Okay, let me try to figure out this problem. So, the question is: 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, alright. So, we need to find all two-digit numbers that end with 6 and are perfect squares. Then, for each of those numbers, note the tens digit B, and count how many distinct Bs there are.

First, let's recall that a two-digit number ranges from 10 to 99. Since the number ends with 6, it must be the square of a number that ends with either 4 or 6 because when you square numbers ending with 4 or 6, the result ends with 6. Let me verify that:

- 4 squared is 16, which ends with 6.
- 6 squared is 36, which ends with 6.
- 14 squared is 196, which ends with 6.
- 16 squared is 256, which ends with 6.
- 24 squared is 576, ends with 6.
- 26 squared is 676, ends with 6.
- 34 squared is 1156, ends with 6.
- 36 squared is 1296, ends with 6.
- 44 squared is 1936, ends with 6.
- 46 squared is 2116, ends with 6.
- 54 squared is 2916, ends with 6.
- 56 squared is 3136, ends with 6.
- 64 squared is 4096, ends with 6.
- 66 squared is 4356, ends with 6.
- 74 squared is 5476, ends with 6.
- 76 squared is 5776, ends with 6.
- 84 squared is 7056, ends with 6.
- 86 squared is 7396, ends with 6.
- 94 squared is 8836, ends with 6.
- 96 squared is 9216, ends with 6.

Wait, but these are all numbers ending with 4 or 6, right? Because when you square a number, the last digit of the result depends only on the last digit of the original number. So, numbers ending with 4 or 6 will have squares ending with 6. Therefore, the possible candidates for the square roots are numbers ending with 4 or 6.

But since we are dealing with two-digit numbers (10 to 99), the square roots must be single-digit or two-digit numbers. Let me check the square roots of two-digit numbers ending with 6.

First, let's list all two-digit numbers ending with 6: 16, 36, 56, 76, 96. Wait, but 16 is a two-digit number, but its square root is 4, which is a single-digit number. But the problem states that the original number is a two-digit number, but the square root can be any positive integer. However, the two-digit number must be the square, so the square root can be a single or two-digit number. But since we are looking for two-digit numbers, their square roots can be from 4 (since 4 squared is 16) up to 9 (since 9 squared is 81) and 10 (10 squared is 100, which is three digits). So, the square roots can be from 4 to 9, but in our case, since the square numbers must end with 6, the square roots must end with 4 or 6. Let's see:

Numbers between 4 and 9: 4, 5, 6, 7, 8, 9. Among these, numbers ending with 4 or 6 are 4 and 6. So, 4 squared is 16, 6 squared is 36. So, the two-digit numbers 16 and 36. But 16 is a two-digit number ending with 6, and 36 is another. So, that gives us two numbers. Wait, but earlier when I considered square roots up to two digits, I thought of numbers like 14, 16, etc., but those would result in three-digit squares. For example, 14 squared is 196, which is three digits. But the problem states that the two-digit number "B6" is the square. So, the square must be a two-digit number. Therefore, the square roots must be numbers such that their squares are two-digit numbers ending with 6.

But wait, 4 squared is 16 (two-digit), 6 squared is 36 (two-digit), 14 squared is 196 (three-digit), 16 squared is 256 (three-digit), etc. So, only 4 and 6 as square roots would give two-digit squares ending with 6. Therefore, the two-digit numbers are 16 and 36, so B can be 1 or 3. Therefore, there are 2 possible values for B.

But wait, hold on. Let me check again. Are there any other two-digit squares ending with 6 besides 16 and 36?

Let me list all two-digit squares:

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

So, among these, only 16 and 36 end with 6. Therefore, the possible values for B are 1 and 3, so two distinct values.

But wait, the problem says "the square of a positive integer". So, if the square root is a single-digit number, that's okay. But also, if the square root is a two-digit number, but the square is a three-digit number. However, the problem specifies that the square is a two-digit number. Therefore, the square must be between 10 and 99. Therefore, the square roots must be integers between 4 and 9, inclusive, because 4^2=16 and 9^2=81. So, numbers from 4 to 9. Among these, only 4 and 6 have squares ending with 6. Therefore, only 16 and 36. Hence, two values of B: 1 and 3.

But wait, let me check again. Wait, 16 is 4^2, 36 is 6^2. But 16 is a two-digit number, so B=1. 36 is a two-digit number, so B=3. So, two distinct Bs. Therefore, the answer should be 2.

But wait, but the initial problem statement says "the two-digit number "B6," where B is the tens digit, is the square of a positive integer." So, the two-digit number is "B6", which is 10*B + 6. So, 10*B +6 must be a perfect square. So, 10*B +6 = n^2, where n is a positive integer. We need to find all possible B (digits from 1 to 9) such that 10*B +6 is a perfect square.

Therefore, 10*B +6 = n^2. So, n^2 ≡ 6 mod 10. So, n^2 ≡ 6 mod 10. Which numbers squared give 6 mod 10? Let's check:

0^2=0, 1=1, 2=4, 3=9, 4=16≡6, 5=25≡5, 6=36≡6, 7=49≡9, 8=64≡4, 9=81≡1. So, numbers ending with 4 or 6 have squares ending with 6. Therefore, n must end with 4 or 6.

Now, since n^2 is a two-digit number, n must be such that n^2 is between 10 and 99. Therefore, n is between 4 (since 4^2=16) and 9 (9^2=81). So, n can be 4,5,6,7,8,9. Among these, n must end with 4 or 6. Therefore, n can be 4 or 6. Therefore, n=4 gives 16, n=6 gives 36. Therefore, B is 1 and 3. So, two values.

Therefore, the answer is 2.

But wait, let me check if there's any other number. For example, n=14: 14^2=196, which is three digits. So, not applicable. Similarly, n=16: 256. So, no. So, only 4 and 6 in the range 4-9. Therefore, only two numbers.

Hence, the number of distinct possible values for B is 2.

But wait, hold on. Let's verify again. Let's check all two-digit numbers ending with 6 and see if they are perfect squares.

Numbers ending with 6 between 10 and 99:

16, 36, 56, 76, 96.

Check which of these are perfect squares:

16: 4^2=16, yes.

36: 6^2=36, yes.

56:  not a perfect square. 7^2=49, 8^2=64, 56 is in between.

76: 8^2=64, 9^2=81, 76 is in between.

96: 9^2=81, 10^2=100, 96 is in between.

Therefore, only 16 and 36 are perfect squares. Therefore, B can be 1 or 3. Hence, two distinct values.

Therefore, the answer is 2.

But wait, the problem says "positive integer", so n can be any positive integer, but since the square must be a two-digit number, n is between 4 and 9. So, n=4,6 are the only ones. Therefore, 2 values.

Therefore, the answer is 2.

**Final Answer**
The number of distinct possible values for \( B \) is \boxed{2}.
Therefore, the number of distinct possible values for $B$ is $\boxed{2}$.
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