Write a c program to check Armstrong number using functions?

Answer 1

TO PRINT ALPHABET PYRAMID

ABCDEFGGFEDCBA

ABCDEFFEDCBA

ABCDEEDCBA

ABCDDCBA

ABCCBA

ABAB

A

#include

#incldue

int main( )

{

int r,c,askey,sp;

clrscr( );

for( r=7; r>=1; r-- )

{

for(sp=6; sp>=r; sp--)

printf(" "); //2 spaces

askey=65;

for(c=1; c<=r; c++ )

printf("%2c", askey++ );

for(c=r-1; c>=1; c-- )

printf("%2c", --askey);

printf("\n");

}

getch();

return 0;

}

by BRAMHA VARDHAN

Answer 2

#include
main()
{
int number, temp, digit1, digit2, digit3;

printf("Printing all Armstrong numbers between 1 and 500:\n\n");

number = 001;

while (number <= 500)
{
digit1 = number - ((number / 10) * 10);

digit2 = (number / 10) - ((number / 100) * 10);

digit3 = (number / 100) - ((number / 1000) * 10);

temp = (digit1*digit1*digit1) + (digit2*digit2*digit2) + (digit3*digit3*digit3);

if (temp == number)
{
printf("\nAmstrong Number:%d", temp);

}

number++;
}
}

Answer 3

#include<stdio.h> #include<conio.h> #include<math.h> int armstrong(int num) { int i,c=0,arm=0; int ar[100]; for(i=0;i<100;i++) { ar[i]=0; } while(num>0) { ar[c]=num%10; c++; num/=10; } for(i=1;i<c;i++) { arm+=pow(ar[i],3); } return arm; } void main() { clrscr(); int n,a=0; printf("\n Enter the number: "); scanf("%d",&n); a=armstrong(n); printf("\n The number %d is",n); if(n==a) printf(" armstrong number"); else printf(" not an armstrong number"); getch(); }

Answer 4

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