 50 (6)
51: 1, 3, 17, 51 (4)
52: 1, 2, 4, 13, 26, 52 (6)
53: 1, 53 (2)
54: 1, 2, 3, 6, 9, 18, 27, 54 (8)
55: 1, 5, 11, 55 (4)
56: 1, 2, 4, 7, 8, 14, 28, 56 (8)
57: 1, 3, 19, 57 (4)
58: 1, 2, 29, 58 (4)
59: 1, 59 (2)
60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 (12)
...
So, up to 49, the numbers with exactly three divisors are 4, 9, 25, 49. So the fourth number is 49. Then the next one would be 11 squared, which is 121. Let me check 121.

Divisors of 121: 1, 11, 121. Yep, three divisors. So that's the fifth number.

But just to make sure there isn't a number between 49 and 121 that I missed. Let me check numbers between 49 and 121. For example, 50 to 120.

Looking at numbers after 49:

50: 6 divisors
51: 4
52: 6
53: 2
54: 8
55: 4
56: 8
57: 4
58: 4
59: 2
60: 12
61: 2
62: 4
63: 6 (divisors: 1, 3, 7, 9, 21, 63)
64: 7 divisors (1, 2, 4, 8, 16, 32, 64)
65: 4
66: 8
67: 2
68: 6
69: 4
70: 8
71: 2
72: 12
73: 2
74: 4
75: 6
76: 6
77: 4
78: 8
79: 2
80: 10
81: 5 (divisors: 1, 3, 9, 27, 81)
82: 4
83: 2
84: 12
85: 4
86: 4
87: 4
88: 8
89: 2
90: 12
91: 4
92: 6
93: 4
94: 4
95: 4
96: 12
97: 2
98: 6 (divisors: 1, 2, 7, 14, 49, 98)
99: 6
100: 9 (divisors: 1, 2, 4, 5, 10, 20, 25, 50, 100)
101: 2
102: 8
103: 2
104: 8
105: 8
106: 4
107: 2
108: 12
109: 2
110: 8
111: 4
112: 10
113: 2
114: 8
115: 4
116: 6
117: 6
118: 4
119: 4
120: 16

So, between 49 and 121, there are no numbers with exactly three divisors. The next square of a prime after 7²=49 is 11²=121. Therefore, the fifth number is indeed 121.

Therefore, the answer is 121.

But just to make sure, let me check if there's a prime between 7 and 11. The primes are 2, 3, 5, 7, 11, 13... So yes, 11 is the next prime after 7. Therefore, 11 squared is 121.

Alternatively, maybe I can think of the list of numbers with exactly three divisors as the squares of primes. So the first few primes are 2, 3, 5, 7, 11, 13, 17,... so their squares are 4, 9, 25, 49, 121, 169,... So fifth term is 121.

Therefore, I think that's correct.

But just to be thorough, let me check if there's a number that is not a square of a prime but still has three divisors. Suppose someone says, for example, 16. Divisors of 16 are 1, 2, 4, 8, 16. That's five divisors. How about 8? Divisors are 1, 2, 4, 8. Four divisors. How about 9? 1, 3, 9. Three divisors. So that's a square of prime. How about 25? Same thing.

What about 49? 1, 7, 49. Yep. So, seems like only squares of primes.

Another way to think: if a number has exactly three divisors, then it must be a square of a prime. Because if it's not a square, then it's either a product of two distinct primes (which has four divisors) or a prime power with exponent higher than 2 (which would have more divisors).

So, for example, if a number is p^2, then divisors are 1, p, p^2. If it's p^3, divisors are 1, p, p^2, p^3. So four divisors. If it's p*q, divisors are 1, p, q, pq. So four divisors. So indeed, only squares of primes have exactly three divisors.

Therefore, the list is the squares of primes in order. Therefore, the fifth number is 11 squared, which is 121.

So, I think that's the answer.

**Final Answer**
The fifth number listed is \boxed{121}.

To determine the fifth number in the list of positive integers that have **exactly three positive divisors**, we begin by analyzing the structure of such numbers.

---

### Step 1: Understanding the Number of Divisors

A number $ n $ with prime factorization
$$
n = p_1^{a_1} \cdot p_2^{a_2} \cdot \dots \cdot p_k^{a_k}
$$
has
$$
(a_1 + 1)(a_2 + 1)\dots(a_k + 1)
$$
positive divisors.

We are looking for numbers with **exactly three** positive divisors. Since 3 is a **prime number**, the only way to express it as a product of integers greater than 1 is:
$$
3 = 3 \times 1
$$
This implies that the number must be of the form $ p^2 $, where $ p $ is a **prime number**. This is because:
- The number of divisors of $ p^2 $ is $ (2 + 1) = 3 $, which are: $ 1, p, p^2 $.

Any other form (e.g., $ p \cdot q $, $ p^3 $, etc.) would result in more than three divisors.

---

### Step 2: Generate the List of Such Numbers

We now list the squares of the first few prime numbers in ascending order:

- $ 2^2 = 4 $
- $ 3^2 = 9 $
- $ 5^2 = 25 $
- $ 7^2 = 49 $
- $ 11^2 = 121 $

So, the numbers with exactly three positive divisors, in ascending order, are:
$$
4, 9, 25, 49, 121, \dots
$$

---

### Step 3: Identify the Fifth Number

From the list above, the fifth number is:
$$
\boxed{121}
$$

---

### Final Answer

$$
\boxed{121}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the fifth number in the list of positive integers that have exactly three positive divisors. Let me start by recalling some number theory concepts.

First, I remember that the number of divisors a number has is related to its prime factorization. If a number n can be expressed as $ n = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k} $, where the $ p_i $ are prime numbers, then the number of positive divisors of n is $ (a_1 + 1)(a_2 + 1)\dots(a_k + 1) $.

So, if a number has exactly three positive divisors, then the product $ (a_1 + 1)(a_2 + 1)\dots(a_k + 1) = 3 $. Since 3 is a prime number, the only way to express it as a product of integers greater than 1 is 3 itself. That means there must be only one term in the product, so $ a_1 + 1 = 3 $, which implies $ a_1 = 2 $. Therefore, the number must be the square of a prime number.

Wait, let me check that again. If the number has exactly three divisors, then it must be of the form $ p^2 $, where p is a prime. Because if it's a square of a prime, then its divisors are 1, p, and $ p^2 $. That's exactly three divisors.

But hold on, could there be another case? For example, if the number had two different prime factors? Let me think. Suppose a number is $ p \times q $, where p and q are distinct primes. Then the number of divisors would be (1+1)(1+1) = 4. So that's four divisors. So that's not three. If the number was $ p^3 $, then the number of divisors would be 3 + 1 = 4. So that's four again. So, no, the only way to get exactly three divisors is to have a square of a prime.

Therefore, the numbers with exactly three positive divisors are squares of primes. So, to list them in ascending order, I need to list the squares of primes in order.

So, let me list the primes in order and then square them. The primes are 2, 3, 5, 7, 11, 13, etc. So their squares are:

- 2² = 4
- 3² = 9
- 5² = 25
- 7² = 49
- 11² = 121
- 13² = 169
- and so on.

