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 up to the nearest integer. So ceiling of 1.0472 is 2. Therefore, f(π/3) = 2. Let me check that again. π/3 is about 1.047, so yes, the ceiling is 2. Okay, that seems right.

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 take the floor of √45. The floor function gives the greatest integer less than or equal to the number. So floor(6.708) is 6. Therefore, f(√45) = 6. Let me confirm that. Yes, 6.708 is between 6 and 7, so floor is 6. Correct.

Third, let's compute f(8^(2/3)). Let me figure out what 8^(2/3) is. Remember that exponents can be tricky. Let me recall that 8 is 2³. So 8^(2/3) can be rewritten as (2³)^(2/3). When you raise a power to a power, you multiply the exponents. So 3*(2/3) = 2. Therefore, 8^(2/3) = 2² = 4. Wait, so 8^(2/3) is exactly 4? Let me check that again. Alternatively, 8^(1/3) is the cube root of 8, which is 2. Then raising that to the 2nd power gives 2² = 4. Yes, that's correct. So 8^(2/3) is exactly 4. Therefore, since n is equal to 4, which is the boundary case. The function says if n ≥ 4, use floor(n). So floor(4) is 4. Therefore, f(8^(2/3)) = 4.

So putting it all together: f(π/3) is 2, f(√45) is 6, and f(8^(2/3)) is 4. Therefore, adding them up: 2 + 6 + 4 = 12.

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

First, 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.

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

Third, 8^(2/3): As I calculated, that's 4. So since n is exactly 4, floor(4) is 4. Correct.

Adding them up: 2 + 6 is 8, plus 4 is 12. So the total is 12.

Wait, but let me check if there's any chance that 8^(2/3) might be something else? Let me think. Another way to compute 8^(2/3): 2/3 exponent can be thought of as taking the cube root first, then squaring. Cube root of 8 is 2, square is 4. Alternatively, square first: 8 squared is 64, then cube root of 64 is 4. Either way, same result. So yes, 8^(2/3) is 4.

Another check: maybe I made a mistake with the function definition? Let me check again. The function f(n) is floor(n) if n ≥ 4, else ceiling(n). So for numbers less than 4, ceiling. For numbers greater or equal to 4, floor. So for n=4, floor(4) is 4. Correct.

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

But just to be thorough, let me check if there's any possible mistake in evaluating the numbers.

For π/3: π is approximately 3.14159265, divided by 3 is approximately 1.04719755. So ceiling of that is indeed 2.

For √45: √45 is √(9*5) = 3√5 ≈ 3*2.23607 ≈ 6.7082. So floor of that is 6.

For 8^(2/3): As above, 4.

So adding 2 + 6 + 4 = 12.

Therefore, I think the answer is 12.

But just to make sure, let me check if there's any alternative interpretation of the function. For example, is there a possibility that the function is defined differently? The problem says:

f(n) is floor(n) if n ≥ 4, else ceiling(n). So for n exactly equal to 4, floor(n) is 4. So that's 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 for example, if n is 4, floor(4) = 4, ceiling(4) = 4. So in that case, regardless of which one you use, you get the same result. So for n=4, it's 4.

So all steps are correct.

Therefore, I think 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 values of $ n $ **less than 4**, we apply the **ceiling function**, and for values **greater than or equal to 4**, we apply the **floor function**.

---

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

We know that:
$$
\pi \approx 3.1416 \Rightarrow \frac{\pi}{3} \approx 1.0472
$$
Since $ \frac{\pi}{3} < 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:
$$
\sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5} \approx 3 \cdot 2.236 = 6.708
$$
Since $ \sqrt{45} > 4 $, we apply the floor function:
$$
f(\sqrt{45}) = \lfloor 6.708 \rfloor = 6
$$

---

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

We simplify the exponent:
$$
8^{2/3} = (2^3)^{2/3} = 2^{(3 \cdot \frac{2}{3})} = 2^2 = 4
$$
Since $ 8^{2/3} = 4 $, and $ 4 \geq 4 $, we apply the floor function:
$$
f(8^{2/3}) = \lfloor 4 \rfloor = 4
$$

---

### Final Step: Add the Results

Now, summing the three values:
$$
f\left(\frac{\pi}{3}\right) + f(\sqrt{45}) + f(8^{2/3}) = 2 + 6 + 4 = \boxed{12}
$$

[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 up to the nearest integer. So ceiling of 1.0472 is 2. Therefore, f(π/3) = 2. Let me check that again. π/3 is about 1.047, so yes, the ceiling is 2. Okay, that seems right.

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 take the floor of √45. The floor function gives the greatest integer less than or equal to the number. So floor(6.708) is 6. Therefore, f(√45) = 6. Let me confirm that. Yes, 6.708 is between 6 and 7, so floor is 6. Correct.

Third, let's compute f(8^(2/3)). Let me figure out what 8^(2/3) is. Remember that exponents can be tricky. Let me recall that 8 is 2³. So 8^(2/3) can be rewritten as (2³)^(2/3). When you raise a power to a power, you multiply the exponents. So 3*(2/3) = 2. Therefore, 8^(2/3) = 2² = 4. Wait, so 8^(2/3) is exactly 4? Let me check that again. Alternatively, 8^(1/3) is the cube root of 8, which is 2. Then raising that to the 2nd power gives 2² = 4. Yes, that's correct. So 8^(2/3) is exactly 4. Therefore, since n is equal to 4, which is the boundary case. The function says if n ≥ 4, use floor(n). So floor(4) is 4. Therefore, f(8^(2/3)) = 4.

So putting it all together: f(π/3) is 2, f(√45) is 6, and f(8^(2/3)) is 4. Therefore, adding them up: 2 + 6 + 4 = 12.

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

First, 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.

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

Third, 8^(2/3): As I calculated, that's 4. So since n is exactly 4, floor(4) is 4. Correct.

Adding them up: 2 + 6 is 8, plus 4 is 12. So the total is 12.

