primelib

Project Euler/Problem 01/sol1.py

@@ -5,8 +5,8 @@
 Find the sum of all the multiples of 3 or 5 below N.
 '''
 n = int(raw_input().strip())
-sum=0;
+sum=0
 for a in range(3,n):
     if(a%3==0 or a%5==0):
         sum+=a
-print sum;
+print sum

Project Euler/Problem 01/sol3.py

@@ -8,8 +8,8 @@
 This solution is based on the pattern that the successive numbers in the series follow: 0+3,+2,+1,+3,+1,+2,+3.
 '''
 n = int(raw_input().strip())
-sum=0;
-num=0;
+sum=0
+num=0
 while(1):
     num+=3
     if(num>=n):
@@ -39,4 +39,4 @@
     if(num>=n):
         break
     sum+=num
-print sum;
+print sum

Project Euler/Problem 02/sol1.py

@@ -8,11 +8,13 @@
 '''
 
 n = int(raw_input().strip())
-i=1; j=2; sum=0
+i=1
+j=2 
+sum=0
 while(j<=n):
     if((j&1)==0): #can also use (j%2==0)
         sum+=j
     temp=i
     i=j
     j=temp+i
-print sum
+print sum

Project Euler/Problem 03/sol1.py

@@ -17,7 +17,7 @@ def isprime(no):
             return False
     return True
 
-max=0
+maxNumber = 0
 n=int(input())
 if(isprime(n)):
     print n
@@ -31,8 +31,8 @@ def isprime(no):
         for i in range(3,n1,2):
             if(n%i==0):
                 if(isprime(n/i)):
-                    max=n/i
+                    maxNumber = n/i
                     break
                 elif(isprime(i)):
-                    max=i
-        print max
+                    maxNumber = i
+        print maxNumber

Project Euler/Problem 29/solution.py

@@ -0,0 +1,34 @@
+def main():
+    """ 
+    Consider all integer combinations of ab for 2 <= a <= 5 and 2 <= b <= 5:
+
+    22=4, 23=8, 24=16, 25=32
+    32=9, 33=27, 34=81, 35=243
+    42=16, 43=64, 44=256, 45=1024
+    52=25, 53=125, 54=625, 55=3125
+    If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
+
+    4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
+
+    How many distinct terms are in the sequence generated by ab for 2 <= a <= 100 and 2 <= b <= 100?
+    """
+
+    collectPowers = set()
+
+    currentPow = 0
+
+    N = 101     # maximum limit
+
+    for a in range(2,N):
+
+        for b in range (2,N):
+
+            currentPow = a**b   # calculates the current power
+            collectPowers.add(currentPow)   # adds the result to the set
+
+
+    print "Number of terms ", len(collectPowers)
+
+
+if __name__ == '__main__':
+    main()