Therefore, the numbers with exactly three divisors in ascending order are 4, 9, 25, 49, 121, 169,... So the fifth number would be 121? Wait, let me count again.

First number: 4 (which is 2²)
Second: 9 (3²)
Third: 25 (5²)
Fourth: 49 (7²)
Fifth: 121 (11²)

Yes, so the fifth number is 121. Therefore, the answer should be 121.

But wait, let me make sure there are no other numbers with exactly three divisors that I might have missed. For example, could there be a composite number that isn't a square of a prime but still has exactly three divisors? Let me think. Suppose there is a number that is a product of two primes, but with exponents... Wait, if it's a product of two distinct primes, then as I said before, it has four divisors. If it's a cube of a prime, then it has four divisors. If it's a product of a prime squared and another prime, then the number of divisors would be (2+1)(1+1) = 6. So that's six divisors.

Alternatively, if a number is 1, but 1 has only one divisor. So no. So, indeed, only squares of primes have exactly three divisors.

Therefore, the list is indeed the squares of primes in order. Therefore, the fifth one is 11 squared, which is 121.

But let me check the first few numbers to make sure. Let me list numbers starting from 1 and check how many divisors they have.

1: 1 divisor (1)
2: divisors 1, 2 (2)
3: 1, 3 (2)
4: 1, 2, 4 (3) → first number
5: 1, 5 (2)
6: 1, 2, 3, 6 (4)
7: 1, 7 (2)
8: 1, 2, 4, 8 (4)
9: 1, 3, 9 (3) → second number
10: 1, 2, 5, 10 (4)
11: 1, 11 (2)
12: 1, 2, 3, 4, 6, 12 (6)
13: 1, 13 (2)
14: 1, 2, 7, 14 (4)
15: 1, 3, 5, 15 (4)
16: 1, 2, 4, 8, 16 (5)
17: 1, 17 (2)
18: 1, 2, 3, 6, 9, 18 (6)
19: 1, 19 (2)
20: 1, 2, 4, 5, 10, 20 (6)
21: 1, 3, 7, 21 (4)
22: 1, 2, 11, 22 (4)
23: 1, 23 (2)
24: 1, 2, 3, 4, 6, 8, 12, 24 (8)
25: 1, 5, 25 (3) → third number
26: 1, 2, 13, 26 (4)
27: 1, 3, 9, 27 (4)
28: 1, 2, 4, 7, 14, 28 (6)
29: 1, 29 (2)
30: 1, 2, 3, 5, 6, 10, 15, 30 (8)
31: 1, 31 (2)
32: 1, 2, 4, 8, 16, 32 (6)
33: 1, 3, 11, 33 (4)
34: 1, 2, 17, 34 (4)
35: 1, 5, 7, 35 (4)
36: 1, 2, 3, 4, 6, 9, 12, 18, 36 (9)
37: 1, 37 (2)
38: 1, 2, 19, 38 (4)
39: 1, 3, 13, 39 (4)
40: 1, 2, 4, 5, 8, 10, 20, 40 (8)
41: 1, 41 (2)
42: 1, 2, 3, 6, 7, 14, 21, 42 (8)
43: 1, 43 (2)
44: 1, 2, 4, 11, 22, 44 (6)
45: 1, 3, 5, 9, 15, 45 (6)
46: 1, 2, 23, 46 (4)
47: 1, 47 (2)
48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 (10)
49: 1, 7, 49 (3) → fourth number
50: 1, 2, 5, 10, 25, 50 (6)
51: 1, 3, 17, 51 (4)
52: 1, 2, 4, 13, 26, 52 (6)
53: 1, 53 (2)
54: 1, 2, 3, 6, 9, 18, 27, 54 (8)
55: 1, 5, 11, 55 (4)
56: 1, 2, 4, 7, 8, 14, 28, 56 (8)
57: 1, 3, 19, 57 (4)
58: 1, 2, 29, 58 (4)
59: 1, 59 (2)
60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 (12)
...
So, up to 49, the numbers with exactly three divisors are 4, 9, 25, 49. So the fourth number is 49. Then the next one would be 11 squared, which is 121. Let me check 121.

Divisors of 121: 1, 11, 121. Yep, three divisors. So that's the fifth number.

But just to make sure there isn't a number between 49 and 121 that I missed. Let me check numbers between 49 and 121. For example, 50 to 120.

Looking at numbers after 49:

50: 6 divisors
51: 4
52: 6
53: 2
54: 8
55: 4
56: 8
57: 4
58: 4
59: 2
60: 12
61: 2
62: 4
63: 6 (divisors: 1, 3, 7, 9, 21, 63)
64: 7 divisors (1, 2, 4, 8, 16, 32, 64)
65: 4
66: 8
67: 2
68: 6
69: 4
70: 8
71: 2
72: 12
73: 2
74: 4
75: 6
76: 6
77: 4
78: 8
79: 2
80: 10
81: 5 (divisors: 1, 3, 9, 27, 81)
82: 4
83: 2
84: 12
85: 4
86: 4
87: 4
88: 8
89: 2
90: 12
91: 4
92: 6
93: 4
94: 4
95: 4
96: 12
97: 2
98: 6 (divisors: 1, 2, 7, 14, 49, 98)
99: 6
100: 9 (divisors: 1, 2, 4, 5, 10, 20, 25, 50, 100)
101: 2
102: 8
103: 2
104: 8
105: 8
106: 4
107: 2
108: 12
109: 2
110: 8
111: 4
112: 10
113: 2
114: 8
115: 4
116: 6
117: 6
118: 4
119: 4
120: 16

So, between 49 and 121, there are no numbers with exactly three divisors. The next square of a prime after 7²=49 is 11²=121. Therefore, the fifth number is indeed 121.

Therefore, the answer is 121.

But just to make sure, let me check if there's a prime between 7 and 11. The primes are 2, 3, 5, 7, 11, 13... So yes, 11 is the next prime after 7. Therefore, 11 squared is 121.

Alternatively, maybe I can think of the list of numbers with exactly three divisors as the squares of primes. So the first few primes are 2, 3, 5, 7, 11, 13, 17,... so their squares are 4, 9, 25, 49, 121, 169,... So fifth term is 121.

Therefore, I think that's correct.

But just to be thorough, let me check if there's a number that is not a square of a prime but still has three divisors. Suppose someone says, for example, 16. Divisors of 16 are 1, 2, 4, 8, 16. That's five divisors. How about 8? Divisors are 1, 2, 4, 8. Four divisors. How about 9? 1, 3, 9. Three divisors. So that's a square of prime. How about 25? Same thing.

What about 49? 1, 7, 49. Yep. So, seems like only squares of primes.

Another way to think: if a number has exactly three divisors, then it must be a square of a prime. Because if it's not a square, then it's either a product of two distinct primes (which has four divisors) or a prime power with exponent higher than 2 (which would have more divisors).

