<Question>
The sum of the first $N$ positive odd numbers is 121. What is the value of $N$?
</Question>

<think>
<Plan_1>
Okay, so I need to find the value of N such that the sum of the first N positive odd numbers is 121. Hmm, let me think about how to approach this.

First, I remember that there's a formula for the sum of the first N odd numbers. Wait, what is it exactly? Let me recall. I think the sum of the first N odd numbers is N squared. Is that right? Let me check with a small example. The first odd number is 1, which is 1².
The sum of the first two odd numbers is 1 + 3 = 4, which is 2². First three: 1 + 3 + 5 = 9, which is 3². Yeah, that seems to hold. So the formula is Sum = N².

So if the sum is 121, then N² = 121. Therefore, N would be the square root of 121.
The square root of 121 is 11, because 11*11 is 121. So N should be 11.

But let me make sure I'm not missing anything here. Let me verify again.

<prompt_1.1> Recheck the formula for the sum of the first N odd numbers and confirm the solution. </prompt_1.1>

But wait, maybe I should write out the first few terms to confirm. Let's see:

<prompt_1.2> Write out the first few terms of the sequence of odd numbers and verify the sum. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
The first odd number is 1, then 3, 5, 7, etc. So adding them up: 1+3=4, 4+5=9, 9+7=16, 9+11=25, 25+13=38, 38+15=53, 53+17=70, 70+19=90, 90+21=111, 111+23=134, 134+25=159... Wait, hold on, that's going too far. Wait, maybe I should do it step by step up to 11 terms.

Wait, let's list the first 11 odd numbers and add them:

1st odd number: 1

2nd: 3

3rd: 5

4th: 7

5th: 9

6th: 11

7th: 13

8th: 15

9th: 17

10th: 19

11th: 21

Now, sum them up:

1 + 3 = 4

4 + 5 = 9

9 + 7 = 16

16 + 9 = 25

25 + 11 = 36

36 + 13 = 49

49 + 15 = 64

64 + 17 = 81

81 + 19 = 100

100 + 21 = 121

Wait, that's exactly 121 after 11 terms! So that checks out.
</execution_1.1>

<execution_1.2>
Term 1: 1 (sum = 1)
Term 2: 1 + 3 = 4
Term 3: 1 + 3 + 5 = 9
Term 4: 1 + 3 + 5 + 7 = 16
Term 5: 1 + 3 + 5 + 7 + 9 = 25
Term 6: 25 + 11 = 36
Term 7: 36 + 13 = 49
Term 8: 49 + 15 = 64
Term 9: 64 + 17 = 81
Term 10: 81 + 19 = 100
Term 11: 100 + 21 = 121
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Therefore, the answer should be 11.

But let me think again if there's any other way this could be approached, maybe in case the formula is not recalled.

<prompt_2.1> Consider alternative methods to solve the problem and confirm the final answer. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
For example, suppose someone forgets the formula and decides to use another method. Let's say they know that the nth odd number is given by 2n - 1. So the sum of the first N odd numbers would be the sum from n=1 to N of (2n - 1). Let's compute that sum:

Sum = Σ(2n - 1) from n=1 to N

This can be split into two sums:

= 2Σn from 1 to N - Σ1 from 1 to N

= 2*(N(N + 1)/2) - N

Simplify:

= N(N + 1) - N

= N² + N - N

= N²

So that confirms the formula again. Therefore, Sum = N² = 121, so N = 11.

Alternatively, if someone didn't know the formula, they might also recognize that the sum of consecutive odd numbers forms a sequence of perfect squares. For example:

Sum of 1 odd number: 1 = 1²

Sum of 2 odd numbers: 1 + 3 = 4 = 2²

Sum of 3 odd numbers: 1 + 3 + 5 = 9 = 3²

... and so on. So each time, the sum is just the square of the number of terms. Therefore, if the sum is 121, which is 11², then N must be 11.

Another way to approach it without recalling the formula is to realize that each odd number can be written as 2k - 1 where k is from 1 to N. So summing them up:

Sum = Σ(2k - 1) = 2Σk - Σ1 = 2*(N(N + 1)/2) - N = N(N + 1) - N = N² + N - N = N².

So regardless of the method, the result is the same. Therefore, N² = 121 => N = 11.

I think that's thorough. I considered different methods to arrive at the same conclusion, checked with examples, verified the formula through summation, and even listed out the terms to confirm.
</execution_2.1>
</Execution_2>

<Final_answer>
Therefore, the value of $N$ is $\boxed{11}$.
</Final_answer>
</think>