Home > Programming, Technology > Project euler problem 21: Amicable Number

Project euler problem 21: Amicable Number

Amicable Number:

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 ab, 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:

This problem is also easy. Without any optimization it takes considerable amount of time. Hence, I solved it by caching the intermediate results into an array.

```#include <iostream>

#include <stdio.h>

using namespace std;
long ami;
int amicable(int num)
{
long sum = 0;
if (ami[num - 1] != 0)
return ami[num - 1];
else
{
for (int i = 1; i <= num /2; i++)  {
if (num % i == 0)
sum += i;
}
ami[num - 1] = sum;
return sum;
}

}

int main()
{
long long sum = 0;

for(int i = 1; i <= 10000; i++ ) {
ami[i] = 0;
}
for(int i = 1; i <= 10000; i++ ) {
//        cout<<amicable(i)<<endl;
if (amicable(i) != i && i == amicable(amicable(i))) {
//          cout<<i<<endl;
sum += i;
}
}
cout<<"Sum of all amicable numbers: "<<sum;
return 0;
}

```

Time taken for the above implementation:

Sum of all amicable numbers: 31626
real 0m0.185s
user 0m0.140s
sys 0m0.046s

Categories: Programming, Technology
1. April 7, 2015 at 5:31 am

static long ami; can replace the loop to initialize ami[i] with zero for 10000 times. ideone.com gave me 0.48 secs for real time execution.

• April 7, 2015 at 12:43 pm

Thanks Hemanth for the comment. I’m planning to try a better mathematical approach.

• April 8, 2015 at 4:44 pm

Here is an optimized version of the above post. Please have a look. https://vasanthexperiments.wordpress.com/2015/04/07/optimized-solution-for-project-euler-problem-21/