So, for example, if a number is p^2, then divisors are 1, p, p^2. If it's p^3, divisors are 1, p, p^2, p^3. So four divisors. If it's p*q, divisors are 1, p, q, pq. So four divisors. So indeed, only squares of primes have exactly three divisors.

Therefore, the list is the squares of primes in order. Therefore, the fifth number is 11 squared, which is 121.

So, I think that's the answer.

**Final Answer**
The fifth number listed is \boxed{121}.

To determine the fifth number in the list of positive integers that have **exactly three positive divisors**, we begin by analyzing the structure of such numbers.

---

### Step 1: Understanding the Number of Divisors

A number $ n $ with prime factorization
$$
n = p_1^{a_1} \cdot p_2^{a_2} \cdot \dots \cdot p_k^{a_k}
$$
has
$$
(a_1 + 1)(a_2 + 1)\dots(a_k + 1)
$$
positive divisors.

We are looking for numbers with **exactly three** positive divisors. Since 3 is a **prime number**, the only way to express it as a product of integers greater than 1 is:
$$
3 = 3 \times 1
$$
This implies that the number must be of the form $ p^2 $, where $ p $ is a **prime number**. This is because:
- The number of divisors of $ p^2 $ is $ (2 + 1) = 3 $, which are: $ 1, p, p^2 $.

Any other form (e.g., $ p \cdot q $, $ p^3 $, etc.) would result in more than three divisors.

---

### Step 2: Generate the List of Such Numbers

We now list the squares of the first few prime numbers in ascending order:

- $ 2^2 = 4 $
- $ 3^2 = 9 $
- $ 5^2 = 25 $
- $ 7^2 = 49 $
- $ 11^2 = 121 $

So, the numbers with exactly three positive divisors, in ascending order, are:
$$
4, 9, 25, 49, 121, \dots
$$

---

### Step 3: Identify the Fifth Number

From the list above, the fifth number is:
$$
\boxed{121}
$$

---

### Final Answer

$$
\boxed{121}
$$

Solution:
If an integer has at least two different prime factors, say, $p$ and $q$, then it must have at least four positive divisors: $1$, $p$, $q$, and $pq$. So, for a number to have exactly three positive divisors, it must be a power of a single prime number. The positive divisors of $p^n$ are $1,p,p^2,p^3,\cdots,p^{n-1},p^n$. Therefore, $p^n$ has $n+1$ different positive divisors, and the only positive integers with exactly three positive divisors are the squares of prime numbers.

The five smallest such integers are, in ascending order, $2^2$, $3^2$, $5^2$, $7^2$, and $11^2$. The fifth number listed is $11^2=\boxed{121}$.


Explanation:
Okay, so I need to figure out how many integer values of t satisfy the condition that the square root of t is greater than 2 and less than 3.5. Let me start by writing down what's given.

The problem says √t > 2 and √t < 3.5. So, I need to find all integers t such that when you take the square root of t, it's between 2 and 3.5. Hmm, okay. Let me think about how to approach this.

First, maybe I should solve the inequalities for t. Since square roots can sometimes be tricky with inequalities, I need to remember that squaring both sides of an inequality is only safe if both sides are non-negative, which they are here because square roots are always non-negative. So, let me square all parts of the inequality to get rid of the square root.

Starting with √t > 2. If I square both sides, that becomes t > 2², which is t > 4. Similarly, for √t < 3.5, squaring both sides gives t < (3.5)². Let me calculate that. 3.5 squared... Well, 3 squared is 9, 0.5 squared is 0.25, and then the cross term is 2*3*0.5 = 3. So, (3 + 0.5)² = 9 + 3 + 0.25 = 12.25. So, t < 12.25.

So putting those together, t has to be greater than 4 and less than 12.25. But t has to be an integer. So, the integer values of t that satisfy 4 < t < 12.25. Let me write that down.

So, t must be integers greater than 4 and less than 12.25. Since t has to be an integer, the smallest integer greater than 4 is 5, and the largest integer less than 12.25 is 12. So, t can be 5, 6, 7, ..., up to 12. Now, I need to count how many integers are in that range.

To find the number of integers from 5 to 12 inclusive, I can subtract the lower bound from the upper bound and add 1. So, 12 - 5 + 1. Let me compute that. 12 - 5 is 7, plus 1 is 8. So, there are 8 integers. Wait, let me check that again. Let me list them out to make sure.

Starting at 5: 5, 6, 7, 8, 9, 10, 11, 12. Let me count them. 5 is 1, 6 is 2, 7 is 3, 8 is 4, 9 is 5, 10 is 6, 11 is 7, 12 is 8. Yep, that's 8 numbers. So, the answer should be 8?

Wait, but hold on. Let me double-check my initial step. The original inequalities were √t > 2 and √t < 3.5. So, when I squared them, I got t > 4 and t < 12.25. But is there a possibility that squaring could introduce any issues? For example, if the original inequalities had negative numbers, but since square roots are always non-negative, and 2 and 3.5 are positive, squaring should be safe here. So, no issues there.

But let me check with some numbers. Let's take t=5. √5 is approximately 2.236, which is greater than 2 and less than 3.5. That works. What about t=12? √12 is approximately 3.464, which is still less than 3.5. So that's okay. What about t=13? √13 is about 3.606, which is greater than 3.5, so that's outside the upper bound. So, 13 is excluded. Similarly, t=4: √4 is 2, which is not greater than 2, so it's excluded. So, t=4 is not included. So, the integers from 5 to 12 inclusive are all valid. So, 8 numbers. So, the answer is 8?

Wait, but let me check if there's another way to approach this. Maybe by considering the square roots. The square root of t is between 2 and 3.5. So, t is between 2² and 3.5², which is 4 and 12.25. So, same as before. So, t must be integers strictly greater than 4 and strictly less than 12.25. Therefore, integers from 5 to 12 inclusive. Which is 8 numbers. So, same result.

But let me think again. Maybe I need to check if 3.5 squared is exactly 12.25. Yes, because 3.5 is 7/2, so (7/2)^2 is 49/4, which is 12.25. So, that's correct. So, t must be less than 12.25, so the maximum integer t can be is 12. So, that's correct.

Alternatively, if I think about the square roots. The square root of t is greater than 2, so t is greater than 4. The square root of t is less than 3.5, so t is less than 12.25. Therefore, t is in (4, 12.25). Since t is an integer, t can be 5,6,7,8,9,10,11,12. So, 8 integers. So, that seems consistent.

But let me check if there's a possibility that maybe the original problem had inclusive or exclusive boundaries? Wait, the problem says "the square root of t is greater than 2 and less than 3.5". So, it's strictly greater than 2 and strictly less than 3.5. Therefore, t must be strictly greater than 4 and strictly less than 12.25. Therefore, t cannot be 4 or 12.25. Since t is integer, 12.25 is not an integer, so the upper bound is 12. So, that's correct.

Alternatively, if the problem had said "greater than or equal to 2 and less than or equal to 3.5", then t would be greater than or equal to 4 and less than or equal to 12.25, so t could be 4, 5,...,12. But since it's strictly greater and strictly less, t cannot be 4 or 12.25. But since 12.25 is not integer, only 4 is excluded. So, the answer is still 8.

