<Question>
A group of $N$ students, where $N < 50$, is on a field trip. If their teacher puts them in groups of 8, the last group has 5 students. If their teacher instead puts them in groups of 6, the last group has 3 students. What is the sum of all possible values of $N$?
</Question>

<think>
<Plan_1>
Okay, let's try to figure out this problem step by step. So, we have a number of students N, less than 50. When the teacher divides them into groups of 8, the last group has 5 students. That means if we divide N by 8, the remainder is 5. Similarly, when divided into groups of 6, the remainder is 3. We need to find all possible N less than 50 that satisfy both these conditions and then sum them up.

First, let's translate these conditions into mathematical equations.

<prompt_1.1> Translate the given conditions into mathematical equations. </prompt_1.1>

Let me list the numbers less than 50 that satisfy each condition separately first, and then find the common numbers.

Starting with the first condition: N ≡ 5 mod 8. So, numbers that when divided by 8 leave a remainder of 5.
The general form would be N = 8k + 5, where k is a non-negative integer. Let's list them:

<prompt_1.2> List all numbers less than 50 that satisfy N ≡ 5 mod 8. </prompt_1.2>

Now the second condition: N ≡ 3 mod 6. So, numbers that when divided by 6 leave a remainder of 3.
Numbers can be written as N = 6m + 3. Let's list them:

<prompt_1.3> List all numbers less than 50 that satisfy N ≡ 3 mod 6. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
When N is divided by 8, the remainder is 5. So, mathematically, that can be written as:
N ≡ 5 (mod 8)

Similarly, when N is divided by 6, the remainder is 3:
N ≡ 3 (mod 6)

Our goal is to find all N < 50 that satisfy both congruences and then sum them.
</execution_1.1>

<execution_1.2>
For k=0: 5
k=1: 13
k=2: 21
k=3: 29
k=4: 37
k=5: 45
k=6: 53 (which is over 50, so stop)
So possible Ns from first condition: 5,13,21,29,37,45.
</execution_1.2>

<execution_1.3>
For m=0: 3, 9, 15, 21, 27, 33, 39, 45, 51,... But since N <50, we'll stop at 45.

Now we need to find numbers that appear in both lists. Let's compare:

First list: 5,13,21,29,37,45
Second list: 3,9,15,21,27,33,39,45

The common numbers are 21 and 45.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.2, execution_1.3:
Wait, let me check again.

<prompt_2.1> Verify the common numbers found in both lists. </prompt_2.1>
-----

Based on execution_1.2:
But wait, let me verify if these numbers actually satisfy both conditions.

Take N=5:

<prompt_2.2> Verify if each number in the intersection satisfies both conditions. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
First list: 5,13,21,29,37,45

Second list: 3,9,15,21,27,33,39,45

So 21 and 45 are common.
</execution_2.1>

<execution_2.2>
Divided by 8: 5/8=0 with remainder 5. Good. Divided by 6: 5/6=0 with remainder 5. But the problem says the last group has 3 students, so remainder 3. Wait, 5 isn't 3. So N=5 doesn't satisfy the second condition. Hmm, that's a problem. So maybe my initial approach is wrong?
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.2:
Wait, N=5: 5 divided by 6 is 0 groups with remainder 5. But the problem states that when divided into groups of 6, the last group has 3 students. So remainder should be 3. Therefore, N=5 is invalid. So my mistake was not checking if the numbers actually satisfy both conditions.

Let's check each common number.

<prompt_3.1> Correct the mistake by checking if the common numbers satisfy both conditions. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
First, N=21:

21 divided by 8: 21 ÷8=2*8=16, remainder 5. Correct, since 21-16=5. So first condition satisfied.

21 divided by 6: 6*3=18, remainder 3. Correct. 21-18=3. So second condition satisfied. So 21 is good.

Next, N=45:

45 ÷8=5*8=40, remainder 5. Correct.

45 ÷6=7*6=42, remainder 3. Correct. So 45 is also good.

So the possible Ns are 21 and 45. Therefore, the sum is 21+45=66.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But wait, let me check if there are any other numbers. Wait, in the lists above, the next common number after 45 would be 69, but that's over 50, so no. So only 21 and 45.

But hold on, let me check if there's another way to approach this problem, maybe using the Chinese Remainder Theorem or something. Since 8 and 6 are not coprime, the CRT doesn't directly apply, but maybe we can find a common solution.

<prompt_4.1> Consider and apply alternative methods to solve the problem. </prompt_4.1>

But let me check if there's another number between 50 and, say, 60. But since N<50, we don't need to.

