In this program, we are trying to check whether the two given numbers by the user through console, are friendly pair or not?
Example
If sum of all divisors of number1 is equal to number1 and sum of all divisors of number2 is equal to number2, then we can say, those two numbers are abundant numbers.
The logic that we used to find friendly pairs is as follows −
For the sum of all divisors of number 1.
for(i=1;i<number1;i++){
if(number1 % i == 0){
result1= result1 +i;
}
}对于数字2的所有除数的总和。
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for(i=1;i<number2;i++){
if(number2 % i == 0){
result2=result2+i;
}
}For the friendly pairs.
if(result1==number1 && result2==number2)
If this condition is satisfied, then they are abundant pairs, otherwise they are not.
Example
Following is the C program to find whether the given numbers are abundant pairs or not −
Live Demo
#include<stdio.h>
int main(){
int number1,number2,i;
printf("Enter two numbers:");
scanf("%d%d",&number1,&number2);
int result1=0,result2=0;
for(i=1;i<number1;i++){
if(number1 % i == 0){
result1= result1 +i;
}
}
for(i=1;i<number2;i++){
if(number2 % i == 0){
result2=result2+i;
}
}
if(result1==number1 && result2==number2)
printf("Abundant Pairs");
else
printf("Not abundant Pairs");
return 0;
}输出
输出如下 −
Enter two numbers:6 28 Abundant Pairs