Wait, but let me check if there's any chance that 8^(2/3) might be something else? Let me think. Another way to compute 8^(2/3): 2/3 exponent can be thought of as taking the cube root first, then squaring. Cube root of 8 is 2, square is 4. Alternatively, square first: 8 squared is 64, then cube root of 64 is 4. Either way, same result. So yes, 8^(2/3) is 4.

Another check: maybe I made a mistake with the function definition? Let me check again. The function f(n) is floor(n) if n ≥ 4, else ceiling(n). So for numbers less than 4, ceiling. For numbers greater or equal to 4, floor. So for n=4, floor(4) is 4. Correct.

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

But just to be thorough, let me check if there's any possible mistake in evaluating the numbers.

For π/3: π is approximately 3.14159265, divided by 3 is approximately 1.04719755. So ceiling of that is indeed 2.

For √45: √45 is √(9*5) = 3√5 ≈ 3*2.23607 ≈ 6.7082. So floor of that is 6.

For 8^(2/3): As above, 4.

So adding 2 + 6 + 4 = 12.

Therefore, I think the answer is 12.

But just to make sure, let me check if there's any alternative interpretation of the function. For example, is there a possibility that the function is defined differently? The problem says:

f(n) is floor(n) if n ≥ 4, else ceiling(n). So for n exactly equal to 4, floor(n) is 4. So that's 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 for example, if n is 4, floor(4) = 4, ceiling(4) = 4. So in that case, regardless of which one you use, you get the same result. So for n=4, it's 4.

So all steps are correct.

Therefore, I think 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 values of $ n $ **less than 4**, we apply the **ceiling function**, and for values **greater than or equal to 4**, we apply the **floor function**.

---

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

We know that:
$$
\pi \approx 3.1416 \Rightarrow \frac{\pi}{3} \approx 1.0472
$$
Since $ \frac{\pi}{3} < 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:
$$
\sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5} \approx 3 \cdot 2.236 = 6.708
$$
Since $ \sqrt{45} > 4 $, we apply the floor function:
$$
f(\sqrt{45}) = \lfloor 6.708 \rfloor = 6
$$

---

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

We simplify the exponent:
$$
8^{2/3} = (2^3)^{2/3} = 2^{(3 \cdot \frac{2}{3})} = 2^2 = 4
$$
Since $ 8^{2/3} = 4 $, and $ 4 \geq 4 $, we apply the floor function:
$$
f(8^{2/3}) = \lfloor 4 \rfloor = 4
$$

---

### Final Step: Add the Results

Now, summing the three values:
$$
f\left(\frac{\pi}{3}\right) + f(\sqrt{45}) + f(8^{2/3}) = 2 + 6 + 4 = \boxed{12}
$$

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 primes, but with exponents such that the product of exponents plus one equals three? But since 3 is prime, the only way is one exponent plus one equals three, so only one prime factor with exponent 2. So yes, numbers with exactly three divisors are squares of primes.

Therefore, the problem reduces to listing the squares of primes in ascending order and finding the fifth one.

Let me list the primes in order and then square them:

The first prime is 2. $ 2^2 = 4 $.

Second prime is 3. $ 3^2 = 9 $.

Third prime is 5. $ 5^2 = 25 $.

Fourth prime is 7. $ 7^2 = 49 $.

Fifth prime is 11. $ 11^2 = 121 $.

Wait, so the fifth number would be 121? But let me check if there are any numbers with exactly three divisors that are not squares of primes. Maybe I missed something?

Suppose there is a number that is a product of two distinct primes, say p and q. Then the number of divisors would be (1+1)(1+1) = 4. So that's four divisors. So numbers that are products of two distinct primes have four divisors. Therefore, they are not in our list.

What about numbers with more than two prime factors? For example, if a number is $ p \times q \times r $, then the number of divisors is (1+1)(1+1)(1+1) = 8. So that's eight divisors. If a number is $ p^3 $, then the number of divisors is 3 + 1 = 4. So that's four divisors as well.

So indeed, only squares of primes have exactly three divisors. Therefore, the numbers we are looking for are squares of primes.

So, the list of numbers with exactly three divisors is 4, 9, 25, 49, 121, 169, etc. So the fifth one is 121.

But wait, let me check if there are any numbers between, say, 4 and 121 that I might have missed. For example, let me list them step by step:

First, check numbers starting from 1:

1 has 1 divisor. Not three.

2 has divisors 1, 2. Two divisors.

3 has 1, 3. Two divisors.

4 has 1, 2, 4. Three divisors. So that's the first one.

5: 1, 5. Two divisors.

6: 1, 2, 3, 6. Four divisors.

7: 1, 7. Two divisors.

8: 1, 2, 4, 8. Four divisors.

9: 1, 3, 9. Three divisors. Second number.

10: 1, 2, 5, 10. Four divisors.

11: 1, 11. Two divisors.

12: 1, 2, 3, 4, 6, 12. Six divisors.

13: 1, 13. Two divisors.

14: 1, 2, 7, 14. Four divisors.

15: 1, 3, 5, 15. Four divisors.

16: 1, 2, 4, 8, 16. Five divisors.

17: 1, 17. Two divisors.

18: 1, 2, 3, 6, 9, 18. Six divisors.

19: 1, 19. Two divisors.

20: 1, 2, 4, 5, 10, 20. Six divisors.

21: 1, 3, 7, 21. Four divisors.

22: 1, 2, 11, 22. Four divisors.

23: 1, 23. Two divisors.

24: 1, 2, 3, 4, 6, 8, 12, 24. Eight divisors.

25: 1, 5, 25. Three divisors. Third number.

26: 1, 2, 13, 26. Four divisors.

27: 1, 3, 9, 27. Four divisors.

28: 1, 2, 4, 7, 14, 28. Six divisors.

29: 1, 29. Two divisors.

30: 1, 2, 3, 5, 6, 10, 15, 30. Eight divisors.

31: 1, 31. Two divisors.

32: 1, 2, 4, 8, 16, 32. Six divisors.

33: 1, 3, 11, 33. Four divisors.

34: 1, 2, 17, 34. Four divisors.

35: 1, 5, 7, 35. Four divisors.

36: 1, 2, 3, 4, 6, 9, 12, 18, 36. Nine divisors.

37: 1, 37. Two divisors.

