i have solve this program in college days it easy to get result

Write a program on reverse a number and test whether it is palindrome or not

l

o/p:

If you enter 'n' value as 101 the output is reverse is 101 and palindrome

Find sum of the digits repeatedly until we get single digit number that is

i. 199 want to become as 1

ii. 30006 want to become as 9

iii. 3456 want to become as 9

iv. 32567 want to become as 5

o/p:

if you enter 'x' value as 199 output is 1

l

Code: [Select]

`ong reverse( long n)`

{

long sum = 0;

int rem;

while( n > 0)

{

rem = n% 10;

sum = sum * 10 + rem; // multiply sum by 10 and add remainder

n = n / 10;

}

return ( sum );

}

char * palin( long n)

{

long reverse( long );

if( n == reverse( n )) // if the given number and reverse number are equal , 'n' is palind'me

return( " palin");

else

return( " not palindrome");

}

main()

{

long n , reverse( long);

char * palin( long);

printf( " enter any number");

scanf("%d",&n);

printf( " reverse number is %ld", reverse( n));

printf(" %s", palin( n));

getch();

}

o/p:

If you enter 'n' value as 101 the output is reverse is 101 and palindrome

Find sum of the digits repeatedly until we get single digit number that is

i. 199 want to become as 1

ii. 30006 want to become as 9

iii. 3456 want to become as 9

iv. 32567 want to become as 5

Code: [Select]

`#include <stdio.h>`

#include <conio.h>

int single-digit( int n)

{

if( n % 9 == 0)

return (9); // if 'n' is divisible by'9' return '9' itself

else

return ( n % 9); // if 'n' is not divisible by '9' return n% 9

}

main()

{

int n, single-digit(int);

printf(" enter any number");

scanf("%d", &x);

printf(" single digit is %d", single-digit ( x));

getch();

}

o/p:

if you enter 'x' value as 199 output is 1

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