Amicable numbers : Problem 21
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.
Problem Source : Euler Project
Python Code
def amicableNumber():
totalSum = 0
def sumOfDivisors(numb):
sumD = 0
for d in range(numb/2):
d = d+1
if numb%d ==0:
sumD = sumD + d
return sumD
for i in range(10000-1):
i = i+1
if i == sumOfDivisors(i):
continue
if i == sumOfDivisors(sumOfDivisors(i)):
print i
totalSum = totalSum + i
print "Evaluate the sum of all the amicable numbers under 10000:",totalSum
amicableNumber()
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.
Problem Source : Euler Project
Python Code
def amicableNumber():
totalSum = 0
def sumOfDivisors(numb):
sumD = 0
for d in range(numb/2):
d = d+1
if numb%d ==0:
sumD = sumD + d
return sumD
for i in range(10000-1):
i = i+1
if i == sumOfDivisors(i):
continue
if i == sumOfDivisors(sumOfDivisors(i)):
print i
totalSum = totalSum + i
print "Evaluate the sum of all the amicable numbers under 10000:",totalSum
amicableNumber()
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