38: 1, 2, 19, 38. Four divisors.

39: 1, 3, 13, 39. Four divisors.

40: 1, 2, 4, 5, 8, 10, 20, 40. Eight divisors.

41: 1, 41. Two divisors.

42: 1, 2, 3, 6, 7, 14, 21, 42. Eight divisors.

43: 1, 43. Two divisors.

44: 1, 2, 4, 11, 22, 44. Six divisors.

45: 1, 3, 5, 9, 15, 45. Six divisors.

46: 1, 2, 23, 46. Four divisors.

47: 1, 47. Two divisors.

48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Ten divisors.

49: 1, 7, 49. Three divisors. Fourth number.

50: 1, 2, 5, 10, 25, 50. Six divisors.

51: 1, 3, 17, 51. Four divisors.

52: 1, 2, 4, 13, 26, 52. Six divisors.

53: 1, 53. Two divisors.

54: 1, 2, 3, 6, 9, 18, 27, 54. Eight divisors.

55: 1, 5, 11, 55. Four divisors.

56: 1, 2, 4, 7, 8, 14, 28, 56. Eight divisors.

57: 1, 3, 19, 57. Four divisors.

58: 1, 2, 29, 58. Four divisors.

59: 1, 59. Two divisors.

60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. Twelve divisors.

61: 1, 61. Two divisors.

62: 1, 2, 31, 62. Four divisors.

63: 1, 3, 7, 9, 21, 63. Six divisors.

64: 1, 2, 4, 8, 16, 32, 64. Seven divisors.

65: 1, 5, 13, 65. Four divisors.

66: 1, 2, 3, 6, 11, 22, 33, 66. Eight divisors.

67: 1, 67. Two divisors.

68: 1, 2, 4, 17, 34, 68. Six divisors.

69: 1, 3, 23, 69. Four divisors.

70: 1, 2, 5, 7, 10, 14, 35, 70. Eight divisors.

71: 1, 71. Two divisors.

72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. Twelve divisors.

73: 1, 73. Two divisors.

74: 1, 2, 37, 74. Four divisors.

75: 1, 3, 5, 15, 25, 75. Six divisors.

76: 1, 2, 4, 19, 38, 76. Six divisors.

77: 1, 7, 11, 77. Four divisors.

78: 1, 2, 3, 6, 13, 26, 39, 78. Eight divisors.

79: 1, 79. Two divisors.

80: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80. Ten divisors.

81: 1, 3, 9, 27, 81. Five divisors.

82: 1, 2, 41, 82. Four divisors.

83: 1, 83. Two divisors.

84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84. Twelve divisors.

85: 1, 5, 17, 85. Four divisors.

86: 1, 2, 43, 86. Four divisors.

87: 1, 3, 29, 87. Four divisors.

88: 1, 2, 4, 8, 11, 22, 44, 88. Eight divisors.

89: 1, 89. Two divisors.

90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. Twelve divisors.

91: 1, 7, 13, 91. Four divisors.

92: 1, 2, 4, 23, 46, 92. Six divisors.

93: 1, 3, 31, 93. Four divisors.

94: 1, 2, 47, 94. Four divisors.

95: 1, 5, 19, 95. Four divisors.

96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96. Twelve divisors.

97: 1, 97. Two divisors.

98: 1, 2, 7, 14, 49, 98. Six divisors.

99: 1, 3, 9, 11, 33, 99. Six divisors.

100: 1, 2, 4, 5, 10, 20, 25, 50, 100. Nine divisors.

Okay, so up to 100, the numbers with exactly three divisors are 4, 9, 25, 49. So that's four numbers. Then the next one would be 11 squared, which is 121. Let me check if there are any numbers between 49 and 121 that I might have missed.

Looking at numbers between 50 and 100, I checked up to 100 and didn't find any other numbers with exactly three divisors. So the next one after 49 is 121. Therefore, the fifth number is 121.

But let me confirm once again. Let me check 121. Its divisors are 1, 11, 121. Exactly three. So that's correct.

But just to be thorough, let me check if there's a number between 49 and 121 that is a square of a prime. The primes after 7 are 11, 13, 17, etc. So 11 squared is 121, which is the next one. The primes between 7 and 11 are... well, 7 is the fourth prime, then 11 is the fifth prime. Wait, primes are 2, 3, 5, 7, 11, 13, 17... So the squares would be 4 (2²), 9 (3²), 25 (5²), 49 (7²), 121 (11²), 169 (13²), etc. So yes, the fifth one is 121.

Therefore, the fifth number in the list is 121.

But just to make sure I didn't skip any primes. Let me list the primes in order:

1st prime: 2

2nd: 3

3rd: 5

4th: 7

5th: 11

6th: 13

So their squares are 4, 9, 25, 49, 121, 169... So fifth is 121.

Therefore, the answer is 121.

But wait, let me check if there's a number that is not a square of a prime but still has exactly three divisors. Suppose someone says, for example, 16. Wait, 16 is 2^4. Its divisors are 1, 2, 4, 8, 16. That's five divisors. Not three. How about 8? Divisors are 1, 2, 4, 8. Four divisors.

What about 1? Divisors are just 1. Not three.

How about 0? Not positive.

So no, there are no other numbers with exactly three divisors except squares of primes.

Therefore, the fifth number is 121.

But let me check once again the list up to 100. The numbers with exactly three divisors are:

4 (2²), 9 (3²), 25 (5²), 49 (7²). So that's four numbers. Then the next one is 11²=121. Therefore, fifth is 121.

Yes, that seems correct.

Alternatively, maybe I can think of the list of numbers with exactly three divisors as the squares of primes. So the sequence is 4, 9, 25, 49, 121, 169, 289, 361, 529, etc. So fifth term is 121.

Therefore, the answer is 121.

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

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

---

### Step 1: Understanding the Number of Divisors

A number $ n $ has a number of positive divisors determined by its prime factorization. If:

$$
n = p_1^{a_1} \cdot p_2^{a_2} \cdot \dots \cdot p_k^{a_k}
$$

then the total number of positive divisors of $ n $ is:

$$
(a_1 + 1)(a_2 + 1)\dots(a_k + 1)
$$

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 have **one prime factor raised to the power of 2**, i.e., $ n = p^2 $, where $ p $ is a prime number. The divisors of $ p^2 $ are:

$$
1, p, p^2
$$

So, **only the squares of prime numbers** have exactly three positive divisors.

