Answer / vinay k bharadwaj

dear frd your ans is wrong.


#include<stdio.h>
#include<conio.h>
void main()
{
int sum=0,n,m,x;
clrscr();
scanf("%d",&n);
m=n;
while(n>0)
{
x=n%10;
n=n/10;
sum=sum+(x*x*x);
}
if(sum==m)
printf("number is armstrong");
else
printf("number is not armstrong");
getch();
}

Answer / saravanan

#include<stdio.h>
#include<conio.h>
void main()
{
int sum=0,n,m;
clrscr();
scanf("%d",&n);
m=n;
while(n>0)
{
m=n%10;
sum=sum+(m*m*m);
n=n/10;
}
if(m==sum)
{
printf("the given number is amstrong");
else
printf("the given number is not amstrong");
}
getch();
}

Answer / roshan patel

#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int x,y,d;
cout<<"\nEnter the no : ";
cin>>x;
y=x;
int sum=0;
while(y>0)
{
d=y%10;
y/=10;
sum=sum+d*d*d;
}
if(x==sum)
cout<<"\n "<<x<<" is Armstrong ";
else
cout<<"\n"<<x<<" is not Armstrong";
getch();
}

Answer / rahul puri

#include<stdio.h>
#include<iostream.h>
#include<conio.h>
void main()
{
int sum=0,n,m;
clrscr();
cin>>n;
m=n;
while(n>0)
{
m=n%10;
sum=sum+(m*m*m);
n=n/10;
}
if(m==sum)
{
cout<<"the given number is amstrong";
}
else
{
cout<<"the given number is not amstrong";
}
getch();
}

Answer / astitva srivastava

#include<iostream.h>
#include<conio.h>
void main()
{
int n,r,s=0;
cout<<"enter the no.";
cin>>n;
int b=n;
while(n>0)
{
r=n%10;
n=n/10;
s=s+(r*r*r);
}
if(b==s)
{
cout<<"the no. is armstrong"<<"\n";
}
else
{
cout<<"the no. is not a armstrong no.";
}
}

Answer / ashish ranjan

#include<iostream.h>
#include<conio.h>
void main()
{
int sum=0,n,m;
clrscr();
cin>>n;
m=n;
while(n>0)
{
m=n%10;
sum=sum+(m*m*m);
n=n/10;
}
if(m==sum)
{
cout<<"the given number is amstrong";
}
else
{
cout<<"the given number is not amstrong";
}
getch();
}

Answer / sarang deshpande

#include<iostream.h>
#include<conio.h>
void main()
{
int n,n1,sum,rem;
clrscr();
cout<<" \n Enter a num ";
cin>>n;
n1=n;
sum=0;
while(n>0)
{
rem=n%10;
sum=sum+(rem*rem*rem);
n=n/10;
}
if(n1==sum)
{
cout<<"\n The number is armstrong "<<n1;
}
else
{
cout<<"\n The number is not armstrong "<<n1;
}
getch();
}

Answer / tapojit roy

#include<stdio.h>
#include<conio.h>
void main()
{
int n,m,x,sum=0;
clrscr();
printf ("the armstrong no. from 0 to 500 are\n");
for (n=1;n<=500;n++){
m=n;
while (m>0)
{
x=m%10;
sum=sum+(x*x*x);
m=m/10;
}
if (sum==n)
{
printf("%d\n",n);
}
sum=0;
}
getch();
}

Answer / ramesh

#include<iostream.h>
#include<conio.h>
void main()
{
int sum=0,n,m;
clrscr();
cin>>n;
m=n;
while(n>0)
{
m=n%10;
sum=sum+(m*m*m);
n=n/10;
}
if(m==sum)
{
cout<<"the given number is amstrong";
}
else
{
cout<<"the given number is not amstrong";
}
getch();
}

Answer / hitesh

Armstrong Program in C++

Armstrong number is a number that is the sum of its own digits each raised to the power of the number of digits is equal to the number itself.
For example 153 is armstrong number, 132 is not prime number. <a href="http://www.tutorial4us.com/cpp-program/cpp-armstrong-program" rel="nofollow">Armstrong program in c++</a> is very simple and easy to write.

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