Wait, but let me check if maybe there's a mistake in the calculation of the number of integers. For example, if I have numbers from a to b inclusive, the number of integers is b - a + 1. So, if a is 5 and b is 12, then 12 - 5 + 1 = 8. So, that's correct. Let me count again:

5,6,7,8,9,10,11,12. That's 8 numbers. So, yes. So, the answer is 8. Therefore, the number of integer values of t is 8.

But just to be thorough, let me check each value:

t=5: √5 ≈ 2.236, which is between 2 and 3.5. Good.

t=6: √6 ≈ 2.449. Good.

t=7: ≈2.645. Good.

t=8: ≈2.828. Good.

t=9: 3. Good, since 3 is less than 3.5.

t=10: ≈3.162. Good.

t=11: ≈3.316. Good.

t=12: ≈3.464. Good.

All of these are between 2 and 3.5. So, all 8 values are valid. If I check t=13, √13 ≈3.606, which is more than 3.5, so that's excluded. Similarly, t=4: √4=2, which is not greater than 2, so excluded. So, yes, only 5 through 12 inclusive. So, 8 integers. Therefore, the answer is 8.

But wait, hold on. Let me check if there's a different interpretation of the original problem. The problem says "the square root of t is greater than 2 and less than 3.5". So, is that √t > 2 and √t < 3.5? Yes. So, combining those two inequalities, 2 < √t < 3.5. Which leads to 4 < t < 12.25. So, same as before.

Alternatively, if someone misreads it as √t > 2 and √t < 3.5, but maybe they think of it as √t is between 2 and 3.5, which is the same thing. So, no problem.

Alternatively, maybe someone might think that the square root of t is between 2 and 3.5, so t is between 2 and 3.5 squared? Wait, but that's exactly what we did. So, no problem.

Alternatively, maybe someone might think that the square root of t is greater than 2 and less than 3.5, so t is between 2^2 and 3.5^2, which is 4 and 12.25. So, same thing. So, no problem.

Alternatively, maybe someone might think that the square root of t is greater than 2 and less than 3.5, so t is between 2 and 3.5, but that would be incorrect because t is under the square root. So, that would be a mistake. But since we correctly squared the inequalities, we get the correct range for t.

Therefore, I think my reasoning is correct. The answer is 8. Therefore, the number of integer values of t is 8.

**Final Answer**
The number of integer values of $ t $ that satisfy the condition is \boxed{8}.

To determine how many integer values of $ t $ satisfy the condition that the square root of $ t $ is greater than 2 and less than 3.5, we begin by translating the inequality into a more manageable form.

---

### Step 1: Translate the Inequality

We are given:

$$
2 < \sqrt{t} < 3.5
$$

To eliminate the square root, we square all parts of the inequality. Since all values involved are positive, squaring is valid and preserves the inequality:

$$
2^2 < t < 3.5^2
$$

$$
4 < t < 12.25
$$

---

### Step 2: Identify Integer Values in the Range

We are looking for **integer values of $ t $** that lie strictly between 4 and 12.25. That is:

$$
t \in \{5, 6, 7, 8, 9, 10, 11, 12\}
$$

These are the integers strictly greater than 4 and strictly less than 12.25.

---

### Step 3: Count the Number of Valid Integers

To count the number of integers in the range from 5 to 12 inclusive, we use the formula:

$$
\text{Number of integers} = \text{Last} - \text{First} + 1
$$

$$
12 - 5 + 1 = 8
$$

---

### Step 4: Verify the Values

Let’s verify that each of these values satisfies the original condition:

- $ \sqrt{5} \approx 2.236 $ → between 2 and 3.5 ✅
- $ \sqrt{6} \approx 2.449 $ → between 2 and 3.5 ✅
- $ \sqrt{7} \approx 2.646 $ → between 2 and 3.5 ✅
- $ \sqrt{8} \approx 2.828 $ → between 2 and 3.5 ✅
- $ \sqrt{9} = 3 $ → between 2 and 3.5 ✅
- $ \sqrt{10} \approx 3.162 $ → between 2 and 3.5 ✅
- $ \sqrt{11} \approx 3.316 $ → between 2 and 3.5 ✅
- $ \sqrt{12} \approx 3.464 $ → between 2 and 3.5 ✅

All values are valid.

---

### Final Answer

$$
\boxed{8}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many integer values of t satisfy the condition that the square root of t is greater than 2 and less than 3.5. Let me start by writing down what's given.

The problem says √t > 2 and √t < 3.5. So, I need to find all integers t such that when you take the square root of t, it's between 2 and 3.5. Hmm, okay. Let me think about how to approach this.

First, maybe I should solve the inequalities for t. Since square roots can sometimes be tricky with inequalities, I need to remember that squaring both sides of an inequality is only safe if both sides are non-negative, which they are here because square roots are always non-negative. So, let me square all parts of the inequality to get rid of the square root.

Starting with √t > 2. If I square both sides, that becomes t > 2², which is t > 4. Similarly, for √t < 3.5, squaring both sides gives t < (3.5)². Let me calculate that. 3.5 squared... Well, 3 squared is 9, 0.5 squared is 0.25, and then the cross term is 2*3*0.5 = 3. So, (3 + 0.5)² = 9 + 3 + 0.25 = 12.25. So, t < 12.25.

So putting those together, t has to be greater than 4 and less than 12.25. But t has to be an integer. So, the integer values of t that satisfy 4 < t < 12.25. Let me write that down.

So, t must be integers greater than 4 and less than 12.25. Since t has to be an integer, the smallest integer greater than 4 is 5, and the largest integer less than 12.25 is 12. So, t can be 5, 6, 7, ..., up to 12. Now, I need to count how many integers are in that range.

To find the number of integers from 5 to 12 inclusive, I can subtract the lower bound from the upper bound and add 1. So, 12 - 5 + 1. Let me compute that. 12 - 5 is 7, plus 1 is 8. So, there are 8 integers. Wait, let me check that again. Let me list them out to make sure.

Starting at 5: 5, 6, 7, 8, 9, 10, 11, 12. Let me count them. 5 is 1, 6 is 2, 7 is 3, 8 is 4, 9 is 5, 10 is 6, 11 is 7, 12 is 8. Yep, that's 8 numbers. So, the answer should be 8?

Wait, but hold on. Let me double-check my initial step. The original inequalities were √t > 2 and √t < 3.5. So, when I squared them, I got t > 4 and t < 12.25. But is there a possibility that squaring could introduce any issues? For example, if the original inequalities had negative numbers, but since square roots are always non-negative, and 2 and 3.5 are positive, squaring should be safe here. So, no issues there.

But let me check with some numbers. Let's take t=5. √5 is approximately 2.236, which is greater than 2 and less than 3.5. That works. What about t=12? √12 is approximately 3.464, which is still less than 3.5. So that's okay. What about t=13? √13 is about 3.606, which is greater than 3.5, so that's outside the upper bound. So, 13 is excluded. Similarly, t=4: √4 is 2, which is not greater than 2, so it's excluded. So, t=4 is not included. So, the integers from 5 to 12 inclusive are all valid. So, 8 numbers. So, the answer is 8?