---

### Step 2: Generate the Sequence

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 $

Thus, the sequence of numbers with exactly three positive divisors is:

$$
4, 9, 25, 49, 121, \dots
$$

---

### 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 primes, but with exponents such that the product of exponents plus one equals three? But since 3 is prime, the only way is one exponent plus one equals three, so only one prime factor with exponent 2. So yes, numbers with exactly three divisors are squares of primes.

Therefore, the problem reduces to listing the squares of primes in ascending order and finding the fifth one.

Let me list the primes in order and then square them:

The first prime is 2. $ 2^2 = 4 $.

Second prime is 3. $ 3^2 = 9 $.

Third prime is 5. $ 5^2 = 25 $.

Fourth prime is 7. $ 7^2 = 49 $.

Fifth prime is 11. $ 11^2 = 121 $.

Wait, so the fifth number would be 121? But let me check if there are any numbers with exactly three divisors that are not squares of primes. Maybe I missed something?

Suppose there is a number that is a product of two distinct primes, say p and q. Then the number of divisors would be (1+1)(1+1) = 4. So that's four divisors. So numbers that are products of two distinct primes have four divisors. Therefore, they are not in our list.

What about numbers with more than two prime factors? For example, if a number is $ p \times q \times r $, then the number of divisors is (1+1)(1+1)(1+1) = 8. So that's eight divisors. If a number is $ p^3 $, then the number of divisors is 3 + 1 = 4. So that's four divisors as well.

So indeed, only squares of primes have exactly three divisors. Therefore, the numbers we are looking for are squares of primes.

So, the list of numbers with exactly three divisors is 4, 9, 25, 49, 121, 169, etc. So the fifth one is 121.

But wait, let me check if there are any numbers between, say, 4 and 121 that I might have missed. For example, let me list them step by step:

First, check numbers starting from 1:

1 has 1 divisor. Not three.

2 has divisors 1, 2. Two divisors.

3 has 1, 3. Two divisors.

4 has 1, 2, 4. Three divisors. So that's the first one.

5: 1, 5. Two divisors.

6: 1, 2, 3, 6. Four divisors.

7: 1, 7. Two divisors.

8: 1, 2, 4, 8. Four divisors.

9: 1, 3, 9. Three divisors. Second number.

10: 1, 2, 5, 10. Four divisors.

11: 1, 11. Two divisors.

12: 1, 2, 3, 4, 6, 12. Six divisors.

13: 1, 13. Two divisors.

14: 1, 2, 7, 14. Four divisors.

15: 1, 3, 5, 15. Four divisors.

16: 1, 2, 4, 8, 16. Five divisors.

17: 1, 17. Two divisors.

18: 1, 2, 3, 6, 9, 18. Six divisors.

19: 1, 19. Two divisors.

20: 1, 2, 4, 5, 10, 20. Six divisors.

21: 1, 3, 7, 21. Four divisors.

22: 1, 2, 11, 22. Four divisors.

23: 1, 23. Two divisors.

24: 1, 2, 3, 4, 6, 8, 12, 24. Eight divisors.

25: 1, 5, 25. Three divisors. Third number.

26: 1, 2, 13, 26. Four divisors.

27: 1, 3, 9, 27. Four divisors.

28: 1, 2, 4, 7, 14, 28. Six divisors.

29: 1, 29. Two divisors.

30: 1, 2, 3, 5, 6, 10, 15, 30. Eight divisors.

31: 1, 31. Two divisors.

32: 1, 2, 4, 8, 16, 32. Six divisors.

33: 1, 3, 11, 33. Four divisors.

34: 1, 2, 17, 34. Four divisors.

35: 1, 5, 7, 35. Four divisors.

36: 1, 2, 3, 4, 6, 9, 12, 18, 36. Nine divisors.

37: 1, 37. Two divisors.

38: 1, 2, 19, 38. Four divisors.

39: 1, 3, 13, 39. Four divisors.

40: 1, 2, 4, 5, 8, 10, 20, 40. Eight divisors.

41: 1, 41. Two divisors.

42: 1, 2, 3, 6, 7, 14, 21, 42. Eight divisors.

43: 1, 43. Two divisors.

44: 1, 2, 4, 11, 22, 44. Six divisors.

45: 1, 3, 5, 9, 15, 45. Six divisors.

46: 1, 2, 23, 46. Four divisors.

47: 1, 47. Two divisors.

48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Ten divisors.

49: 1, 7, 49. Three divisors. Fourth number.

50: 1, 2, 5, 10, 25, 50. Six divisors.

51: 1, 3, 17, 51. Four divisors.

52: 1, 2, 4, 13, 26, 52. Six divisors.

53: 1, 53. Two divisors.

54: 1, 2, 3, 6, 9, 18, 27, 54. Eight divisors.

55: 1, 5, 11, 55. Four divisors.

56: 1, 2, 4, 7, 8, 14, 28, 56. Eight divisors.

57: 1, 3, 19, 57. Four divisors.

58: 1, 2, 29, 58. Four divisors.

59: 1, 59. Two divisors.

60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. Twelve divisors.

61: 1, 61. Two divisors.

62: 1, 2, 31, 62. Four divisors.

63: 1, 3, 7, 9, 21, 63. Six divisors.

64: 1, 2, 4, 8, 16, 32, 64. Seven divisors.

65: 1, 5, 13, 65. Four divisors.

66: 1, 2, 3, 6, 11, 22, 33, 66. Eight divisors.

67: 1, 67. Two divisors.

68: 1, 2, 4, 17, 34, 68. Six divisors.

69: 1, 3, 23, 69. Four divisors.

70: 1, 2, 5, 7, 10, 14, 35, 70. Eight divisors.

71: 1, 71. Two divisors.

72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. Twelve divisors.

73: 1, 73. Two divisors.

