<Question>
The two-digit number $``B6,''$ where $B$ is the tens digit, is the square of a positive integer. How many distinct possible values are there for $B$?
</Question>

<think>
<Plan_1>
Okay, so I have this problem here: The two-digit number "B6," where B is the tens digit, is the square of a positive integer. How many distinct possible values are there for B?

Hmm, let me try to figure this out step by step. First, I need to understand what the problem is asking. We have a two-digit number that ends with 6, and the tens digit is B. So the number looks like B6, where B can be any digit from 1 to 9 (since it's a two-digit number, the tens digit can't be 0). The question is asking how many different values of B make this number a perfect square.

Alright, so my goal is to find all possible values of B such that B6 is a square number. Let me think about how to approach this.

First, let's recall that a two-digit number ending with 6 must be a square of a number that ends with either 4 or 6. Because when you square numbers, the last digit of the result depends on the last digit of the original number. For example:

<prompt_1.1> Determine the possible last digits of numbers whose squares end with 6. </prompt_1.1>

But wait, we need to check which squares of numbers ending with 4 or 6 result in a two-digit number ending with 6. Let's list all two-digit squares and see which ones end with 6.

<prompt_1.2> List all two-digit squares and identify those ending with 6. </prompt_1.2>
-----

But let me check again if there are other possibilities. Wait, maybe numbers larger than two digits?

<prompt_1.3> Verify if there are other possibilities for numbers larger than two digits. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
- 4 squared is 16, which ends with 6.
- 6 squared is 36, which also ends with 6.
- Numbers ending with 1: 1 squared is 1, 11 squared is 121, which ends with 1.
- Numbers ending with 2: 2 squared is 4, 12 squared is 144, ends with 4.
- Numbers ending with 3: 13 squared is 169, ends with 9.
- Numbers ending with 5: 15 squared is 225, ends with 5.
- Numbers ending with 7: 17 squared is 289, ends with 9.
- Numbers ending with 8: 18 squared is 324, ends with 4.
- Numbers ending with 9: 19 squared is 361, ends with 1.

So indeed, only numbers ending with 4 or 6 will have squares ending with 6. Therefore, the original number (B6) must be the square of a number ending with 4 or 6.
</execution_1.1>

<execution_1.2>
The squares of numbers from 4 to 9 (since 10^2 is 100, which is three digits):

- 4^2 = 16
- 5^2 = 25
- 6^2 = 36
- 7^2 = 49
- 8^2 = 64
- 9^2 = 81

Then from 10 to 31 (since 32^2 is 1024, which is four digits):

But maybe I can focus on numbers whose squares are two-digit numbers. Wait, the problem says "the two-digit number B6," so we need squares that are two-digit numbers. So the squares must be between 16 and 99, inclusive. So let's list all squares between 16 and 99:

16 (4^2), 25 (5^2), 36 (6^2), 49 (7^2), 64 (8^2), 81 (9^2). Then 100 is already three digits, so we can stop here.

From these, which ones end with 6? Let's see:

16 ends with 6.
36 ends with 6.
64 ends with 4.
81 ends with 1.

Wait, 16 is 4 squared, 36 is 6 squared, 64 is 8 squared. So 16, 36, 64. Each of these ends with a 6. So the possible numbers are 16, 36, 64. Therefore, the tens digit B in each case is 1, 3, and 6.
</execution_1.2>

<execution_1.3>
But the problem specifies a two-digit number, so we can't go beyond 99. So, 16, 36, 56, 76, 96. Only 16 and 36 are perfect squares. Therefore, B can be 1 or 3, so two values.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.2:
Wait, but 64 ends with 4, not 6. Wait, no, 64 ends with 4.

<prompt_2.1> Verify which squares end with 6. </prompt_2.1>
-----

Based on execution_1.1, execution_1.2:
Let me check if those squares are correct:

<prompt_2.2> Verify the squares and their last digits. </prompt_2.2>

Therefore, the possible values for B are 1, 3, and 6, which are three distinct values.

Wait, but hold on. Let me verify if there are any other two-digit squares ending with 6.

<prompt_2.3> Re-evaluate the list of two-digit squares to confirm the values of B. </prompt_2.3>

But wait, let me check if there are other numbers whose squares end with 6. For example, 14^2 is 196, which is three digits, but 14 is two digits. 14 squared is 196, which ends with 6. But 196 is a three-digit number, so it's outside our consideration here.

