diff --git a/Algorithms/NumberTheory/FindingAllPrimes/SieveofEratosthenes.cpp b/Algorithms/NumberTheory/FindingAllPrimes/SieveofEratosthenes.cpp
new file mode 100644
index 00000000..2afefb3d
--- /dev/null
+++ b/Algorithms/NumberTheory/FindingAllPrimes/SieveofEratosthenes.cpp
@@ -0,0 +1,33 @@
+// Sieve Of Eratosthenes
+// Time Complexity O(nlog(log(n))) whereas following brute force will end you up with O(n*sqrt(n))
+#include <iostream>
+#include<math.h>
+using namespace std;
+
+int main() {
+    int n;cin>>n;
+    int prime[n+1];//prime[i]=1 will imply that i is prime & prime[i]=0 will imply that i is not prime
+    for(int i=0;i<=n;i++)
+        prime[i]=1;//considering all as primes as of now.
+    prime[0]=0;
+    prime[1]=0;//since 0 & 1 are not prime ofcourse
+    
+    /*
+    Here we will be running are first loop only till sqrt(n) instead of n because we know that
+    if a number is prime it does not have any factor till its square root so we
+    can use it for optimization.
+    */
+    for(int i=2;i<=sqrt(n);i++){
+        if(prime[i]==1){//if i is prime
+            for(int j=2;j*i<=n;j++){
+                prime[i*j]=0;//setting the multiples of i to 0 because they can't be prime ofcourse
+            }
+        }
+    }
+    for(int i=0;i<=n;i++){
+        if(prime[i]!=0){
+            cout<<i<<endl;
+        }
+    }
+	return 0;
+}
\ No newline at end of file
diff --git a/Problem Solutions/SPOJ/thelastdigit.cpp b/Problem Solutions/SPOJ/thelastdigit.cpp
new file mode 100644
index 00000000..ae794a1b
--- /dev/null
+++ b/Problem Solutions/SPOJ/thelastdigit.cpp	
@@ -0,0 +1,24 @@
+#include <iostream>
+using namespace std;
+int mod(int,int);//modular function
+int main() {
+	//will be using modular exponentiation (check wikipedia for details over the algo)
+	int t;cin>>t;
+	while(t--){
+	    int a,b;cin>>a>>b;//we need to find last digit of a^b
+	    cout<<mod(a,b);
+	    cout<<endl;
+	}
+	return 0;
+}
+int mod(int a,int b){
+    if(b==0)//power is 0
+        return 1;
+    else if(b%2==0){//power is even
+        int y = mod(a,b/2);
+        return (y*y)%10;
+    }
+    else{//power is odd
+        return ((a%10)*(mod(a,b-1)%10))%10;
+    }
+}