<Question>
When the positive integers with exactly three positive divisors are listed in ascending order, what is the fifth number listed?
</Question>

<think>
<Plan_1>
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 understanding the problem step by step.

First, what does it mean for a number to have exactly three positive divisors? Usually, a number has divisors in pairs. For example, if a number is prime, it has exactly two divisors: 1 and itself. But if a number has three divisors, that's different. Let me think. If a number has three divisors, then one of those divisors must be repeated in the pair.
Wait, but divisors come in pairs, right? So if a number has an odd number of divisors, that usually means it's a perfect square because the square root is repeated. For example, 4 has divisors 1, 2, 4. Here, 2 is the square root and only counts once. So maybe numbers with exactly three divisors are squares of prime numbers?

Let me check that.

Let's take a prime number, say 2.
The square of 2 is 4. The divisors of 4 are 1, 2, 4. That's three divisors. Similarly, take 3.

<prompt_1.1> Verify the hypothesis that squares of prime numbers have exactly three divisors by checking examples with prime numbers 2, 3, 5, and 7. </prompt_1.1>

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

Let me confirm this. Let's take another number that's not a square of a prime. For example, 6.

<prompt_1.2> Confirm that non-square numbers have an odd number of divisors only if they are perfect squares and check if they have more than three divisors. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
The square is 9. Divisors of 9 are 1, 3, 9. That's three. Similarly, 5 squared is 25. Divisors are 1, 5, 25. So that works. So it seems like the numbers with exactly three divisors are squares of primes.
</execution_1.1>

<execution_1.2>
Its divisors are 1, 2, 3, 6. That's four divisors. How about 8? Divisors are 1, 2, 4, 8. That's four again. How about 9? Divisors are 1, 3, 9. That's three. 9 is 3 squared. 25 is 5 squared, divisors 1, 5, 25. So that's three. So it seems that the numbers with exactly three divisors are squares of primes. Because if a number is a square of a prime, then its divisors are 1, the prime, and the square. So that's three. If the number is not a square of a prime, then it would have either more than three divisors or an odd number of divisors but not three.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2:
Therefore, the list of numbers with exactly three divisors in ascending order would be the squares of primes in order. So first, list the primes in order: 2, 3, 5, 7, 11, 13, etc. Then their squares are 4, 9, 25, 49, 121, 169, and so on. Therefore, the fifth number in this list would be the square of the fifth prime.

<prompt_2.1> List the primes in order and calculate their squares to determine the fifth number with exactly three divisors. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Wait, let's check:

First prime: 2, square is 4 (first number).

Second prime: 3, square is 9 (second number).

Third prime: 5, square is 25 (third number).

Fourth prime: 7, square is 49 (fourth number).

Fifth prime: 11, square is 121 (fifth number).

So the fifth number listed would be 121.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
But let me make sure I'm not missing anything here. Are there any other numbers that have exactly three divisors which are not squares of primes? Let's think. Suppose a number has three divisors. Let’s denote the number as n. If n has exactly three divisors, then those divisors must be 1, a, and n, where a is a prime number. Wait, why?

<prompt_3.1> Analyze whether numbers with three divisors must be squares of primes by considering their prime factorization. </prompt_3.1>
-----

But wait a second. Let me make sure that between 4, 9, 25, 49, and 121, there are no other numbers that might have three divisors. For example, is 16 a possible candidate? 16 is 2^4. Its divisors are 1, 2, 4, 8, 16.

<prompt_3.2> Check numbers like 16, 64, and 81 to ensure they do not have three divisors. </prompt_3.2>

Let me list the squares of primes up to the fifth prime to confirm:

<prompt_3.3> List the squares of primes up to the fifth prime to confirm the fifth number in the sequence. </prompt_3.3>

But just to be thorough, let's check the numbers between these squares to ensure there are no other numbers with three divisors. For example, after 4 (2²), next is 6, which we saw has four divisors.
Then 7 is prime, 8 has four divisors, 9 is 3². Then between 9 and 25: 10 has four divisors, 11 is prime, 12 has six divisors, 13 is prime, 14 has four, 15 has four, 16 has five, 17 prime, 18 has six, 19 prime, 20 has six, 21 has four, 22 has four, 23 prime, 24 has eight, 25 is 5². So yes, no other numbers between 4 and 25 except 9, which is next. Then similarly, after 25, the next square is 49. Let's see: 26 has four divisors, 27 has four, 28 has six, 29 prime, 30 has eight, 31 prime, 32 has six, 33 has four, 34 has four, 35 has four, 36 has nine, 37 prime, 38 has four, 39 has four, 40 has eight, 41 prime, 42 has eight, 43 prime, 44 has six, 45 has six, 46 has four, 47 prime, 48 has ten, 49 is 7². So yes, 49 is next after 25. Then 121 is the fifth. So that seems correct.
</Plan_3>