<prompt_2.4> Consider numbers larger than two digits and verify if they affect the solution. </prompt_2.4>

But let me check once more to make sure I haven't missed any numbers. Let's list all numbers ending with 4 or 6, square them, and see if any two-digit squares result:

<prompt_2.5> Perform a final verification of numbers ending with 4 or 6 to ensure no other solutions are missed. </prompt_2.5>
</Plan_2>

<Execution_2>
<execution_2.1>
So 16 and 36 end with 6. Wait, 36 is 6 squared, which is 6, so 6 squared is 36. So 36 ends with 6. Then 8 squared is 64, which ends with 4. Then 10 squared is 100, which is three digits. So among the two-digit squares, only 16 and 36 end with 6. Therefore, B6 can be 16 or 36, so B is 1 or 3. Therefore, two possible values.
</execution_2.1>

<execution_2.2>
4^2 = 16: yes, ends with 6. So B is 1 here.

6^2 = 36: ends with 6. So B is 3.

8^2 = 64: ends with 4. Not 6. So that's out.

10^2 = 100: too big.

Wait, but what about numbers like 14^2 = 196, but that's three digits. So in two-digit numbers, only 4^2, 6^2, and 8^2. So 16, 36, 64. Only 16 and 36 end with 6. Wait, 64 ends with 4. So only 16 and 36. Therefore, B is 1 and 3. That would be two values.

But wait, hold on. Wait, 6^2 is 36, which ends with 6. So 36 is a two-digit square ending with 6. So B is 3. Then 16 is 4^2, B is 1. Then 8^2 is 64, which ends with 4, so not 6. So only 16 and 36.
</execution_2.2>

<execution_2.3>
Wait, 4^2 is 16, which ends with 6. 6^2 is 36, which ends with 6. Then 14^2 is 196, which is three digits. 16^2 is 256. So in two-digit numbers, only 4 and 6. So their squares are 16 and 36. Then 14 and 16 are beyond two digits. So yes, only 16 and 36. Therefore, B is 1 and 3. But wait, 64 is 8^2, which is 64, a two-digit number. 64 ends with 4, not 6. So 64 is not ending with 6. So that's not a candidate. So 16 and 36 are the only two-digit squares ending with 6. Therefore, B is 1 and 3. So that's two values. But wait, the original problem says "the two-digit number B6, where B is the tens digit, is the square of a positive integer". So B6 is the square. So B6 must be 16, 36, or 64? Wait, 64 ends with 4, so it's not B6. So only 16 and 36. Therefore, B is 1 and 3. So two values. But wait, the answer I initially thought was three, but now I'm getting two. There's a discrepancy here.
</execution_2.3>

<execution_2.4>
Similarly, 24^2 is 576, which is also three digits. So in the two-digit range, only 4 and 6 as the roots give squares ending with 6. Wait, 4^2 is 16, 6^2 is 36. Then 14^2 is 196, which is three digits. So in the two-digit range, only 16 and 36. Wait, but 8^2 is 64, which ends with 4, and 10^2 is 100. So, no, in the two-digit squares, only 16 and 36 end with 6.
</execution_2.4>

<execution_2.5>
Starting with numbers ending with 4:

14^2 = 196 (three digits)
4^2 = 16 (two digits)
24^2 = 576 (three digits)
So only 4^2 is a two-digit square ending with 6.

Now numbers ending with 6:

6^2 = 36 (two digits)
16^2 = 256 (three digits)
26^2 = 676 (three digits)
So only 6^2 is a two-digit square ending with 6.

Thus, the only two-digit squares ending with 6 are 16 and 36. Therefore, the possible B values are 1 and 3. Wait, but earlier I thought 64 is a square ending with 4, but 64 is 8^2, which ends with 4, not 6. So that's not relevant here. So actually, only 16 and 36. Therefore, B can be 1 or 3. So that's two distinct values.

But wait, hold on. Wait, 64 is 8 squared, which is 64. So 64 is a two-digit number ending with 4. But the problem is about numbers ending with 6. So 64 is not considered here. So only 16, 36, and 64? Wait, no, 64 ends with 4. So only 16 and 36 end with 6. Therefore, B6 can be 16 or 36, so B is 1 or 3. Therefore, two possible values.
</execution_2.5>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So B could be 1 or 3. Therefore, two possible values. But wait, the problem says "how many distinct possible values are there for B"? So the answer would be 2? But I feel like I might be missing something here. Let me go back through the reasoning again.

