Add more solutions

Added solution for problems 102 and 124 in C, for problem 104
in python and for problem 145 in both C and python.

Python/p104.py

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+# The Fibonacci sequence is defined by the recurrence relation:
+#
+# F_n = F_n−1 + F_n−2, where F_1 = 1 and F_2 = 1.
+# It turns out that F_541, which contains 113 digits, is the first Fibonacci number for which the last nine digits are 1-9 pandigital
+# (contain all the digits 1 to 9, but not necessarily in order). And F_2749, which contains 575 digits, is the first Fibonacci number
+# for which the first nine digits are 1-9 pandigital.
+#
+# Given that F_k is the first Fibonacci number for which the first nine digits AND the last nine digits are 1-9 pandigital, find k.
+
+#!/usr/bin/python
+
+from mpmath import matrix, mp
+
+from timeit import default_timer
+from projecteuler import is_pandigital
+
+def main():
+    start = default_timer()
+
+    found = 0
+    fib1 = 1
+    fib2 = 1
+    i = 2
+
+    while not found:
+#       Calculate the next Fibonacci number modulo 10^9 and check if the result is 1-9 pandigital.
+        fibn = (fib1 + fib2) % 1000000000
+        fib1 = fib2
+        fib2 = fibn
+        i = i + 1
+
+#       If the last 9 digits of fib_n are pandigital, calculate the ith Fibonacci number using
+#       the matrix representation, with sufficient precision so that the at least the first
+#       9 digits are correct (we don't need the whole number. If the first 9 digits are also
+#       pandigital, we found the solution.
+        if is_pandigital(fibn, 9):
+            fib_matrix = matrix(2)
+            fib_matrix[0, 0] = 1
+            fib_matrix[0, 1] = 1
+            fib_matrix[1, 0] = 1
+            fib_matrix[1, 1] = 0
+
+            fib = fib_matrix ** i
+            fib = int(fib[0, 1])
+
+            if is_pandigital(int(str(fib)[:9]), 9):
+                found = 1
+ 
+    end = default_timer()
+
+    print('Project Euler, Problem 104')
+    print('Answer: {}'.format(i))
+
+    print('Elapsed time: {:.9f} seconds'.format(end - start))
+
+if __name__ == '__main__':
+    main()

Python/p145.py

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+#!/usr/bin/python
+
+# Some positive integers n have the property that the sum [ n + reverse(n) ] consists entirely of odd (decimal) digits.
+# For instance, 36 + 63 = 99 and 409 + 904 = 1313. We will call such numbers reversible; so 36, 63, 409, and 904 are reversible.
+# Leading zeroes are not allowed in either n or reverse(n).
+#
+# There are 120 reversible numbers below one-thousand.
+#
+# How many reversible numbers are there below one-billion (109)?
+
+from timeit import default_timer
+
+def main():
+    start = default_timer()
+
+    N = 10000000
+    
+    count = 0
+
+#   Brute force approach, sum each number and their reverse and
+#   check if there are only odd digits.
+    for i in range(11, N):
+        if i % 10 != 0:
+            s = str(i + int(''.join(reversed(str(i)))))
+
+            if not '0' in s and not '2' in s and not '4' in s and\
+                    not '6' in s and not '8' in s:
+                count = count + 1
+
+    end = default_timer()
+
+    print('Project Euler, Problem 145')
+    print('Answer: {}'.format(count))
+
+    print('Elapsed time: {:.9f} seconds'.format(end - start))
+
+if __name__ == '__main__':
+    main()