diff --git a/c_programs/pattern2.C b/c_programs/pattern2.C
new file mode 100644
index 0000000..ee5efbc
--- /dev/null
+++ b/c_programs/pattern2.C
@@ -0,0 +1,38 @@
+#include <stdio.h>
+ 
+int main()
+{
+  int n, c, row, num = 1, space;
+ 
+  scanf("%d", &n);
+ 
+  space = n - 1;
+ 
+  for (row = 1; row <= n; row++)
+  {
+    num = row;
+ 
+  for (c = 1; c <= space; c++)
+    printf(" ");
+ 
+  space--;
+ 
+  for (c = 1; c <= row; c++)
+  {
+    printf("%d", num);
+    num++;
+  }
+ 
+  num = num - 2;
+ 
+  for (c = 1 ; c < row; c++)
+  {
+    printf("%d", num);
+    num--;
+  }
+ 
+   printf("\n");
+  }
+ 
+  return 0;
+}
diff --git a/c_programs/prime_number.C b/c_programs/prime_number.C
new file mode 100644
index 0000000..bcca21c
--- /dev/null
+++ b/c_programs/prime_number.C
@@ -0,0 +1,32 @@
+// C program to check if a 
+// number is prime 
+  
+#include <stdio.h> 
+  
+int main() 
+{ 
+    int n, i, flag = 1; 
+  
+    // Ask user for input 
+    printf("Enter a number: \n"); 
+  
+    // Store input number in a variable 
+    scanf("%d", &n); 
+  
+    // Iterate from 2 to n/2, since numbers greater than n/2 can obviously not be a factor of n ;)
+    for (i = 2; i <= n/2; i++) {   // you can improvre the time complexity by runnin the loop only till sqrt(n), since for any factor after
+        if (n % i == 0) {          // that u would already have a factor before the sqrt(n)
+            flag = 0; 
+            break; 
+        } 
+    } 
+  
+    if (flag == 1) { 
+        printf("%d is a prime number", n); 
+    } 
+    else { 
+        printf("%d is not a prime number", n); 
+    } 
+  
+    return 0; 
+} 
diff --git a/c_programs/sum_of_n_digit_no_C b/c_programs/sum_of_n_digit_no_C
new file mode 100644
index 0000000..1443f49
--- /dev/null
+++ b/c_programs/sum_of_n_digit_no_C
@@ -0,0 +1,21 @@
+#include<iostream.h>
+
+inrt main()
+{
+  int n;
+  printf("Enter the number");
+  scanf("%d",&n);      //number is accepted from the user
+  
+  int sum=0;
+  
+  while(n)
+  {
+    int dig=n%10;      //finds the last digit, say n=123, 123%10 gives, dig=3
+    n/=10;             //number n is divided by 10
+    sum=sum+dig;       // did is now added in a sum vaiable initially assigned 0, next dig=2 & sum=3+2=5, next dig=1 & sum=5=1=6, n=0,breaks
+  }
+  
+  printf("The sum of the digits is %d",sum);
+
+  return 0;
+}