problem Description : 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.
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.
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| /// Solved By Shohanur Rahaman | |
| #include<stdio.h> | |
| #include<math.h> | |
| int sumofdiv(int n); | |
| int main() | |
| { | |
| int i,j; | |
| int n,count=0,div1=1,tmp=0,div2=0; | |
| int sum=0; | |
| n=1; | |
| while(n<=10000){ | |
| int root=sqrt(n); | |
| for(i=2;i<=root;i++){ | |
| if(n%i==0){ | |
| div1=div1+i; | |
| if(i!=n/i) | |
| div1=div1+(n/i); | |
| } | |
| } | |
| tmp=sumofdiv(div1); | |
| if(tmp==n && tmp!=div1){ | |
| sum=sum+div1; | |
| printf("Div1 : %d Tmp: %d \n",div1,tmp); | |
| } | |
| div1=1; | |
| n++; | |
| } | |
| printf("\n sum : %d\n",sum); | |
| return 0; | |
| } | |
| int sumofdiv(int n) | |
| { | |
| int i,div=1; | |
| int root=sqrt(n); | |
| for(i=2;i<=root;i++){ | |
| if(n%i==0){ | |
| div=div+i; | |
| if(i!=n/i){ | |
| div=div+n/i; | |
| } | |
| } | |
| } | |
| return div; | |
| } | |
