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+# Dimik - Summation
+
+In this problem, you will be given `T` testcases. Each line of the testcase consists of a 5 digit integer `n`. We just have to print the summation of leftmost and rightmost digit of an integer `n`.
+
+### Solution
+We can find the solution by dividing the integer `n` by 10000 to get the leftmost digit and finding the remainder of `n` being divided by 10, we get the rightmost digit. Once we obtain both the digits, we can add them and print them in the format `Sum = summation of both digits`.
+
+### C++
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+int main()
+{
+    int t;
+    cin >> t;
+    for (int k = 1; k <= t; k++)
+    {
+        int n;
+        cin >> n;
+        int sum = 0;
+        sum += n / 10000;
+        sum += n % 10;
+        cout << "Sum = " << sum << endl;
+    }
+}
+```