Wait, but let me check if there's another way to approach this. Maybe by considering the square roots. The square root of t is between 2 and 3.5. So, t is between 2² and 3.5², which is 4 and 12.25. So, same as before. So, t must be integers strictly greater than 4 and strictly less than 12.25. Therefore, integers from 5 to 12 inclusive. Which is 8 numbers. So, same result.

But let me think again. Maybe I need to check if 3.5 squared is exactly 12.25. Yes, because 3.5 is 7/2, so (7/2)^2 is 49/4, which is 12.25. So, that's correct. So, t must be less than 12.25, so the maximum integer t can be is 12. So, that's correct.

Alternatively, if I think about the square roots. The square root of t is greater than 2, so t is greater than 4. The square root of t is less than 3.5, so t is less than 12.25. Therefore, t is in (4, 12.25). Since t is an integer, t can be 5,6,7,8,9,10,11,12. So, 8 integers. So, that seems consistent.

But let me check if there's a possibility that maybe the original problem had inclusive or exclusive boundaries? Wait, the problem says "the square root of t is greater than 2 and less than 3.5". So, it's strictly greater than 2 and strictly less than 3.5. Therefore, t must be strictly greater than 4 and strictly less than 12.25. Therefore, t cannot be 4 or 12.25. Since t is integer, 12.25 is not an integer, so the upper bound is 12. So, that's correct.

Alternatively, if the problem had said "greater than or equal to 2 and less than or equal to 3.5", then t would be greater than or equal to 4 and less than or equal to 12.25, so t could be 4, 5,...,12. But since it's strictly greater and strictly less, t cannot be 4 or 12.25. But since 12.25 is not integer, only 4 is excluded. So, the answer is still 8.

Wait, but let me check if maybe there's a mistake in the calculation of the number of integers. For example, if I have numbers from a to b inclusive, the number of integers is b - a + 1. So, if a is 5 and b is 12, then 12 - 5 + 1 = 8. So, that's correct. Let me count again:

5,6,7,8,9,10,11,12. That's 8 numbers. So, yes. So, the answer is 8. Therefore, the number of integer values of t is 8.

But just to be thorough, let me check each value:

t=5: √5 ≈ 2.236, which is between 2 and 3.5. Good.

t=6: √6 ≈ 2.449. Good.

t=7: ≈2.645. Good.

t=8: ≈2.828. Good.

t=9: 3. Good, since 3 is less than 3.5.

t=10: ≈3.162. Good.

t=11: ≈3.316. Good.

t=12: ≈3.464. Good.

All of these are between 2 and 3.5. So, all 8 values are valid. If I check t=13, √13 ≈3.606, which is more than 3.5, so that's excluded. Similarly, t=4: √4=2, which is not greater than 2, so excluded. So, yes, only 5 through 12 inclusive. So, 8 integers. Therefore, the answer is 8.

But wait, hold on. Let me check if there's a different interpretation of the original problem. The problem says "the square root of t is greater than 2 and less than 3.5". So, is that √t > 2 and √t < 3.5? Yes. So, combining those two inequalities, 2 < √t < 3.5. Which leads to 4 < t < 12.25. So, same as before.

Alternatively, if someone misreads it as √t > 2 and √t < 3.5, but maybe they think of it as √t is between 2 and 3.5, which is the same thing. So, no problem.

Alternatively, maybe someone might think that the square root of t is between 2 and 3.5, so t is between 2 and 3.5 squared? Wait, but that's exactly what we did. So, no problem.

Alternatively, maybe someone might think that the square root of t is greater than 2 and less than 3.5, so t is between 2^2 and 3.5^2, which is 4 and 12.25. So, same thing. So, no problem.

Alternatively, maybe someone might think that the square root of t is greater than 2 and less than 3.5, so t is between 2 and 3.5, but that would be incorrect because t is under the square root. So, that would be a mistake. But since we correctly squared the inequalities, we get the correct range for t.

Therefore, I think my reasoning is correct. The answer is 8. Therefore, the number of integer values of t is 8.

**Final Answer**
The number of integer values of $ t $ that satisfy the condition is \boxed{8}.

To determine how many integer values of $ t $ satisfy the condition that the square root of $ t $ is greater than 2 and less than 3.5, we begin by translating the inequality into a more manageable form.

---

### Step 1: Translate the Inequality

We are given:

$$
2 < \sqrt{t} < 3.5
$$

To eliminate the square root, we square all parts of the inequality. Since all values involved are positive, squaring is valid and preserves the inequality:

$$
2^2 < t < 3.5^2
$$

$$
4 < t < 12.25
$$

---

### Step 2: Identify Integer Values in the Range

We are looking for **integer values of $ t $** that lie strictly between 4 and 12.25. That is:

$$
t \in \{5, 6, 7, 8, 9, 10, 11, 12\}
$$

These are the integers strictly greater than 4 and strictly less than 12.25.

---

### Step 3: Count the Number of Valid Integers

To count the number of integers in the range from 5 to 12 inclusive, we use the formula:

$$
\text{Number of integers} = \text{Last} - \text{First} + 1
$$

$$
12 - 5 + 1 = 8
$$

---

### Step 4: Verify the Values

Let’s verify that each of these values satisfies the original condition:

- $ \sqrt{5} \approx 2.236 $ → between 2 and 3.5 ✅
- $ \sqrt{6} \approx 2.449 $ → between 2 and 3.5 ✅
- $ \sqrt{7} \approx 2.646 $ → between 2 and 3.5 ✅
- $ \sqrt{8} \approx 2.828 $ → between 2 and 3.5 ✅
- $ \sqrt{9} = 3 $ → between 2 and 3.5 ✅
- $ \sqrt{10} \approx 3.162 $ → between 2 and 3.5 ✅
- $ \sqrt{11} \approx 3.316 $ → between 2 and 3.5 ✅
- $ \sqrt{12} \approx 3.464 $ → between 2 and 3.5 ✅

All values are valid.

---

### Final Answer

$$
\boxed{8}
$$

Solution: We have: $2 < \sqrt{t} < \frac{7}{2}$ so squaring the inequality (which we can do because all the terms in it are positive) gives us  $4 < t <\frac{49}{4}=12.25$.  Therefore, $t$ is an integer between 5 and 12 inclusive, which leaves us with $\boxed{8}$ possible integer values of $t$.


Explanation:
Okay, so I need to find the domain of the function f(x) = (2 - x) / log(2 - log(x - 2)). Alright, let me recall that the domain of a function consists of all real numbers x for which the function is defined. Since this function is a rational expression involving a logarithm, I need to consider the restrictions imposed by both the denominator and the arguments of the logarithms.

First, let me note that the function has two logarithms here: one in the denominator, log(2 - log(x - 2)), and another inside that logarithm, which is log(x - 2). So, I need to make sure that both logarithms are defined, and also that the denominator isn't zero because division by zero is undefined.

Let me break this down step by step.