Wait, but let me check another way. Let's list numbers that are 3 mod6: 3,9,15,21,27,33,39,45. Then check which of these are 5 mod8.

<prompt_4.2> Verify the solution using an alternative method by listing numbers that are 3 mod 6 and checking if they are 5 mod 8. </prompt_4.2>

So the answer is 21+45=66. But let me check once more.

<prompt_4.3> Perform a final verification and conclude the solution by checking if the numbers satisfy both conditions. </prompt_4.3>

Wait, but hold on. Let me check if there are any other numbers. Wait, maybe when k and m are different. Let me see.

<prompt_4.4> Re-evaluate the solution to ensure no other possible values are missed. </prompt_4.4>

But wait, let me check if there is a number smaller than 21. For example, N=5.

<prompt_4.5> Confirm the solution by checking all possible numbers and ensuring no mistakes were made. </prompt_4.5>
</Plan_4>

<Execution_4>
<execution_4.1>
We have:

N ≡5 mod8

N≡3 mod6

We can write N=8k+5. Substitute this into the second equation:

8k +5 ≡3 mod6

So 8k ≡3 -5 mod6

8k ≡-2 mod6

But -2 mod6 is 4, so 8k ≡4 mod6

But 8 mod6 is 2, so 2k ≡4 mod6

Divide both sides by 2: k ≡2 mod3

Therefore, k=3m +2, where m is a non-negative integer.

Therefore, substituting back into N=8k +5:

N=8*(3m +2)+5=24m +16 +5=24m +21

So N=24m +21. Now, since N<50, let's find possible m:

For m=0: N=21

m=1: 45

m=2: 69 (too big)

So N=21 and 45. Therefore, same result. So the sum is 21+45=66.
</execution_4.1>

<execution_4.2>
3 mod8=3≠5. 9 mod8=1≠5. 15 mod8=7≠5. 21 mod8=5. Yes. 27 mod8=3. 33 mod8=1. 39 mod8=7. 45 mod8=5. So yes, 21 and 45. So that's correct.

Therefore, the sum is 21+45=66.
</execution_4.2>

<execution_4.3>
Wait, let me check N=5 again. Maybe there's a mistake in the listing.

Wait, the first list is numbers congruent to 5 mod8: 5,13,21,29,37,45. The second list is numbers congruent to 3 mod6: 3,9,15,21,27,33,39,45. The intersection is indeed 21 and 45. So sum is 66.

But wait, let me check N=5. If N=5, when divided into groups of 8, the last group has 5 students. That's correct. But when divided into groups of 6, the last group has 3 students. But 5 divided by 6 is 0 groups with remainder 5, which is not 3. So N=5 is invalid. Similarly, check N=13.

13 divided by 6: 6*2=12, remainder 1. Not 3. So N=13 is invalid. Similarly, N=21: 21 divided by 6 is 3*6=18, remainder 3. Correct. So 21 is valid. N=29: 29 divided by 6: 6*4=24, remainder 5. Not 3. So invalid. N=37: 37 divided by 6: 6*6=36, remainder 1. Not 3. N=45: As before, valid. So indeed only 21 and 45. So sum is 66.

But wait, let me check if there are other numbers between 45 and 50. Next number after 45 in the first list is 53, which is over 50. So no. So only 21 and 45.
</execution_4.3>

<execution_4.4>
Let me take numbers congruent to 3 mod6 and check if they are congruent to 5 mod8.

3 mod8: 3. Not 5. So no.

9 mod8: 1. No.

15 mod8: 7. No.

21 mod8: 5. Yes.

27 mod8: 3. No.

33 mod8: 1. No.

39 mod8: 7. No.

45 mod8: 5. Yes.

So only 21 and 45.
</execution_4.4>

<execution_4.5>
We saw N=5: 5 mod6 is 5, not 3. So no. Next, N=13. 13 mod6 is 1 (since 6*2=12, 13-12=1). Not 3. So no. Then N=21: 21 mod6=3. Yes. So 21 is the first. Then N=29: 29 mod6=5. No. N=37: 37 mod6=1. No. N=45: 45 mod6=3. Yes. So indeed, only 21 and 45. So sum is 66.

Therefore, the answer is 66.
</execution_4.5>
</Execution_4>

<Plan_5>
Based on execution_4.3, execution_4.5:
**Final Answer**
The sum of all possible values of \( N \) is \boxed{66}.
</Plan_5>

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

<Final_answer>
Therefore, the sum of all possible values of $N$ is $\boxed{66}$.
</Final_answer>
</think>