74: 1, 2, 37, 74. Four divisors.

75: 1, 3, 5, 15, 25, 75. Six divisors.

76: 1, 2, 4, 19, 38, 76. Six divisors.

77: 1, 7, 11, 77. Four divisors.

78: 1, 2, 3, 6, 13, 26, 39, 78. Eight divisors.

79: 1, 79. Two divisors.

80: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80. Ten divisors.

81: 1, 3, 9, 27, 81. Five divisors.

82: 1, 2, 41, 82. Four divisors.

83: 1, 83. Two divisors.

84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84. Twelve divisors.

85: 1, 5, 17, 85. Four divisors.

86: 1, 2, 43, 86. Four divisors.

87: 1, 3, 29, 87. Four divisors.

88: 1, 2, 4, 8, 11, 22, 44, 88. Eight divisors.

89: 1, 89. Two divisors.

90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. Twelve divisors.

91: 1, 7, 13, 91. Four divisors.

92: 1, 2, 4, 23, 46, 92. Six divisors.

93: 1, 3, 31, 93. Four divisors.

94: 1, 2, 47, 94. Four divisors.

95: 1, 5, 19, 95. Four divisors.

96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96. Twelve divisors.

97: 1, 97. Two divisors.

98: 1, 2, 7, 14, 49, 98. Six divisors.

99: 1, 3, 9, 11, 33, 99. Six divisors.

100: 1, 2, 4, 5, 10, 20, 25, 50, 100. Nine divisors.

Okay, so up to 100, the numbers with exactly three divisors are 4, 9, 25, 49. So that's four numbers. Then the next one would be 11 squared, which is 121. Let me check if there are any numbers between 49 and 121 that I might have missed.

Looking at numbers between 50 and 100, I checked up to 100 and didn't find any other numbers with exactly three divisors. So the next one after 49 is 121. Therefore, the fifth number is 121.

But let me confirm once again. Let me check 121. Its divisors are 1, 11, 121. Exactly three. So that's correct.

But just to be thorough, let me check if there's a number between 49 and 121 that is a square of a prime. The primes after 7 are 11, 13, 17, etc. So 11 squared is 121, which is the next one. The primes between 7 and 11 are... well, 7 is the fourth prime, then 11 is the fifth prime. Wait, primes are 2, 3, 5, 7, 11, 13, 17... So the squares would be 4 (2²), 9 (3²), 25 (5²), 49 (7²), 121 (11²), 169 (13²), etc. So yes, the fifth one is 121.

Therefore, the fifth number in the list is 121.

But just to make sure I didn't skip any primes. Let me list the primes in order:

1st prime: 2

2nd: 3

3rd: 5

4th: 7

5th: 11

6th: 13

So their squares are 4, 9, 25, 49, 121, 169... So fifth is 121.

Therefore, the answer is 121.

But wait, let me check if there's a number that is not a square of a prime but still has exactly three divisors. Suppose someone says, for example, 16. Wait, 16 is 2^4. Its divisors are 1, 2, 4, 8, 16. That's five divisors. Not three. How about 8? Divisors are 1, 2, 4, 8. Four divisors.

What about 1? Divisors are just 1. Not three.

How about 0? Not positive.

So no, there are no other numbers with exactly three divisors except squares of primes.

Therefore, the fifth number is 121.

But let me check once again the list up to 100. The numbers with exactly three divisors are:

4 (2²), 9 (3²), 25 (5²), 49 (7²). So that's four numbers. Then the next one is 11²=121. Therefore, fifth is 121.

Yes, that seems correct.

Alternatively, maybe I can think of the list of numbers with exactly three divisors as the squares of primes. So the sequence is 4, 9, 25, 49, 121, 169, 289, 361, 529, etc. So fifth term is 121.

Therefore, the answer is 121.

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

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

---

### Step 1: Understanding the Number of Divisors

A number $ n $ has a number of positive divisors determined by its prime factorization. If:

$$
n = p_1^{a_1} \cdot p_2^{a_2} \cdot \dots \cdot p_k^{a_k}
$$

then the total number of positive divisors of $ n $ is:

$$
(a_1 + 1)(a_2 + 1)\dots(a_k + 1)
$$

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 have **one prime factor raised to the power of 2**, i.e., $ n = p^2 $, where $ p $ is a prime number. The divisors of $ p^2 $ are:

$$
1, p, p^2
$$

So, **only the squares of prime numbers** have exactly three positive divisors.

---

### Step 2: Generate the Sequence

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 $

Thus, the sequence of numbers with exactly three positive divisors is:

$$
4, 9, 25, 49, 121, \dots
$$

---

### 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, t must be greater than 4 and less than 12.25. Therefore, integers from 5 to 12 inclusive. Which is 8 numbers. So, 8 integer values.

But let me check with specific numbers. Let's take t=5. √5 is approximately 2.236, which is greater than 2 and less than 3.5. That works. How about t=12? √12 is approximately 3.464, which is 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, 5,6,7,8,9,10,11,12. That's 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 between 4 and 12.25. So, same as before. So, integers t where 4 < t < 12.25. So, same conclusion. So, 5 to 12 inclusive. 8 numbers. So, seems consistent.

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 3.5 squared. So, same thing. So, yeah, seems like 8 integers.

But let me check if there's a possibility that t could be a non-integer but the question is about integer t. Wait, the question says "how many integer values of t satisfy this condition". So, t must be integer, but the square root of t can be non-integer. So, for example, t=5 is okay, even though √5 is irrational. So, that's fine.

Alternatively, maybe I need to check if there are any other constraints? For example, t has to be non-negative because you can't take the square root of a negative number. But since t is an integer and we have t > 4, which is already positive, so that's covered.

So, I think my reasoning is correct. The integer values of t are from 5 to 12 inclusive, which is 8 numbers. Therefore, the answer is 8. Wait, but let me check once more by listing them:

5, 6, 7, 8, 9, 10, 11, 12. That's 8 numbers. Let me count again:

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

Yep, 8 numbers. So, I think that's correct. Therefore, the number of integer values of t is 8.

