diff --git a/35 b/35
index 1f28e1c..a869275 100755
Binary files a/35 and b/35 differ
diff --git a/35.c b/35.c
index 4b92268..bfe1064 100644
--- a/35.c
+++ b/35.c
@@ -7,14 +7,14 @@
 int main () {
     int bound = 1000000;
 
-    int* primes;
+    long* primes;
     bool* primesTable;
-    int numPrimes = listOfPrimes (bound, &primes, &primesTable);
+    long numPrimes = listOfPrimes (bound, &primes, &primesTable);
 
     bool* circularPrimesTable = calloc (bound + 1, sizeof (bool));
-    for (int i = 0; i < numPrimes; i++) {
+    for (long i = 0; i < numPrimes; i++) {
         if (circularPrimesTable[primes[i]] != true) {
-            int n = primes[i];
+            long n = primes[i];
             int numOfDigits = floor (log10 (n)) + 1;
             int* digits = malloc (sizeof (int) * numOfDigits);
             for (int d = 0; d < numOfDigits; d++) {
diff --git a/37 b/37
index d68fea7..e79a293 100755
Binary files a/37 and b/37 differ
diff --git a/37.c b/37.c
index 76f42fa..cf0172b 100644
--- a/37.c
+++ b/37.c
@@ -25,13 +25,13 @@ bool isRightTrunc (int n) {
 }
 
 int main () {
-    int* primes;
-    int numOfPrimes = listOfPrimes (1000000, &primes, &primesTable);
+    long* primes;
+    long numOfPrimes = listOfPrimes (1000000, &primes, &primesTable);
     // unfortunately I couldn't find a deterministic way to find
     // the upper-bound of all truncatable primes... 1 mil will do
     
-    int sum = 0;
-    for (int i = 0; i < numOfPrimes; i++) {
+    long sum = 0;
+    for (long i = 0; i < numOfPrimes; i++) {
         if (isLeftTrunc (primes[i]) &&
             isRightTrunc (primes[i]) &&
             primes[i] > 10) {