program for printing amstrong numbers within 1000 using c language?

Question

give complete program details using c language

Answer 1

PROGRAM ArmstrongNumber
IMPLICIT NONE

INTEGER :: a, b, c ! the three digits
INTEGER :: abc, a3b3c3 ! the number and its cubic sum
INTEGER :: Count ! a counter

Count = 0
DO a = 0, 9 ! for the left most digit
DO b = 0, 9 ! for the middle digit
DO c = 0, 9 ! for the right most digit
abc = a*100 + b*10 + c ! the number
a3b3c3 = a**3 + b**3 + c**3 ! the sum of cubes
IF (abc == a3b3c3) THEN ! if they are equal
Count = Count + 1 ! count and display it
WRITE(*,*) 'Armstrong number ', Count, ': ', abc
END IF
END DO
END DO
END DO

END PROGRAM ArmstrongNumber

Answer 2

here is the program to print armstrong numbers from 1 to 1000

#include
main()
{
int a,b,c,d;
for(a=1;a<=1000;a++)
{
c=a;
d=0;
while(a!=0)
{
b=a%10;
a=a/10;
d=(b*b*b)+d;
}
a=c;
if(d==c)
printf("\narmstrong number %d\n",c);

}

Answer 3

#include

void main()
{
int i=0, a=0,b=0,c=0,sum=0;

for(i=0;i<999;i++)
{
sum=0;
a=i%10; /* Units Digit */
b=(i%100-a)/10; /* Tenths digit */
c=(i%1000-(b*10+a))/100; /*100ths Digit*/
sum= (a*a*a) + (b*b*b) + (c*c*c);
if(i==sum)
printf("Armstong Number: %d\n",sum);
}

}

Answer 4

it truly is a pgm to print unusual form b/w 3 and one thousand no longer best form.best form is a form it truly is divisible by one and the type itself to discover best form do following int isPrime(int num){ static int i=2; if(i<=num/2){ if(nump.c.i==0) go back 0; else{ i++; isPrime(num); } } go back a million; }

Answer 5

The asker asked for it in C.

Check out this link - it has a version in C

Answer 6

int cube(int n)
{
return n*n*n;
}

main()
{
int i;
for(i=0;i<1000;i++)
{
if ( i == cube(i%10) + cube((i/10)%10) + cube((i/100)%10))
{
printf("%d\n", i);
}
}
}