First, let's consider the inner logarithm: log(x - 2). For this logarithm to be defined, its argument must be positive. That means:

x - 2 > 0
=> x > 2

So, the first condition is that x must be greater than 2. That's straightforward.

Next, looking at the denominator: log(2 - log(x - 2)). For the logarithm to be defined, its argument must also be positive. Therefore:

2 - log(x - 2) > 0
=> log(x - 2) < 2

Since log here is base 10, log(x - 2) < 2 implies that:

x - 2 < 10^2
=> x - 2 < 100
=> x < 102

So, combining this with the previous condition x > 2, we have that x must be in (2, 102). However, we also need to ensure that the denominator is not zero. So, log(2 - log(x - 2)) ≠ 0.

Let me handle that. The denominator is zero when:

log(2 - log(x - 2)) = 0

Which implies:

2 - log(x - 2) = 10^0 = 1

Therefore:

2 - log(x - 2) = 1
=> log(x - 2) = 2 - 1 = 1
=> x - 2 = 10^1 = 10
=> x = 10 + 2 = 12

So, the denominator is zero when x = 12. Therefore, we need to exclude x = 12 from the domain.

So, putting it all together, the domain is all real numbers x such that 2 < x < 102, except x = 12. Therefore, in interval notation, that would be (2, 12) union (12, 102).

Wait, but let me check if there are any other restrictions. Let me verify each step again.

First, log(x - 2) requires x > 2. Then, log(2 - log(x - 2)) requires that 2 - log(x - 2) > 0, which leads to log(x - 2) < 2, so x - 2 < 100, so x < 102. So, combining x > 2 and x < 102 gives (2, 102). Then, the denominator is zero when x = 12, so we remove that point. Therefore, the domain is (2, 102) without 12. So, (2, 12) union (12, 102). That seems right.

But let me check if there are any other points where the denominator could be undefined or zero. For example, is there a case where log(2 - log(x - 2)) is undefined? Well, we already considered that 2 - log(x - 2) must be positive, so that's already covered by the condition log(x - 2) < 2. Therefore, as long as x is in (2, 102), the argument of the outer logarithm is positive, so the outer logarithm is defined. And we also excluded x = 12 where the denominator is zero.

But let me check if there are any other points where log(x - 2) might be problematic. For example, if log(x - 2) is negative? Wait, log(x - 2) can be negative or positive. For example, if x - 2 is between 0 and 1, then log(x - 2) is negative. But that's okay, as long as 2 - log(x - 2) is positive. Let me check with an example. Suppose x = 3, then x - 2 = 1, so log(1) = 0. Then 2 - 0 = 2, so log(2) is defined. If x = 2.5, then x - 2 = 0.5, so log(0.5) is negative, so 2 - log(0.5) is 2 - (-0.3010...) ≈ 2.3010, which is positive. So, that's okay. So, even if log(x - 2) is negative, as long as 2 - log(x - 2) is positive, which is already covered by x < 102. Therefore, there are no other restrictions.

Wait, but let me check if there's a case where log(x - 2) is undefined? But we already have x > 2, so x - 2 is positive, so log(x - 2) is defined. Therefore, the only restrictions are x > 2, x < 102, and x ≠ 12.

Therefore, the domain is (2, 102) without 12. So, written in interval notation, that's (2, 12) ∪ (12, 102). Let me check if there's any other possible issues.

Wait, let me check if x = 12 is the only point where the denominator is zero. Let me verify:

log(2 - log(x - 2)) = 0 implies 2 - log(x - 2) = 1, so log(x - 2) = 1, so x - 2 = 10^1 = 10, so x = 12. So, yes, only x = 12. Therefore, that's the only point to exclude.

Therefore, the domain is (2, 12) ∪ (12, 102). Let me check if there are any other points where the denominator could be undefined. For example, if 2 - log(x - 2) is negative, but we already have that 2 - log(x - 2) > 0, so that's already excluded. Therefore, the domain is indeed (2, 12) ∪ (12, 102).

But let me check with specific numbers. Let me pick x = 3. Then, x - 2 = 1, log(1) = 0, so denominator is log(2 - 0) = log(2) ≈ 0.3010, which is fine. So, f(3) is (2 - 3)/log(2) = (-1)/log(2), which is defined. How about x approaching 2 from the right? Let's say x approaches 2+, then x - 2 approaches 0+, so log(x - 2) approaches -infty. Therefore, 2 - log(x - 2) approaches 2 - (-infty) = +infty, so log(2 - log(x - 2)) approaches log(+infty) = +infty. Therefore, denominator approaches infinity, so f(x) approaches (2 - x)/infty, which is 0. So, no problem there. Similarly, as x approaches 102 from the left, x - 2 approaches 100, so log(x - 2) approaches log(100) = 2. Therefore, 2 - log(x - 2) approaches 2 - 2 = 0+. Therefore, log(2 - log(x - 2)) approaches log(0+) which is -infty. Therefore, denominator approaches -infty, so f(x) approaches (2 - x)/(-infty). Since 2 - x approaches 2 - 102 = -100, so (-100)/(-infty) approaches 0. So, again, no problem. At x = 12, denominator is log(2 - log(10)) = log(2 - 1) = log(1) = 0, so division by zero, which is undefined. So, that's the only point to exclude.

Therefore, I think my conclusion is correct. The domain is (2, 12) union (12, 102). Let me just check if there's any other possible mistake here.

Wait, another thought: when dealing with logarithms, sometimes there are more restrictions. For example, if the argument of the logarithm is negative, but we already considered that. Let me check again:

For log(x - 2) to be defined, x - 2 > 0 => x > 2. For log(2 - log(x - 2)) to be defined, 2 - log(x - 2) > 0 => log(x - 2) < 2 => x - 2 < 100 => x < 102. So, that's correct. So, combining x > 2 and x < 102 gives 2 < x < 102. Then, the denominator is zero when x = 12, so we remove that. Therefore, the domain is (2, 102) without 12. So, written as (2, 12) ∪ (12, 102).

Therefore, I think that's the answer. Let me check once more with another value. Let me pick x = 10. Then, x - 2 = 8, log(8) ≈ 0.9031. Then, 2 - log(8) ≈ 2 - 0.9031 ≈ 1.0969. Then, log(1.0969) ≈ 0.04, so denominator is approximately 0.04, so f(10) is (2 - 10)/0.04 = (-8)/0.04 = -200, which is defined. So, that's okay. How about x = 12? Then, denominator is log(2 - log(10)) = log(2 - 1) = log(1) = 0, so undefined. How about x = 100? Then, x - 2 = 98, log(98) ≈ 1.9912, so 2 - log(98) ≈ 2 - 1.9912 ≈ 0.0088, so log(0.0088) ≈ log(8.8 * 10^{-3}) ≈ log(8.8) + log(10^{-3}) ≈ 0.9445 - 3 ≈ -2.0555. Therefore, denominator is approximately -2.0555, so f(100) is (2 - 100)/(-2.0555) ≈ (-98)/(-2.0555) ≈ 47.67, which is defined. So, that's okay.

