Project Euler Problem 21

By
Administrator
Published: December 31, 2009Posted in: Project Euler, Python, Ruby and JavaTags:
Project Euler Problem 21

The problem:

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.

The solution:

?View Code PYTHON
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38

def properDivisor(n):
    listProperDivisor = []
 
    for i in range(1, (int(n/2)+1)):
        if n%i == 0:
            listProperDivisor.append(i)
 
    return listProperDivisor
 
def sumList(aList):
 
    total = 0
 
    for number in aList:
        total = total + number
    return total
 
def checkAmicablePair (n,m):
    if (sumList(properDivisor(n)) == m) and (sumList(properDivisor(m)) == n) and (n <> m):
        return 1
 
listAmicablePairs = []
counterAmicablePairs = 0
sumAmicablePairs = 0
 
for x in range (1,10001):
    tempList = properDivisor(x)
    tempVar = sumList(tempList)
    if checkAmicablePair(x, tempVar) == 1:     
        listAmicablePairs.append([x, tempVar])
        counterAmicablePairs = counterAmicablePairs + 1
        sumAmicablePairs = sumAmicablePairs + x + tempVar
 
print "TOTAL AMICABLE PAIRS ", counterAmicablePairs
print listAmicablePairs
print
 
print "Evaluation of exclusive sum = ", sumAmicablePairs / 2

About the Author