If a number, which when read in both forward and backward way is same, then such a number is called a palindrome number.
Example:
121, 23532, 12321 etc.
#include <stdio.h>
int main() {
int n, n1, rev = 0, rem;
printf("Enter any number: ");
scanf("%d", &n);
n1 = n;
/* logic */ while (n > 0){
rem = n % 10;
rev = rev * 10 + rem;
n = n / 10;
}
if (n1 == rev){
printf("Given number is a palindromic number \n");
}
else{
printf("Given number is not a palindromic number \n");
}
return 0;
} <b> Output 1</b>
Enter any number: 1221
Given number is a palindromic number
Press any key to continue . . .
<b> Output 2</b>
Enter any number: 1232
Given number is not a palindromic number
Press any key to continue . . .
Consider a numbern=121, reverse=0 and remainder;
number=121
now the while loop is executed /* the condition (n>0) is satisfied */
/* calculate remainder */
remainder of 121 divided by 10=(121%10)=1;
now reverse=(reverse*10)+remainder
=(0*10)+1 /* we have initialized reverse=0 */
=1
number=number/10
=121/10
=12
now the number is 12, greater than 0. The above process is repeated for number=12.
remainder=12%10=2;
reverse=(1*10)+2=12;
number=12/10=1;
now the number is 1, greater than 0. The above process is repeated for number=1.
remainder=1%10=1;
reverse=(12*10)+1=121;
number=1/10 /* the condition n>0 is not satisfied,control leaves the while loop */
Program stops here. The given number=121 equals the reverse of the number. Thus the given number is a palindrome number.
First read the algorithm, then study the program code line by line. After that, compare the code with the output and finally go through the explanation. This approach helps learners understand both the logic and the implementation properly.
After understanding this example, try to rewrite the same program without looking at the code. Then change some values or logic and run it again. This helps improve confidence and keeps learners engaged on the page for longer.