<prompt_3.1> Re-evaluate the possible values of B and verify the squares. </prompt_3.1>
-----

Based on execution_2.5:
Wait, but let me check once more. Let's list all squares from 4^2 up to 9^2:

<prompt_3.2> Perform a final confirmation of the two-digit squares and their last digits. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
The problem is asking for the number of possible values of B where B6 is a square. So first, B6 is a two-digit number, so B is from 1 to 9. Then, B6 must be a square. So squares between 16 and 96 (since the maximum two-digit number ending with 6 is 96). Let's list all squares between 16 and 96:

4^2 = 16

5^2 = 25 (ends with 5, not 6)

6^2 = 36

7^2 = 49 (ends with 9)

8^2 = 64 (ends with 4)

9^2 = 81 (ends with 1)

10^2 = 100 (too big)

So in this list, only 16, 36 are squares ending with 6. Therefore, B6 can be 16 or 36, so B is 1 or 3. Therefore, two values. So the answer should be 2.
</execution_3.1>

<execution_3.2>
4^2 = 16 (B=1)
5^2 = 25
6^2 = 36 (B=3)
7^2 = 49
8^2 = 64
9^2 = 81

So among these, only 16 and 36 end with 6. Therefore, B is 1 and 3. So two values. Therefore, the answer should be 2. But wait, hold on. Wait, 64 is 8^2, which is 64. So 64 is a two-digit number, but it ends with 4, not 6. So that's not considered here. So 64 is out. Therefore, only 16 and 36. So B is 1 and 3. Therefore, two possible values. Therefore, the answer is 2. But wait, the problem is presented as a question in a math competition or something, and the answer is expected to be in a box. But I recall that sometimes in these problems, there might be more numbers if you consider higher numbers. Wait, but the problem specifically says "two-digit number," so we can only have numbers from 16, 25, 36, 49, 64, 81. Among these, only 16 and 36 end with 6. Therefore, B is 1 and 3. Therefore, two values.

Wait, but wait a second. Let me check 14^2. 14^2 is 196, which is a three-digit number ending with 6. But since the problem is about two-digit numbers, 196 is out of consideration. Similarly, 16^2 is 256, which is three digits. So in the two-digit range, only 4^2 and 6^2 give squares ending with 6. Therefore, B is 1 and 3. Therefore, two possible values. But wait, the problem is asking for "distinct possible values for B". So 1 and 3, so two values. Therefore, the answer is 2.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
But let me check once more to ensure I haven't missed any numbers. Let me list all squares from 10^2 up to 31^2 to see if any two-digit squares are missed.

<prompt_4.1> Verify the solution by checking a broader range of squares. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Wait, no. Wait, 10^2 is 100, which is three digits. So in the two-digit range, the squares are from 4^2 to 9^2. So 16, 25, 36, 49, 64, 81. So only those. So 16 and 36 are the only ones ending with 6. Therefore, B is 1 and 3. Therefore, two values. Therefore, the answer should be 2. Hmm. But the problem is presented as if it's a bit more complex, so maybe I need to check again.

Wait, maybe the original number is B6, which is a two-digit number, but maybe B can be zero? But no, because B is the tens digit, so if B were zero, the number would be 06, which is just 6, a one-digit number. So B must be from 1 to 9. Therefore, two-digit numbers from 16 to 96. So squares in that range are 16, 25, 36, 49, 64, 81. As before, only 16 and 36 end with 6. Therefore, B is 1 and 3. Therefore, two possible values. So the answer is 2. Therefore, the number of distinct possible values for B is 2.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But wait, let me check 6^2 again. 6^2 is 36, so B is 3. Then 4^2 is 16, so B is 1. Then 8^2 is 64, which is 64, so B would be 6, but 64 ends with 4, so that's not B6. Wait, if B were 6, the number would be 66. Is 66 a square? Let's check.

<prompt_5.1> Verify if 66 is a square and confirm the solution. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Square root of 66 is approximately 8.124, which is not an integer. So 66 is not a square. Therefore, B cannot be 6 here. So even though 8^2 is 64, which is a two-digit number, it's not in the form B6. So 64 is 6 and 4, so it's 64, which is 6 in the tens place and 4 in the units place. So that's not B6. Therefore, 64 is not a candidate. Therefore, only 16 and 36. Therefore, B is 1 and 3. Therefore, two values. Therefore, the answer is 2.
</execution_5.1>
</Execution_5>

<Final_answer>
Therefore, there are $\boxed{2}$ distinct possible values for $B$.
</Final_answer>
</think>