38,39,48 solved, 48 is so easy

solutions/38.py

@@ -4,15 +4,22 @@
 '''
 What is the largest 1 to 9 pandigital 9-digit number that can be formed
 as the concatenated product of an integer with(1,2..n) where n>1
+starting with n=2
+9?*1++9?*2 and 9??*1+9??*2 they will never yield a 9digit number
+but 9???*1+9???*2 will, so our starting search must be a 4 digit number 
+beginnning with 9 and never contain 0
+it should be obvious that any n>2 is impossible
 '''
 def isPandigital(n, s=9):
   n = str(n)
   return len(n)==s and not '1234567890'[:s].strip(n)
 
 def main():
-  for i in range(1000000000, 918273644, -1):
-    if isPandigital(i):
-     # 
+  for i in range(9876, 9123, -1):
+    p = str(i*1)+ str(i*2)
+    if isPandigital(int(p)):
+      print p
+      break
 
 if __name__ == '__main__':
   main()

solutions/39.py

@@ -0,0 +1,22 @@
+#! /usr/bin/python
+# filename: 39.py
+
+'''
+If p is the perimeter of a right angle triangle {a,b,c},which value
+for p<=1000, has the most solutions?
+interesting: a+b+c=p and a**2+b**2=c**2
+'''
+
+def main():
+  t_max=0
+  p_limit = 1000
+  for p in range(p_limit/2, p_limit+1, 2):
+    t = 0
+    for a in range(2, p/4+1):
+      if p*(p-2*a)%(2*(p-a))==0: t+=1
+      if t> t_max: (t_max, p_max) = (t,p)
+
+  print p_max
+
+if __name__ == '__main__':
+  main()