But just to be thorough, let me check the endpoints again. For t=5: √5 ≈ 2.236, which is between 2 and 3.5. For t=12: √12 ≈ 3.464, which is less than 3.5. What about t=12.25? Well, t has to be less than 12.25, so t=12.25 is not allowed, but since t must be integer, the next lower integer is 12. So, that's correct. Similarly, t=4: √4=2, which is not greater than 2, so excluded. So, all the numbers from 5 to 12 inclusive are valid. Therefore, 8 integers.

I think that's solid. I don't see any mistakes in my reasoning. So, the answer should be 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:

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

we begin by eliminating the square root by squaring all parts of the inequality. Since square roots are always non-negative, and both 2 and 3.5 are positive, squaring is valid and preserves the inequality direction.

---

### Step 1: Square the Inequality

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

Squaring all parts:

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

$$
4 < t < 12.25
$$

---

### Step 2: Identify Integer Values of $ t $

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 Integer Values

We count the integers from 5 to 12 inclusive:

$$
12 - 5 + 1 = 8
$$

---

### 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, t must be greater than 4 and less than 12.25. Therefore, integers from 5 to 12 inclusive. Which is 8 numbers. So, 8 integer values.

But let me check with specific numbers. Let's take t=5. √5 is approximately 2.236, which is greater than 2 and less than 3.5. That works. How about t=12? √12 is approximately 3.464, which is 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, 5,6,7,8,9,10,11,12. That's 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 between 4 and 12.25. So, same as before. So, integers t where 4 < t < 12.25. So, same conclusion. So, 5 to 12 inclusive. 8 numbers. So, seems consistent.

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 3.5 squared. So, same thing. So, yeah, seems like 8 integers.

But let me check if there's a possibility that t could be a non-integer but the question is about integer t. Wait, the question says "how many integer values of t satisfy this condition". So, t must be integer, but the square root of t can be non-integer. So, for example, t=5 is okay, even though √5 is irrational. So, that's fine.

Alternatively, maybe I need to check if there are any other constraints? For example, t has to be non-negative because you can't take the square root of a negative number. But since t is an integer and we have t > 4, which is already positive, so that's covered.

So, I think my reasoning is correct. The integer values of t are from 5 to 12 inclusive, which is 8 numbers. Therefore, the answer is 8. Wait, but let me check once more by listing them:

5, 6, 7, 8, 9, 10, 11, 12. That's 8 numbers. Let me count again:

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

Yep, 8 numbers. So, I think that's correct. Therefore, the number of integer values of t is 8.

But just to be thorough, let me check the endpoints again. For t=5: √5 ≈ 2.236, which is between 2 and 3.5. For t=12: √12 ≈ 3.464, which is less than 3.5. What about t=12.25? Well, t has to be less than 12.25, so t=12.25 is not allowed, but since t must be integer, the next lower integer is 12. So, that's correct. Similarly, t=4: √4=2, which is not greater than 2, so excluded. So, all the numbers from 5 to 12 inclusive are valid. Therefore, 8 integers.

I think that's solid. I don't see any mistakes in my reasoning. So, the answer should be 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:

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

we begin by eliminating the square root by squaring all parts of the inequality. Since square roots are always non-negative, and both 2 and 3.5 are positive, squaring is valid and preserves the inequality direction.

---

### Step 1: Square the Inequality

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

Squaring all parts:

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

$$
4 < t < 12.25
$$

---

### Step 2: Identify Integer Values of $ t $

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 Integer Values

We count the integers from 5 to 12 inclusive:

$$
12 - 5 + 1 = 8
$$

---

### 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 start by recalling 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, which is 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, let's look at the denominator of the entire function: log(2 - log(x - 2)). For the logarithm to be defined, its argument must also be positive. Therefore, we need:

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

Since log here is base 10, let me recall that log(a) < 2 is equivalent to a < 10^2, because log(a) = 2 implies a = 10^2. But since log is increasing, if log(a) < 2, then a < 100. Therefore:

log(x - 2) < 2
=> x - 2 < 100
=> x < 102

But remember, we already have the condition that x > 2 from the inner logarithm. So combining these two, we have:

2 < x < 102

But wait, that's not the only condition. Because the denominator is log(2 - log(x - 2)), and we also need to make sure that this denominator is not zero. So, we need:

log(2 - log(x - 2)) ≠ 0

Let me solve this inequality. log(something) ≠ 0 implies that the something ≠ 10^0 = 1. Therefore:

2 - log(x - 2) ≠ 1
=> log(x - 2) ≠ 2 - 1
=> log(x - 2) ≠ 1

Again, since log(x - 2) = 1 implies x - 2 = 10^1 = 10, so x = 12. Therefore, we must 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). But let me check if there are any other restrictions or if I missed something.

Wait, let me verify each step again to make sure.

First, the inner log(x - 2) requires x > 2. Then, the argument of the outer log, which is 2 - log(x - 2), must be positive. So:

2 - log(x - 2) > 0 => log(x - 2) < 2 => x - 2 < 100 => x < 102. So combining with x > 2, we have 2 < x < 102. Then, the denominator cannot be zero, so log(2 - log(x - 2)) ≠ 0. Which implies 2 - log(x - 2) ≠ 1 => log(x - 2) ≠ 1 => x - 2 ≠ 10 => x ≠ 12. So yes, that seems correct.

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 2 - log(x - 2) is negative? Well, we already considered that in the first step, where we required 2 - log(x - 2) > 0. So that's already taken care of. Therefore, the only restrictions are x > 2, x < 102, and x ≠ 12.

Therefore, the domain is (2, 102) without the point 12. So in interval notation, that's (2, 12) union (12, 102). Let me check if there are any other points where the function might not be defined. For example, could there be a case where log(x - 2) is negative? Well, log(x - 2) can be negative if x - 2 is between 0 and 1, which is when x is between 2 and 3. But that's allowed as long as the outer log is defined. Wait, but in that case, 2 - log(x - 2) would be 2 minus a negative number, which would be greater than 2, so positive. Therefore, the outer log is defined. So that's okay.

