Project Euler Problem 21 Solved

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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

p21.py
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#!/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
p21.go
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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.

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