Another check: x = 12. Let me try x = 12. Then, denominator is log(2 - log(10)) = log(2 - 1) = log(1) = 0. So, division by zero, which is undefined. So, x = 12 is excluded.

How about x = 2.5? Then, x - 2 = 0.5, log(0.5) ≈ -0.3010, so 2 - (-0.3010) = 2.3010, log(2.3010) ≈ 0.362, so denominator is 0.362, so f(2.5) is (2 - 2.5)/0.362 ≈ (-0.5)/0.362 ≈ -1.38, which is defined.

Therefore, all these checks seem to confirm that the domain is indeed (2, 12) union (12, 102).

Wait, but let me check if there's a case where log(2 - log(x - 2)) is undefined. For example, if 2 - log(x - 2) is negative? But we already have that 2 - log(x - 2) > 0, so that's already taken care of. Therefore, no issues there.

Therefore, I think I can confidently say that the domain is (2, 12) ∪ (12, 102).

**Final Answer**
The domain of the function is \boxed{(2, 12) \cup (12, 102)}.

To determine the domain of the function
$$
f(x) = \frac{2 - x}{\log(2 - \log(x - 2))},
$$
we must identify all values of $ x $ for which the function is defined. This involves analyzing the conditions under which the logarithmic expressions and the denominator are valid.

---

### Step 1: Domain of the Inner Logarithm

The inner logarithm is $ \log(x - 2) $. For this to be defined, the argument must be positive:

$$
x - 2 > 0 \quad \Rightarrow \quad x > 2
$$

---

### Step 2: Domain of the Outer Logarithm

The outer logarithm is $ \log(2 - \log(x - 2)) $. For this to be defined, the argument must be positive:

$$
2 - \log(x - 2) > 0 \quad \Rightarrow \quad \log(x - 2) < 2
$$

Since the logarithm is base 10, this implies:

$$
x - 2 < 10^2 = 100 \quad \Rightarrow \quad x < 102
$$

Combining this with the previous condition $ x > 2 $, we now have:

$$
2 < x < 102
$$

---

### Step 3: Exclude Points Where the Denominator is Zero

The denominator is $ \log(2 - \log(x - 2)) $. We must ensure that this is **not zero**, because division by zero is undefined.

Set the denominator equal to zero:

$$
\log(2 - \log(x - 2)) = 0
$$

This implies:

$$
2 - \log(x - 2) = 10^0 = 1 \quad \Rightarrow \quad \log(x - 2) = 1
$$

Solving this:

$$
x - 2 = 10^1 = 10 \quad \Rightarrow \quad x = 12
$$

So, the function is **undefined** at $ x = 12 $, and we must **exclude** this value from the domain.

---

### Final Domain

Combining all the conditions:

- $ x > 2 $
- $ x < 102 $
- $ x \ne 12 $

This gives the domain as the union of two intervals:

$$
(2, 12) \cup (12, 102)
$$

---

### Final Answer

$$
\boxed{(2, 12) \cup (12, 102)}
$$

[Assistant]
Explanation:
Okay, so I need to find the domain of the function f(x) = (2 - x) / log(2 - log(x - 2)). Alright, let me recall that the domain of a function consists of all real numbers x for which the function is defined. Since this function is a rational expression involving a logarithm, I need to consider the restrictions imposed by both the denominator and the arguments of the logarithms.

First, let me note that the function has two logarithms here: one in the denominator, log(2 - log(x - 2)), and another inside that logarithm, which is log(x - 2). So, I need to make sure that both logarithms are defined, and also that the denominator isn't zero because division by zero is undefined.

Let me break this down step by step.

First, let's consider the inner logarithm: log(x - 2). For this logarithm to be defined, its argument must be positive. That means:

x - 2 > 0
=> x > 2

So, the first condition is that x must be greater than 2. That's straightforward.

Next, looking at the denominator: log(2 - log(x - 2)). For the logarithm to be defined, its argument must also be positive. Therefore:

2 - log(x - 2) > 0
=> log(x - 2) < 2

Since log here is base 10, log(x - 2) < 2 implies that:

x - 2 < 10^2
=> x - 2 < 100
=> x < 102

So, combining this with the previous condition x > 2, we have that x must be in (2, 102). However, we also need to ensure that the denominator is not zero. So, log(2 - log(x - 2)) ≠ 0.

Let me handle that. The denominator is zero when:

log(2 - log(x - 2)) = 0

Which implies:

2 - log(x - 2) = 10^0 = 1

Therefore:

2 - log(x - 2) = 1
=> log(x - 2) = 2 - 1 = 1
=> x - 2 = 10^1 = 10
=> x = 10 + 2 = 12

So, the denominator is zero when x = 12. Therefore, we need to exclude x = 12 from the domain.

So, putting it all together, the domain is all real numbers x such that 2 < x < 102, except x = 12. Therefore, in interval notation, that would be (2, 12) union (12, 102).

Wait, but let me check if there are any other restrictions. Let me verify each step again.

First, log(x - 2) requires x > 2. Then, log(2 - log(x - 2)) requires that 2 - log(x - 2) > 0, which leads to log(x - 2) < 2, so x - 2 < 100, so x < 102. So, combining x > 2 and x < 102 gives (2, 102). Then, the denominator is zero when x = 12, so we remove that point. Therefore, the domain is (2, 102) without 12. So, (2, 12) union (12, 102). That seems right.

But let me check if there are any other points where the denominator could be undefined or zero. For example, is there a case where log(2 - log(x - 2)) is undefined? Well, we already considered that 2 - log(x - 2) must be positive, so that's already covered by the condition log(x - 2) < 2. Therefore, as long as x is in (2, 102), the argument of the outer logarithm is positive, so the outer logarithm is defined. And we also excluded x = 12 where the denominator is zero.

But let me check if there are any other points where log(x - 2) might be problematic. For example, if log(x - 2) is negative? Wait, log(x - 2) can be negative or positive. For example, if x - 2 is between 0 and 1, then log(x - 2) is negative. But that's okay, as long as 2 - log(x - 2) is positive. Let me check with an example. Suppose x = 3, then x - 2 = 1, so log(1) = 0. Then 2 - 0 = 2, so log(2) is defined. If x = 2.5, then x - 2 = 0.5, so log(0.5) is negative, so 2 - log(0.5) is 2 - (-0.3010...) ≈ 2.3010, which is positive. So, that's okay. So, even if log(x - 2) is negative, as long as 2 - log(x - 2) is positive, which is already covered by x < 102. Therefore, there are no other restrictions.

Wait, but let me check if there's a case where log(x - 2) is undefined? But we already have x > 2, so x - 2 is positive, so log(x - 2) is defined. Therefore, the only restrictions are x > 2, x < 102, and x ≠ 12.

Therefore, the domain is (2, 102) without 12. So, written in interval notation, that's (2, 12) ∪ (12, 102). Let me check if there's any other possible issues.

Wait, let me check if x = 12 is the only point where the denominator is zero. Let me verify:

log(2 - log(x - 2)) = 0 implies 2 - log(x - 2) = 1, so log(x - 2) = 1, so x - 2 = 10^1 = 10, so x = 12. So, yes, only x = 12. Therefore, that's the only point to exclude.