Wait, let me take an example. Suppose x = 3. Then log(x - 2) = log(1) = 0. Then the denominator becomes log(2 - 0) = log(2), which is defined and not zero. So that's okay. If x approaches 2 from the right, then x - 2 approaches 0+, so log(x - 2) approaches -infty, so 2 - log(x - 2) approaches +infty, so log(2 - log(x - 2)) approaches log(+infty) which is +infty. So denominator approaches infinity, so the function approaches 0. So that's okay. Similarly, when x approaches 102 from the left, x - 2 approaches 100, so log(x - 2) approaches log(100) = 2, so 2 - log(x - 2) approaches 0 from the positive side, so log(2 - log(x - 2)) approaches log(0+) which is -infty. Therefore, the denominator approaches -infty, so the function approaches 0. So that's okay. So the only problem is at x = 12, where the denominator becomes log(2 - log(10)) = log(2 - 1) = log(1) = 0. Therefore, the function is undefined there. So yes, that seems correct.

Therefore, the domain is all real numbers between 2 and 102, not including 2 and 102, and not including 12. So in interval notation, that's (2, 12) union (12, 102). Let me check if there are any other points where the denominator could be zero. For example, suppose log(2 - log(x - 2)) = 0. Then 2 - log(x - 2) = 1, so log(x - 2) = 1, so x - 2 = 10, so x = 12. So that's the only point. Therefore, only x = 12 is excluded. So yes, the domain is (2, 12) union (12, 102).

Wait, but let me check if there are any other restrictions. For example, is there a case where log(x - 2) is undefined? Well, we already considered that x > 2. So that's covered. Also, the numerator is 2 - x, which is a linear function, so it's defined for all real numbers. Therefore, the only restrictions are from the denominator and the inner logarithm. Therefore, the domain is indeed (2, 12) union (12, 102).

But let me check with specific numbers. Let me pick x = 3, which is in (2, 12). Then f(3) = (2 - 3)/log(2 - log(1)) = (-1)/log(2 - 0) = (-1)/log(2). Since log(2) is defined and not zero, so that's okay. How about x = 12? Then f(12) = (2 - 12)/log(2 - log(10)) = (-10)/log(2 - 1) = (-10)/log(1) = (-10)/0, which is undefined. So that's correct. How about x = 100? Then x - 2 = 98, log(98) is log(98) ≈ 1.991, so 2 - log(98) ≈ 2 - 1.991 ≈ 0.009, so log(0.009) is log(9*10^{-3}) = log(9) + log(10^{-3}) ≈ 0.9542 - 3 = -2.0458. Therefore, denominator is approximately -2.0458, so f(100) = (2 - 100)/(-2.0458) ≈ (-98)/(-2.0458) ≈ positive number. So that's defined. How about x approaching 102? Let me take x = 101. Then x - 2 = 99, log(99) ≈ 1.9956, so 2 - log(99) ≈ 2 - 1.9956 ≈ 0.0044, so log(0.0044) ≈ log(4.4*10^{-3}) ≈ log(4.4) + log(10^{-3}) ≈ 0.6435 - 3 ≈ -2.3565. Therefore, denominator is -2.3565, so f(101) = (2 - 101)/(-2.3565) ≈ (-99)/(-2.3565) ≈ positive number. So that's okay. How about x approaching 2 from the right? Let me take x = 2.1. Then x - 2 = 0.1, log(0.1) = -1. Then 2 - (-1) = 3, so log(3) ≈ 0.4771. Therefore, denominator is 0.4771, so f(2.1) = (2 - 2.1)/0.4771 ≈ (-0.1)/0.4771 ≈ -0.2096. So that's defined. So seems okay.

Therefore, I think my reasoning is correct. The domain is (2, 12) union (12, 102). Therefore, in interval notation, that's written as (2, 12) ∪ (12, 102).

But let me check once more if there are any other points where the denominator could be zero. Suppose log(2 - log(x - 2)) = 0. Then 2 - log(x - 2) = 1, so log(x - 2) = 1, so x - 2 = 10^1 = 10, so x = 12. So that's the only solution. Therefore, only x = 12 is excluded. Therefore, the domain is indeed (2, 12) union (12, 102).

Therefore, I think that's the answer.

**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 restrictions imposed by the logarithmic expressions and the denominator.

---

### Step 1: 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: Outer Logarithm in the Denominator

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

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

Since $ \log(x - 2) < 2 $, we can rewrite this using the definition of logarithms:

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

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

$$
2 < x < 102
$$

---

### Step 3: Denominator Cannot Be Zero

The denominator is $ \log(2 - \log(x - 2)) $. For the function to be defined, the denominator must not be zero:

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

This implies:

$$
2 - \log(x - 2) \neq 1 \quad \Rightarrow \quad \log(x - 2) \neq 1
$$

Solving $ \log(x - 2) = 1 $:

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

So, $ x = 12 $ must be excluded 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 start by recalling 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, which is 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, let's look at the denominator of the entire function: log(2 - log(x - 2)). For the logarithm to be defined, its argument must also be positive. Therefore, we need:

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

Since log here is base 10, let me recall that log(a) < 2 is equivalent to a < 10^2, because log(a) = 2 implies a = 10^2. But since log is increasing, if log(a) < 2, then a < 100. Therefore:

log(x - 2) < 2
=> x - 2 < 100
=> x < 102

But remember, we already have the condition that x > 2 from the inner logarithm. So combining these two, we have:

2 < x < 102

But wait, that's not the only condition. Because the denominator is log(2 - log(x - 2)), and we also need to make sure that this denominator is not zero. So, we need:

log(2 - log(x - 2)) ≠ 0

Let me solve this inequality. log(something) ≠ 0 implies that the something ≠ 10^0 = 1. Therefore:

2 - log(x - 2) ≠ 1
=> log(x - 2) ≠ 2 - 1
=> log(x - 2) ≠ 1

Again, since log(x - 2) = 1 implies x - 2 = 10^1 = 10, so x = 12. Therefore, we must 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). But let me check if there are any other restrictions or if I missed something.

Wait, let me verify each step again to make sure.