<Execution_3>
<execution_3.1>
n could be factored into primes. Let's suppose n is a positive integer with exactly three positive divisors. Then the number of divisors function, denoted as d(n), equals 3. The number of divisors function for a number n = p1^a1 * p2^a2 * ... * pk^ak is (a1 + 1)(a2 + 1)... (ak + 1). So if this product is 3, then since 3 is prime, the only way to write 3 as a product of integers greater than 1 is 3 = 3 * 1 * 1 * ... * 1. Therefore, n must be of the form p^2, where p is a prime. Because then the number of divisors would be (2 + 1) = 3. So that confirms that numbers with exactly three divisors are squares of primes. Therefore, the list is indeed the squares of primes in order.
</execution_3.1>

<execution_3.2>
So five divisors. So 16 is out. How about 12? Divisors are 1, 2, 3, 4, 6, 12. That's six. 18: 1, 2, 3, 6, 9, 18. Also six. 20: 1, 2, 4, 5, 10, 20. Six again. 21: 1, 3, 7, 21. Four. 22: 1, 2, 11, 22. Four. 24: multiple. 25: 1, 5, 25. Three. So yes, after 16, the next is 25. So no numbers between 16 and 25 (other than 25) have three divisors.
</execution_3.2>

<execution_3.3>
1. 2² = 4
2. 3² = 9
3. 5² = 25
4. 7² = 49
5. 11² = 121

So the fifth number is 121. Therefore, the answer is 121.
</execution_3.3>
</Execution_3>

<Plan_4>
Based on execution_3.1:
So the numbers with exactly three divisors are 4, 9, 25, 49, 121, etc. So arranged in order, the fifth one is 121.

But let me check if there is any number before 121 that is not a square of a prime but still has three divisors.
For example, let's take 8.

<prompt_4.1> Check numbers like 8, 12, 16, and 18 to ensure they do not have three divisors. </prompt_4.1>
-----

Based on execution_3.3:
But let me check with another approach. Let's list numbers with three divisors:

<prompt_4.2> List numbers with three divisors by checking their factorization and confirm the fifth number is 121. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
Divisors of 8: 1, 2, 4, 8. That's four. 6: 1, 2, 3, 6. Four. 10: 1, 2, 5, 10. Four. 12: more than four. 14: 1, 2, 7, 14. Four. 15: 1, 3, 5, 15. Four. 16: five divisors. 18: six. 20: six. 21: four. 22: four. 24: eight. 25: three. So 25 comes next after 16. Wait, but 16 is 4 squared, which is 2^4. So 16 has five divisors: 1, 2, 4, 8, 16. So that's five. So 16 is not in the list. Then 25 is next. So 25 is the third in the list. Then 49 (7²) is fourth, and 121 (11²) is fifth.
</execution_4.1>

<execution_4.2>
Start from 1 and go up:

1: 1 divisor.

2: two divisors.

3: two.

4: 1,2,4. Three. So first.

5: two.

6: four.

7: two.

8: four.

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

10: four.

11: two.

12: six.

13: two.

14: four.

15: four.

16: five.

17: two.

18: six.

19: two.

20: six.

21: four.

22: four.

23: two.

24: eight.

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

26: four.

27: four.

28: six.

29: two.

30: eight.

31: two.

32: six.

33: four.

34: four.

35: four.

36: nine.

37: two.

38: four.

39: four.

40: eight.

41: two.

42: eight.

43: two.

44: six.

45: six.

46: four.

47: two.

48: ten.

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

50: six.

... Then next would be 121. So yes, the fourth is 49, fifth is 121. So that confirms it. Therefore, the fifth number is 121.

Therefore, the answer is 121. But just to make sure, let's confirm the order:

1st: 4 (2²)

2nd: 9 (3²)

3rd: 25 (5²)

4th: 49 (7²)

5th: 121 (11²)

Yes, that's correct. So 121 is the fifth number in the list. So the answer should be 121.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.2:
**Final Answer**
The fifth number listed is \boxed{121}.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, the fifth number in the list of positive integers with exactly three positive divisors is $\boxed{121}$.
</Final_answer>
</think>