Same commit as before. Forgotten to include new files

e_15.py

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+#!/usr/bin/env python 
+"""
+ ##---
+ # jlengrand
+ #Created on : Mon Jan 16 13:53:07 CET 2012
+ #
+ # DESCRIPTION : Solves problem 15 of Project Euler
+ Starting in the top left corner of a 2x2 grid, there are 6 routes (without backtracking) to the bottom right corner.
+ How many routes are there through a 20x20 grid?
+
+ This problem is linked with the catalan number.
+ To get to the extreme right, 2xn moves have to be performed
+ Moves are either R or D. 
+ There is the same the same number of R and D in the sequence.
+ 
+ Problem is finally : How many strings of size 2xn are there consisting of nxR and nxD? 
+ -> number of ways in which we can choose n positions from 2xn available.
+ binomial :
+ (2n n) = (2n)! / (n! x n!)
+ ##---
+"""
+def square_grid_path(n):
+    """
+    Returns the number of paths from top left to bottom right of a 
+    nxn grid. 
+    """
+    return (fact(2*n) / (fact(n) * fact(n)))
+
+def fact(value):
+    """
+    Returns value!
+    """
+    if value < 2:
+        return value
+    else:
+        return value * fact(value - 1)        
+
+if __name__ == '__main__':
+    print "Answer : %d" % (square_grid_path(20))

e_18.py

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+#!/usr/bin/env python 
+"""
+ ##---
+ # jlengrand
+ #Created on : Mon Jan 16 15:34:46 CET 2012
+ #
+ # DESCRIPTION : Solves problem 18 of Project Euler
+ By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
+    3
+   7 4
+  2 4 6
+ 8 5 9 3
+ That is, 3 + 7 + 4 + 9 = 23.
+
+ Find the maximum total from top to bottom of the triangle below:
+               75
+              95 64
+             17 47 82
+            18 35 87 10
+           20 04 82 47 65
+          19 01 23 75 03 34
+         88 02 77 73 07 63 67
+        99 65 04 28 06 16 70 92
+       41 41 26 56 83 40 80 70 33
+      41 48 72 33 47 32 37 16 94 29
+     53 71 44 65 25 43 91 52 97 51 14
+    70 11 33 28 77 73 17 78 39 68 17 57
+   91 71 52 38 17 14 91 43 58 50 27 29 48
+  63 66 04 68 89 53 67 30 73 16 69 87 40 31  
+ 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23 
+ ##---
+"""
+def fun():
+    """
+    """
+    
+    return 1
+
+if __name__ == '__main__':
+    print "Answer : %d" % (fun())

e_21.py

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+#!/usr/bin/env python 
+"""
+ ##---
+ # jlengrand
+ #Created on : Mon Jan 16 16:22:29 CET 2012
+ #
+ # DESCRIPTION : Solves problem 21 of Project Euler
+ Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
+ If d(a) = b and d(b) = a, where a  b, then a and b are an amicable pair and each of a and b are called amicable numbers.
+
+ For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
+
+ Evaluate the sum of all the amicable numbers under 10000. 
+ ##---
+"""
+def amicable_numbers(max_val):
+    """
+    Returns the sum of amicable numbers under max_val
+    """
+    am_list = []
+    for val in range(1, max_val):
+        am = sum_divisors(val)
+        if am != val and sum_divisors(am) == val : 
+            am_list.append(val)
+    
+    return sum(am_list)
+
+def sum_divisors(value):
+    """
+    Returns the sum of the list of proper divisors of value
+    """
+    div = []
+    for val in range(1, value ):
+        if (value % val == 0): 
+            div.append(val)
+    return sum(div)
+
+if __name__ == '__main__':
+    print "Answer : %d" % (amicable_numbers(10000))

e_25.py

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+#!/usr/bin/env python 
+"""
+ ##---
+ # jlengrand
+ #Created on : Mon Jan 16 14:01:09 CET 2012
+ #
+ # DESCRIPTION : Solves problem 25 of Project Euler
+ The Fibonacci sequence is defined by the recurrence relation:
+
+ Fn = Fn1 + Fn2, where F1 = 1 and F2 = 1.
+ The 12th term, F12, is the first term to contain three digits.
+
+ What is the first term in the Fibonacci sequence to contain 1000 digits?
+ ##---
+"""
+def fibo_dig_count(count):
+    """
+    Returns the first term of Fibonacci sequence to contain count digits
+    """
+    last_val = 1
+    val = 1 # Fibonacci starts with 1
+    inc = 0
+
+    while (len(str(val)) < count):
+        temp = last_val
+        last_val = val
+        val += temp
+        inc += 1
+    
+    return inc + 2
+
+if __name__ == '__main__':
+    print "Answer : %d" % (fibo_dig_count(1000))

e_48.py

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+#!/usr/bin/env python 
+"""
+ ##---
+ # jlengrand
+ #Created on : Mon Jan 16 15:47:00 CET 2012
+ #
+ # DESCRIPTION : Solves problem 48 of Project Euler
+ The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
+ Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
+ ##---
+"""
+def sum_pow(max_val):
+    """
+    Returns the last 10 digits of 1^1 + ... + max_val^max_val
+    """
+    summ = 0
+    for j in range(1, max_val + 1):
+        summ += pow(j, j)
+    
+    return int(str(summ)[-10:])
+
+if __name__ == '__main__':
+    print "Answer : %d" % (sum_pow(1000))