diff --git a/Project Euler/Problem 01/sol1.c b/Project Euler/Problem 01/sol1.c
new file mode 100644
index 00000000..66b0672f
--- /dev/null
+++ b/Project Euler/Problem 01/sol1.c	
@@ -0,0 +1,19 @@
+/*
+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.
+*/
+#include <stdio.h>
+
+int main() {
+	int n = 0;
+	int sum = 0;
+	scanf("%d", &n);
+	for (int a = 0; a < n; a++) {
+		if ((a % 3 == 0) || (a % 5 == 0)) {
+			sum += a;
+		}
+	}
+
+	printf("%d\n", sum);
+}
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diff --git a/Project Euler/Problem 01/sol2.c b/Project Euler/Problem 01/sol2.c
new file mode 100644
index 00000000..af1fb71e
--- /dev/null
+++ b/Project Euler/Problem 01/sol2.c	
@@ -0,0 +1,24 @@
+/*
+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 N.
+'''
+'''
+This solution is based on the pattern that the successive numbers in the series follow: 0+3,+2,+1,+3,+1,+2,+3.
+*/
+#include <stdio.h>
+
+int main() {
+	int n = 0;
+	int sum = 0;
+	scanf("%d", &n);
+
+	int terms = (n - 1) / 3;
+	sum += ((terms)*(6 + (terms - 1) * 3)) / 2; //sum of an A.P.
+	terms = (n - 1) / 5;
+	sum += ((terms)*(10 + (terms - 1) * 5)) / 2;
+	terms = (n - 1) / 15;
+	sum -= ((terms)*(30 + (terms - 1) * 15)) / 2;
+
+	printf("%d\n", sum);
+}
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diff --git a/Project Euler/Problem 01/sol3.c b/Project Euler/Problem 01/sol3.c
new file mode 100644
index 00000000..04c12bf4
--- /dev/null
+++ b/Project Euler/Problem 01/sol3.c	
@@ -0,0 +1,49 @@
+/*
+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 N.
+'''
+'''
+This solution is based on the pattern that the successive numbers in the series follow: 0+3,+2,+1,+3,+1,+2,+3.
+*/
+#include <stdio.h>
+
+int main() {
+	int n = 0;
+	int sum = 0;
+	int num = 0;
+	scanf("%d", &n);
+	
+	while (1) {
+		num += 3;
+		if (num >= n)
+			break;
+		sum += num;
+		num += 2;
+		if (num >= n)
+			break;
+		sum += num;
+		num += 1;
+		if (num >= n) 
+			break;
+		sum += num;
+		num += 3;
+		if (num >= n) 
+			break;
+		sum += num;
+		num += 1;
+		if (num >= n) 
+			break;
+		sum += num;
+		num += 2;
+		if (num >= n) 
+			break;
+		sum += num;
+		num += 3;
+		if (num >= n) 
+				break;
+		sum += num;
+	}
+
+	printf("%d\n", sum);
+}
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