PROCEDURE:-
1.set n=1
2.repeat steps 3 to 6 until n reaches 100 incrementing n by 1 each time
3.set sum to 0
4.start finding modulo value for n with i=1 to n-1 value incrementing i by 1 each time
5.if at any point remainder is 0 add i to sum
6.if sum is same as n then print n
CODE:-
#include<stdio.h>
void main()
{
int n ,i,s;
printf(“the perfect numbers below 100 are \n”);
for(n=1;n<=100;n++)
{
s=0;
for(i=1;i<n;i++)
{
if (n % i = = 0)
s=s+i;
}
if (n = = s)
printf(“%3d “,n);
}
}
Output:-6,28
1.set n=1
2.repeat steps 3 to 6 until n reaches 100 incrementing n by 1 each time
3.set sum to 0
4.start finding modulo value for n with i=1 to n-1 value incrementing i by 1 each time
5.if at any point remainder is 0 add i to sum
6.if sum is same as n then print n
CODE:-
#include<stdio.h>
void main()
{
int n ,i,s;
printf(“the perfect numbers below 100 are \n”);
for(n=1;n<=100;n++)
{
s=0;
for(i=1;i<n;i++)
{
if (n % i = = 0)
s=s+i;
}
if (n = = s)
printf(“%3d “,n);
}
}
Output:-6,28
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