Therefore, the domain is (2, 12) ∪ (12, 102). Let me check if there are any other points where the denominator could be undefined. For example, if 2 - log(x - 2) is negative, but we already have that 2 - log(x - 2) > 0, so that's already excluded. Therefore, the domain is indeed (2, 12) ∪ (12, 102).

But let me check with specific numbers. Let me pick x = 3. Then, x - 2 = 1, log(1) = 0, so denominator is log(2 - 0) = log(2) ≈ 0.3010, which is fine. So, f(3) is (2 - 3)/log(2) = (-1)/log(2), which is defined. How about x approaching 2 from the right? Let's say x approaches 2+, then x - 2 approaches 0+, so log(x - 2) approaches -infty. Therefore, 2 - log(x - 2) approaches 2 - (-infty) = +infty, so log(2 - log(x - 2)) approaches log(+infty) = +infty. Therefore, denominator approaches infinity, so f(x) approaches (2 - x)/infty, which is 0. So, no problem there. Similarly, as x approaches 102 from the left, x - 2 approaches 100, so log(x - 2) approaches log(100) = 2. Therefore, 2 - log(x - 2) approaches 2 - 2 = 0+. Therefore, log(2 - log(x - 2)) approaches log(0+) which is -infty. Therefore, denominator approaches -infty, so f(x) approaches (2 - x)/(-infty). Since 2 - x approaches 2 - 102 = -100, so (-100)/(-infty) approaches 0. So, again, no problem. At x = 12, denominator is log(2 - log(10)) = log(2 - 1) = log(1) = 0, so division by zero, which is undefined. So, that's the only point to exclude.

Therefore, I think my conclusion is correct. The domain is (2, 12) union (12, 102). Let me just check if there's any other possible mistake here.

Wait, another thought: when dealing with logarithms, sometimes there are more restrictions. For example, if the argument of the logarithm is negative, but we already considered that. Let me check again:

For log(x - 2) to be defined, x - 2 > 0 => x > 2. For log(2 - log(x - 2)) to be defined, 2 - log(x - 2) > 0 => log(x - 2) < 2 => x - 2 < 100 => x < 102. So, that's correct. So, combining x > 2 and x < 102 gives 2 < x < 102. Then, the denominator is zero when x = 12, so we remove that. Therefore, the domain is (2, 102) without 12. So, written as (2, 12) ∪ (12, 102).

Therefore, I think that's the answer. Let me check once more with another value. Let me pick x = 10. Then, x - 2 = 8, log(8) ≈ 0.9031. Then, 2 - log(8) ≈ 2 - 0.9031 ≈ 1.0969. Then, log(1.0969) ≈ 0.04, so denominator is approximately 0.04, so f(10) is (2 - 10)/0.04 = (-8)/0.04 = -200, which is defined. So, that's okay. How about x = 12? Then, denominator is log(2 - log(10)) = log(2 - 1) = log(1) = 0, so undefined. How about x = 100? Then, x - 2 = 98, log(98) ≈ 1.9912, so 2 - log(98) ≈ 2 - 1.9912 ≈ 0.0088, so log(0.0088) ≈ log(8.8 * 10^{-3}) ≈ log(8.8) + log(10^{-3}) ≈ 0.9445 - 3 ≈ -2.0555. Therefore, denominator is approximately -2.0555, so f(100) is (2 - 100)/(-2.0555) ≈ (-98)/(-2.0555) ≈ 47.67, which is defined. So, that's okay.

Another check: x = 12. Let me try x = 12. Then, denominator is log(2 - log(10)) = log(2 - 1) = log(1) = 0. So, division by zero, which is undefined. So, x = 12 is excluded.

How about x = 2.5? Then, x - 2 = 0.5, log(0.5) ≈ -0.3010, so 2 - (-0.3010) = 2.3010, log(2.3010) ≈ 0.362, so denominator is 0.362, so f(2.5) is (2 - 2.5)/0.362 ≈ (-0.5)/0.362 ≈ -1.38, which is defined.

Therefore, all these checks seem to confirm that the domain is indeed (2, 12) union (12, 102).

Wait, but let me check if there's a case where log(2 - log(x - 2)) is undefined. For example, if 2 - log(x - 2) is negative? But we already have that 2 - log(x - 2) > 0, so that's already taken care of. Therefore, no issues there.

Therefore, I think I can confidently say that the domain is (2, 12) ∪ (12, 102).

**Final Answer**
The domain of the function is \boxed{(2, 12) \cup (12, 102)}.

To determine the domain of the function
$$
f(x) = \frac{2 - x}{\log(2 - \log(x - 2))},
$$
we must identify all values of $ x $ for which the function is defined. This involves analyzing the conditions under which the logarithmic expressions and the denominator are valid.

---

### Step 1: Domain of the Inner Logarithm

The inner logarithm is $ \log(x - 2) $. For this to be defined, the argument must be positive:

$$
x - 2 > 0 \quad \Rightarrow \quad x > 2
$$

---

### Step 2: Domain of the Outer Logarithm

The outer logarithm is $ \log(2 - \log(x - 2)) $. For this to be defined, the argument must be positive:

$$
2 - \log(x - 2) > 0 \quad \Rightarrow \quad \log(x - 2) < 2
$$

Since the logarithm is base 10, this implies:

$$
x - 2 < 10^2 = 100 \quad \Rightarrow \quad x < 102
$$

Combining this with the previous condition $ x > 2 $, we now have:

$$
2 < x < 102
$$

---

### Step 3: Exclude Points Where the Denominator is Zero

The denominator is $ \log(2 - \log(x - 2)) $. We must ensure that this is **not zero**, because division by zero is undefined.

Set the denominator equal to zero:

$$
\log(2 - \log(x - 2)) = 0
$$

This implies:

$$
2 - \log(x - 2) = 10^0 = 1 \quad \Rightarrow \quad \log(x - 2) = 1
$$

Solving this:

$$
x - 2 = 10^1 = 10 \quad \Rightarrow \quad x = 12
$$

So, the function is **undefined** at $ x = 12 $, and we must **exclude** this value from the domain.

---

### Final Domain

Combining all the conditions:

- $ x > 2 $
- $ x < 102 $
- $ x \ne 12 $

This gives the domain as the union of two intervals:

$$
(2, 12) \cup (12, 102)
$$

---

### Final Answer

$$
\boxed{(2, 12) \cup (12, 102)}
$$

Solution: The inner logarithm is only defined if $x - 2 > 0$, so $x > 2$. Furthermore, the outer logarithm is only defined if $2 - \log(x-2) > 0$, which implies that $2 > \log(x-2)$, so that $100 > x-2$. Thus, $x < 102$. Finally, it must also be true that $\log(2-\log(x-2)) \neq 0$, so that $2 - \log(x-2) \neq 1$. Equivalently, $\log(x-2) \neq 1$, so $x \neq 12$. Thus, the answer is $x \in \boxed{(2,12) \cup (12,102)}$