First, the inner log(x - 2) requires x > 2. Then, the argument of the outer log, which is 2 - log(x - 2), must be positive. So:

2 - log(x - 2) > 0 => log(x - 2) < 2 => x - 2 < 100 => x < 102. So combining with x > 2, we have 2 < x < 102. Then, the denominator cannot be zero, so log(2 - log(x - 2)) ≠ 0. Which implies 2 - log(x - 2) ≠ 1 => log(x - 2) ≠ 1 => x - 2 ≠ 10 => x ≠ 12. So yes, that seems correct.

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 2 - log(x - 2) is negative? Well, we already considered that in the first step, where we required 2 - log(x - 2) > 0. So that's already taken care of. Therefore, the only restrictions are x > 2, x < 102, and x ≠ 12.

Therefore, the domain is (2, 102) without the point 12. So in interval notation, that's (2, 12) union (12, 102). Let me check if there are any other points where the function might not be defined. For example, could there be a case where log(x - 2) is negative? Well, log(x - 2) can be negative if x - 2 is between 0 and 1, which is when x is between 2 and 3. But that's allowed as long as the outer log is defined. Wait, but in that case, 2 - log(x - 2) would be 2 minus a negative number, which would be greater than 2, so positive. Therefore, the outer log is defined. So that's okay.

Wait, let me take an example. Suppose x = 3. Then log(x - 2) = log(1) = 0. Then the denominator becomes log(2 - 0) = log(2), which is defined and not zero. So that's okay. If x approaches 2 from the right, then x - 2 approaches 0+, so log(x - 2) approaches -infty, so 2 - log(x - 2) approaches +infty, so log(2 - log(x - 2)) approaches log(+infty) which is +infty. So denominator approaches infinity, so the function approaches 0. So that's okay. Similarly, when x approaches 102 from the left, x - 2 approaches 100, so log(x - 2) approaches log(100) = 2, so 2 - log(x - 2) approaches 0 from the positive side, so log(2 - log(x - 2)) approaches log(0+) which is -infty. Therefore, the denominator approaches -infty, so the function approaches 0. So that's okay. So the only problem is at x = 12, where the denominator becomes log(2 - log(10)) = log(2 - 1) = log(1) = 0. Therefore, the function is undefined there. So yes, that seems correct.

Therefore, the domain is all real numbers between 2 and 102, not including 2 and 102, and not including 12. So in interval notation, that's (2, 12) union (12, 102). Let me check if there are any other points where the denominator could be zero. For example, suppose log(2 - log(x - 2)) = 0. Then 2 - log(x - 2) = 1, so log(x - 2) = 1, so x - 2 = 10, so x = 12. So that's the only point. Therefore, only x = 12 is excluded. So yes, the domain is (2, 12) union (12, 102).

Wait, but let me check if there are any other restrictions. For example, is there a case where log(x - 2) is undefined? Well, we already considered that x > 2. So that's covered. Also, the numerator is 2 - x, which is a linear function, so it's defined for all real numbers. Therefore, the only restrictions are from the denominator and the inner logarithm. Therefore, the domain is indeed (2, 12) union (12, 102).

But let me check with specific numbers. Let me pick x = 3, which is in (2, 12). Then f(3) = (2 - 3)/log(2 - log(1)) = (-1)/log(2 - 0) = (-1)/log(2). Since log(2) is defined and not zero, so that's okay. How about x = 12? Then f(12) = (2 - 12)/log(2 - log(10)) = (-10)/log(2 - 1) = (-10)/log(1) = (-10)/0, which is undefined. So that's correct. How about x = 100? Then x - 2 = 98, log(98) is log(98) ≈ 1.991, so 2 - log(98) ≈ 2 - 1.991 ≈ 0.009, so log(0.009) is log(9*10^{-3}) = log(9) + log(10^{-3}) ≈ 0.9542 - 3 = -2.0458. Therefore, denominator is approximately -2.0458, so f(100) = (2 - 100)/(-2.0458) ≈ (-98)/(-2.0458) ≈ positive number. So that's defined. How about x approaching 102? Let me take x = 101. Then x - 2 = 99, log(99) ≈ 1.9956, so 2 - log(99) ≈ 2 - 1.9956 ≈ 0.0044, so log(0.0044) ≈ log(4.4*10^{-3}) ≈ log(4.4) + log(10^{-3}) ≈ 0.6435 - 3 ≈ -2.3565. Therefore, denominator is -2.3565, so f(101) = (2 - 101)/(-2.3565) ≈ (-99)/(-2.3565) ≈ positive number. So that's okay. How about x approaching 2 from the right? Let me take x = 2.1. Then x - 2 = 0.1, log(0.1) = -1. Then 2 - (-1) = 3, so log(3) ≈ 0.4771. Therefore, denominator is 0.4771, so f(2.1) = (2 - 2.1)/0.4771 ≈ (-0.1)/0.4771 ≈ -0.2096. So that's defined. So seems okay.

Therefore, I think my reasoning is correct. The domain is (2, 12) union (12, 102). Therefore, in interval notation, that's written as (2, 12) ∪ (12, 102).

But let me check once more if there are any other points where the denominator could be zero. Suppose log(2 - log(x - 2)) = 0. Then 2 - log(x - 2) = 1, so log(x - 2) = 1, so x - 2 = 10^1 = 10, so x = 12. So that's the only solution. Therefore, only x = 12 is excluded. Therefore, the domain is indeed (2, 12) union (12, 102).

Therefore, I think that's the answer.

**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 restrictions imposed by the logarithmic expressions and the denominator.

---

### Step 1: 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: Outer Logarithm in the Denominator

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

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

Since $ \log(x - 2) < 2 $, we can rewrite this using the definition of logarithms:

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

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

$$
2 < x < 102
$$

---

### Step 3: Denominator Cannot Be Zero

The denominator is $ \log(2 - \log(x - 2)) $. For the function to be defined, the denominator must not be zero:

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

This implies:

$$
2 - \log(x - 2) \neq 1 \quad \Rightarrow \quad \log(x - 2) \neq 1
$$

Solving $ \log(x - 2) = 1 $:

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

So, $ x = 12 $ must be excluded 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 $x \in \boxed{(2,12) \cup (12,102)}$