Positively Described Numbers - Part 1

Computer Science Level 3

Two numbers are "amicable" if the sum of the proper factors of one number is equal to the other number. The first pair of "amicable" numbers is $220$ and $284.$ What is the sum of the numbers that make the second pair of "amicable" numbers?

$\textbf{Details and Assumptions}$

The proper factors of a number are the positive factors of the number that are not equal to the number itself. For example, the proper factors of $12$ are $1,$ $2,$ $3,$ $4,$ and $6.$

The proper factors of $220$ are $1,$ $2,$ $4,$ $5,$ $10,$ $11,$ $20,$ $22,$ $44,$ $55,$ and $110.$ The factors of $284$ are $1,$ $2,$ $4,$ $71,$ and $142.$ The numbers $220$ and $284$ are "amicable" because $1+2+4+5+10+11+20+22+44+55+110=284$ and $1+2+4+71+142=220.$

The answer is 2394.

7 solutions

Thaddeus Abiy
Mar 2, 2014

Python:

'Prints first five pairs'
def sum_divisors(n):
    s = 0
    for i in range(1,n):
        if n % i == 0: s += i
    return s

n , Count = 1 , 0
while Count <= 5:
    n += 1
    Val = sum_divisors(n)    
    if n == sum_divisors(sum_divisors(n)):
        print Count,n,Val,n+Val
        Count += 1

Python all the way! ;)

Thomas Luo - 7 years, 3 months ago

Karan Siwach
Oct 30, 2015

include<stdio.h>

include<math.h>

int main()

{

int i,j,m,n;

for(i=285;i<=1800;i++)
{
     int s1=0,s2=0,k=sqrt(i);
        for(j=2;j<=k;j++)
        {
            if(i%j==0) {s1=s1+j+i/j;}}
            s1+=1;
            int l=sqrt(s1);
            for(m=2;m<=l;m++)
            {
                if(s1%m==0){ s2=s2+m+s1/m;}

            } if(s2+1==i && s1!=i) {printf("%d\n",i+s1);
            return 0;}
        }

return 0;

}

Mayssara Moëtez
Apr 16, 2014

My Code in C:

  main(){int i,j,k, sum1=0, sum3=0, sum2=0;
 for(j=285; j<2000; j++)
   {  for(i=1; i<j; i++)
   {  
       if(j%i==0)
          sum1=sum1+i;
   }
    if(sum1!=j) 
    {
   for(k=1; k<sum1; k++)
   {  
       if(sum1%k==0)
          sum2=sum2+k;
   }
    }  
   if(sum2==j)
   {
     sum3=sum1+sum2;
     printf("s3=%d\n", sum3);
     break;  
   }    
    sum1=0;
    sum2=0;   
   }

}

Vaibhav Jain
Apr 12, 2014

Answere in Bacis C language:-

include<stdio.h>

include<stdlib.h>

int profac(int a){

int i,sum=0;

    for(i=1;i<=a/2;i++){

        if(a%i==0){

            //printf("a factor is %d\n",i);

            sum+=i;

        }

    }

    //printf("total sum of proper factor of %d is :-%d\n",a,sum);

    return sum;

} void main(){

int a=7,b;

int flag=1;

while(flag){

    b=profac(a);

    if (a==profac(b) && a!=b && (a!=220&&a!=284)){

        flag=0;

    }

    a+=1;

}

printf("AMICABLE numbers are %d and %d\n",a-1,b);

}

Sanjay Kamath
Apr 9, 2014

Here is my code in C#

        Here is a subroutine to be called for each number, say 1 to 2000

        private void GetFactorDetails(double num, ref double numsum)
       {
            //Positively Described Numbers - Part 1
        double n = num;
        double j = 2;
        double k = 0;
        double fcount = 1;
        ArrayList ar = new ArrayList();

        while (j < n)
        {
            k = (n % j);

            if (n != j)
            {
                if ((int)k == 0)
                {
                    fcount += j; j++; numsum = fcount; continue;

                }
                else
                {
                    j++; continue; 
                }
            }


            n++;
            fcount = 0;
            numsum = 0;
            if ((int)n == 1000) break;
        }//loop

    }

    Here is the calling routine.

         ArrayList ar = new ArrayList();


        double snum = 0;
        double tnum = 0;
        for (int jj = 1; jj < 2000; jj++)
        {

            GetFactorDetails(jj, ref snum);
            GetFactorDetails(snum, ref tnum);

            if ((jj == tnum) && (jj != snum))
            {
                if (!(ar.Contains(jj.ToString())))
                {
                    ar.Add(jj.ToString());
                }

                if (!(ar.Contains(snum.ToString())))
                {
                    ar.Add(snum.ToString());
                }
            }

        }

        for (int jk = 0; jk < ar.Count; jk++)
        {
            MessageBox.Show(ar[jk].ToString());
        }

Souradeep Roy Choudhury
Mar 2, 2014

import java.io.*; class hwe {

int r(int x) { int sum=0; for(int i=1;i<x;i++) { if(x%i==0) sum=sum+i; } return sum; } void am(int n) { for(int i=1;i<=n;i++) { int s1=r(i); int s2=r(s1); if(s2==i && s1>s2 && s2<=n) System.out.println(i+" "+s1); } } } here,void r function is used to generate the sum of proper factors of a number, s1 to find out the sum of proper factors of every number>=1 but<n and s2 to generate the sum of proper factors of s1

Joji Thomas
Mar 1, 2014

check wikipedia

That is definitely NOT how you solve this problem. That is cheating.

Trevor B. - 7 years, 3 months ago

I checked wikipedia as well because I was confused; my code gave "amicable" numbers much lower than 220 and 284, such as 6 and 25: 1+2+3=1+5 -- the sum of the proper factors of 6 is equal to the sum of the proper factors of 25.

Turns out the first sentence of the question mislead me (and then I didn't read Details and Assumptions closely enough): it should say something like "Two numbers are "amicable" if the sum of the proper factors of one number is equal to the other number, and vice versa". Can you fix?

Casey McNamara - 6 years, 10 months ago

Positively Described Numbers - Part 1

The answer is 2394.

7 solutions

include<stdio.h>

include<math.h>

include<stdio.h>

include<stdlib.h>

0 pending reports