diff --git a/dsa-solutions/gfg-solutions/easy/rightmost-different-bit.md b/dsa-solutions/gfg-solutions/easy/rightmost-different-bit.md
new file mode 100644
index 000000000..66b2f3d45
--- /dev/null
+++ b/dsa-solutions/gfg-solutions/easy/rightmost-different-bit.md
@@ -0,0 +1,71 @@
+---
+id: rightmost-different-bit
+title: Rightmost Different Bit
+sidebar_label: 0106-Rightmost Different Bit
+tags:
+  - Bit Magic
+  - Data Structures
+description: "A solution to the problem of finding the rightmost position of the bit which is different in both numbers"
+---
+
+In this page, we will solve the problem of finding the rightmost position of the bit which is different in both numbers.
+
+## Problem Description
+
+Given two numbers M and N. The task is to find the position of the rightmost different bit in the binary representation of numbers. If both M and N are the same then return -1 in this case.
+
+### Examples
+
+**Example 1:**
+
+```plaintext
+Input: 
+M = 11, N = 9
+Output: 
+2
+Explanation: 
+Binary representation of the given numbers are: 1011 and 1001, 2nd bit from right is different.
+```
+
+**Example 2:**
+
+```plaintext
+Input: 
+M = 52, N = 4
+Output: 
+5
+Explanation: 
+Binary representation of the given numbers are: 110100 and 0100, 5th-bit from right is different.
+```
+
+### Constraints
+
+- $1 \leq$ M,N $\leq10^9$
+
+## Solution
+
+### Intuition and Approach
+
+This problem can be solved usnig bitwise operations
+
+### Approach: Inorder Traversal
+
+1. We will first XOR operation (^) between two integers, it will produce a result where each bit is set to 1 if the corresponding bits of m and n are different, and 0 if they are the same and then the result is stored in res variable
+2. If res is 0, it means that m and n are identical and therefore we return -1 to indicate that there is no differing bit.
+3. Now, To identify the position of the rightmost bit that is set to 1, the expression res & -res is used. This operation isolates the rightmost 1 bit in res. This works because -res is the two’s complement of res, which flips all bits of res and adds 1.
+4. Then we use math.log2(res & -res) which computes the base-2 logarithm of the isolated rightmost 1 bit and gives the zero-based index of the rightmost differing bit.
+5. Next we will add 1 converts this index to a one-based position and returned position.
+#### Code in Python
+```python
+
+class Solution:
+  def posOfRightMostDiffBit(self,m,n):
+    res = m^n
+    if res==0: return -1
+    return math.log2(res & -res) + 1
+ ```
+
+#### Complexity Analysis
+
+- **Time Complexity:** $O(1)$
+- **Space Complexity:** $O(1)$