diff --git a/DIRECTORY.md b/DIRECTORY.md
index 8861d9b2..d4d9bcb0 100644
--- a/DIRECTORY.md
+++ b/DIRECTORY.md
@@ -1061,6 +1061,8 @@
     * [Test Perfect Squares](https://github.com/BrianLusina/PythonSnips/blob/master/pymath/perfect_square/test_perfect_squares.py)
   * Power Of I
     * [Test Power Of I](https://github.com/BrianLusina/PythonSnips/blob/master/pymath/power_of_i/test_power_of_i.py)
+  * Power Of Three
+    * [Test Power Of Three](https://github.com/BrianLusina/PythonSnips/blob/master/pymath/power_of_three/test_power_of_three.py)
   * Power Of Two
     * [Test Power Of Two](https://github.com/BrianLusina/PythonSnips/blob/master/pymath/power_of_two/test_power_of_two.py)
   * Rectangle Area
diff --git a/pymath/power_of_i/README.md b/pymath/power_of_i/README.md
index 79e4e73e..743e5a18 100644
--- a/pymath/power_of_i/README.md
+++ b/pymath/power_of_i/README.md
@@ -1,6 +1,6 @@
 # Power of I
 
-iii is the imaginary unit, it is defined by i²=−1i² = -1i²=−1, therefore it is a solution to x²+1=0x² + 1 = 0x²+1=0.
+i is the imaginary unit, defined by i² = −1; thus it is a solution to x² + 1 = 0.
 
-Your Task Complete the function pofi that returns iii to the power of a given non-negative integer in its simplest form,
-as a string (answer may contain iii).
\ No newline at end of file
+Your Task Complete the function pofi that returns i to the power of a given non-negative integer in its simplest form,
+as a string (answer may contain i).
\ No newline at end of file
diff --git a/pymath/power_of_three/README.md b/pymath/power_of_three/README.md
new file mode 100644
index 00000000..8391acec
--- /dev/null
+++ b/pymath/power_of_three/README.md
@@ -0,0 +1,135 @@
+# Power Of Three
+
+Given an integer n, return `true` if it is a power of three. Otherwise, return `false`.
+
+An integer n is a power of three, if there exists an integer x such that `n == 3^x`.
+
+## Examples
+
+Example 1:
+
+```text
+Input: n = 27
+Output: true
+Explanation: 27 = 33
+```
+
+Example 2:
+```text
+Input: n = 0
+Output: false
+Explanation: There is no x where 3x = 0.`
+```
+
+Example 3:
+```text
+Input: n = -1
+Output: false
+Explanation: There is no x where 3x = (-1).
+```
+
+## Constraints
+
+- -2^31 <= n <= 2^31 - 1
+
+## Topics
+
+- Math
+- Recursion
+
+## Solution
+
+### Using Modulo
+
+The key insight is that since 3 is a prime number, any power of 3 will only have 3 as its prime factor. The largest power
+of 3 that fits within a 32-bit signed integer range is `3^19 = 1162261467`. If `n` is a power of 3, then n must be a
+divisor of `1162261467`. Because 3 is prime, the only divisors of 3^19 are 3^0, 3^1, 3^2,...3^19, which are exactly all
+the powers of 3 in range. Therefore, we simply check whether n is positive and whether `1162261467` is evenly divisible
+by `n`. This approach requires no loops or recursion.
+
+Now, let’s look at the solution steps below:
+
+1. Check if n is greater than 0. Any power of 3 must be a positive integer, so if n is 0 or negative, return false
+   immediately.
+2. Compute 1162261467 % n, where 1162261467 is 3^19, the largest power of three within the 32-bit signed integer range.
+   - If the remainder equals 0, then n is a divisor of 3^19, meaning n is itself a power of 3. Return true.
+   - If the remainder is not 0, then n is not a power of 3. Return false.
+
+#### Time Complexity
+
+The time complexity of the solution is O(1) because we perform a single comparison and a single modulo operation, both
+of which take constant time regardless of the input.
+
+#### Space Complexity
+
+The space complexity of the solution is O(1) because we use only a fixed amount of space (no additional data structures)
+and the constant 1162261467 is stored as a single integer.
+
+### Using Loops
+
+To check if a number is a power of three, we need to think about what it means mathematically. If n = 3^x, then we can
+keep dividing n by 3 exactly x times until we reach 1. This is the key insight.
+
+Think of it like peeling layers off an onion. If we have 27 = 3^3, we can divide by 3 three times: 27 ÷ 3 = 9, then
+9 ÷ 3 = 3, then 3 ÷ 3 = 1. We end up with exactly 1, confirming that 27 is indeed a power of three.
+
+On the other hand, if we try this with a non-power like 45: 45 ÷ 3 = 15, then 15 ÷ 3 = 5. Now 5 is not divisible by 3
+(it gives remainder 2), so we know 45 cannot be a power of three.
+
+The pattern becomes clear: a power of three should be repeatedly divisible by 3 until we reach 1, with no remainders
+along the way. If at any point during our division we get a remainder (checked using n % 3), we immediately know the
+number isn't a power of three.
+
+The algorithm naturally emerges from this observation: keep dividing by 3 while checking for remainders. If we
+successfully reduce the number to 1 through repeated division by 3, it's a power of three. If we encounter a remainder
+or end up with a number other than 1, it's not.
