Co-authored-by: sigonasr2 <sigonasr2@gmail.com>

archives/1/src/main.c

@@ -0,0 +1,30 @@
+#include <stdio.h>
+
+/*
+If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
+
+Find the sum of all the multiples of 3 or 5 below 1000.
+
+https://projecteuler.net/problem=1
+*/
+int main(int argc,char**argv) {
+    int counter1=0;
+    int counter2=0;
+    int sum=0;
+    while (counter1<1000||counter2<1000) {
+        counter1+=3;
+        if (counter1!=counter2&&counter1<1000) {
+            sum+=counter1;
+            printf("Sum is %d (+%d) [3]\n",sum,counter1);
+        }
+        if (counter1>counter2) {
+            counter2+=5;
+            if (counter1!=counter2&&counter2<1000) {
+                sum+=counter2;
+                printf("Sum is %d (+%d) [5]\n",sum,counter2);
+            }
+        }
+    }
+    printf("%d",sum);
+    return 0;
+}

src/main.c

@@ -1,5 +1,14 @@
 #include <stdio.h>
 
+/*
+Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
+
+1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
+
+By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
+
+https://projecteuler.net/problem=2
+*/
 int main(int argc,char**argv) {
-    printf("Hello World!");
+    return 0;
 }