Im thinking of a two-digit number. It is a prime number which is 3 less than a perfect square. Also, when it's digits are reversed, it becomes a perfect square. What is the one possible number?

61

http://www.factmonster.com/math/numbers/prime.html

To find the two-digit prime number that meets the given conditions, we can go through the following steps:

Step 1: Start by listing out perfect squares and their corresponding values:
- 1^2 = 1
- 2^2 = 4
- 3^2 = 9
- 4^2 = 16
- 5^2 = 25
- 6^2 = 36
- 7^2 = 49
- 8^2 = 64
- 9^2 = 81

Step 2: Identify the perfect squares that are 3 greater than a prime number:
- 4 + 3 = 7 (which is a prime number)
- 9 + 3 = 12 (which is not a prime number)
- 16 + 3 = 19 (which is a prime number)
- 25 + 3 = 28 (which is not a prime number)
- 36 + 3 = 39 (which is not a prime number)
- 49 + 3 = 52 (which is not a prime number)
- 64 + 3 = 67 (which is a prime number)
- 81 + 3 = 84 (which is not a prime number)

Step 3: We are left with three potential prime numbers: 7, 19, and 67.

Step 4: Check which of these numbers, when its digits are reversed, forms a perfect square:
- Reversing the digits of 7 gives us 7, which is a perfect square (7^2 = 49).
- Reversing the digits of 19 gives us 91, which is not a perfect square.
- Reversing the digits of 67 gives us 76, which is not a perfect square.

Step 5: Thus, the only number that meets all the given conditions is 7.

Therefore, the one possible number that satisfies the given conditions is 7.