+
+The solution implements the trial division method to determine if a number is a power of three. Let's walk through the
+implementation step by step:
+
+- Main Loop Condition: The algorithm starts with a while loop that continues as long as n > 2. This is because the 
+  smallest power of three greater than 2 is 3^1 = 3. Any power of three will be at least 3, so we can stop when n
+  becomes 2 or less.
+- Divisibility Check: Inside the loop, we check if n % 3 is non-zero. The modulo operation n % 3 returns the remainder
+  when n is divided by 3. If there's any remainder (meaning n % 3 evaluates to true), then n is not perfectly divisible
+  by 3, which means it cannot be a power of three. We immediately return False.
+- Division Step: If n is divisible by 3 (passes the modulo check), we perform integer division: n //= 3. This reduces n
+  by a factor of 3 and continues the process. The integer division operator // ensures we're working with whole numbers
+  throughout.
+- Final Check: After the loop exits (when n <= 2), we check if n == 1. If we've successfully divided a power of three by
+  3 repeatedly, we should end up with exactly 1. For example:
+  - 27 → 9 → 3 → 1 (returns True)
+  - 18 → 6 → 2 (returns False since 2 ≠ 1)
+
+#### Time Complexity 
+
+The algorithm uses a while loop that repeatedly divides n by 3 until n becomes less than or equal to 2. In each iteration,
+we perform:
+
+- A modulo operation n % 3 to check divisibility
+- An integer division n //= 3 to reduce n
+
+The number of iterations is determined by how many times we can divide n by 3 before it becomes less than or equal to 2.
+This is equivalent to finding the largest integer k such that 3^k ≤ n, which gives us k ≤ log₃n. Therefore, the loop
+runs at most ⌊log₃n⌋ times, resulting in a time complexity of O(log₃n).
+
+#### Space Complexity
+
+The algorithm only uses a constant amount of extra space:
+
+- The input variable n is modified in place
+- No additional data structures are created
+- The space used does not depend on the size of the input
+
+Therefore, the space complexity is O(1).
+
+### Using Logarithms
+
+A common loop-free math check (for positive integers n):
+
+- Compute x = log₃n = ln(n)/ln(3)
+- Check whether x is an integer: x ≈ k for some integer k (e.g., ∣x−round(x)∣<ε like 10^−10)
+- If yes, then n = 3^k; otherwise it isn’t.
+
+Alternative integer-only (no loops) via factorization: compute whether n has any prime factors other than 3, and whether
+the exponent of 3 is ≥ 0; but that typically requires some factoring method.
diff --git a/pymath/power_of_three/__init__.py b/pymath/power_of_three/__init__.py
new file mode 100644
index 00000000..5fae7b47
--- /dev/null
+++ b/pymath/power_of_three/__init__.py
@@ -0,0 +1,31 @@
+from math import log
+
+
+def is_power_of_three_with_log(n: int) -> bool:
+    if n < 1:
+        return False
+
+    k = log(n, 3)
+    k_round = int(round(k))
+    return 3**k_round == n
+
+
+def is_power_of_three_with_mod(n: int) -> bool:
+    # The largest power of 3 within 32-bit signed integer range is 3^19 = 1162261467
+    # If n is a power of 3, then 1162261467 % n must equal 0
+    # We also need n to be positive since negative numbers and zero can't be powers of 3
+    return n > 0 and 1162261467 % n == 0
+
+
+def is_power_of_three_with_loop(n: int) -> bool:
+    # Keep dividing by 3 while n is greater than 2
+    while n > 2:
+        # If n is not divisible by 3, it's not a power of 3
+        if n % 3 != 0:
+            return False
+        # Divide n by 3 using integer division
+        n //= 3
+
+    # After the loop, n should be 1 if it was a power of 3
+    # (since 3^0 = 1)
+    return n == 1
diff --git a/pymath/power_of_three/test_power_of_three.py b/pymath/power_of_three/test_power_of_three.py
new file mode 100644
index 00000000..0d63a8b7
--- /dev/null
+++ b/pymath/power_of_three/test_power_of_three.py
@@ -0,0 +1,40 @@
+import unittest
+from parameterized import parameterized
+from utils.test_utils import custom_test_name_func
+from pymath.power_of_three import (
+    is_power_of_three_with_log,
+    is_power_of_three_with_mod,
+    is_power_of_three_with_loop,
+)
+
+POWER_OF_THREE_TEST_CASES = [
+    (27, True),
+    (0, False),
+    (-1, False),
+    (9, True),
+    (11, False),
+    (45, False),
+    (243, True),
+    (10, False),
+]
+
+
+class PowerOfThreeTestCases(unittest.TestCase):
+    @parameterized.expand(POWER_OF_THREE_TEST_CASES, name_func=custom_test_name_func)
+    def test_power_of_three_using_logs(self, n: int, expected: bool):
+        actual = is_power_of_three_with_log(n)
+        self.assertEqual(expected, actual)
+
+    @parameterized.expand(POWER_OF_THREE_TEST_CASES, name_func=custom_test_name_func)
+    def test_power_of_three_using_mod(self, n: int, expected: bool):
+        actual = is_power_of_three_with_mod(n)
+        self.assertEqual(expected, actual)
+
+    @parameterized.expand(POWER_OF_THREE_TEST_CASES, name_func=custom_test_name_func)
+    def test_power_of_three_using_loop(self, n: int, expected: bool):
+        actual = is_power_of_three_with_loop(n)
+        self.assertEqual(expected, actual)
+
+
+if __name__ == "__main__":
+    unittest.main()