# Project Euler Problem 21 Solved

## Amicable numbers

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.

## Solution

 `````` 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 `````` ``````#!/usr/bin/python # -*- coding: utf-8 -*- def sum_proper_factors(n): (result, sqrt) = (1, n ** 0.5) (start, step) = n % 2 == 1 and (3, 2) or (2, 1) for i in range(start, int(sqrt) + 1, step): if n % i == 0: result += i + n / i if sqrt == int(sqrt): result -= sqrt return result def main(): result = 0 for i in range(1, 10000): sum1 = sum_proper_factors(i) if sum1 > i: if i == sum_proper_factors(sum1): result += i + sum1 print result if __name__ == '__main__': import time startTime = time.time() main() print time.time() - startTime ``````
 `````` 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 39 40 41 42 `````` ``````package main import ( "fmt" "math" "time" ) func sum_proper_factors(n int) int { sum, sqrt := 1, math.Sqrt(float64(n)) start, step := 2, 1 if n%2 == 1 { start, step = 3, 2 } for i := start; i <= int(sqrt); i += step { if n%i == 0 { sum += i + n/i } } if sqrt == float64(int(sqrt)) { sum -= int(sqrt) } return sum } func main() { result, startTime := 0, time.Now() for i := 1; i < 10000; i++ { iSum := sum_proper_factors(i) if iSum > i { if i == sum_proper_factors(iSum) { result += i + iSum } } } fmt.Println(result, time.Now().Sub(startTime)) } ``````

I’m the 70186th person to have solved this problem.