Given a number, write a algorithm, flowchart, pseudocode to check if it is palindrome or not.
A string is said to be palindrome if reverse of the string is same as string. For example, 1221
is palindrome, but 1223
is not palindrome.
We will follow a method where we get each numbers using mathematical operations.
Logic

Get the number to check and store it in
n
. 
Create empty variables
r = 0
,s = 0
andt = n
. Here,s
will become the reversed number,t
will be a reference to original number,r
will be a temporary variable. 
Start a while loop checking whether the
n != 0
(if n is 0 exit the while loop). 
Inside the while loop do the following operations.
r = n % 10 s = s * 10 + r n = n / 10
First get the last digit of
n
by dividing the given number(n
) by 10 and finding the remainder(the%
operator) and store it inr
. Then, addr
tos
after multiplying it by previous stored value ofs
by 10. And finally dividen
by 10 and store it again inn
. This way the number is reversed and stored ins
. 
After the loop ends, check whether
s == t
(whether the original number is same as reversed number). 
If
s == t
, display thatn
palindrome, else displayn
is not a palindrome.
Flowchart
WIP, have a look here.
graph TD
A((Start)) > B[/Read Number to check as n/];
B > C[Assign s=0, t = n];
C > D{ While n!=0 };
D  True > E["r = n % 10 <br>s = s * 10 + r<br>n = n / 10"] > D
D  False > F{Is n == t?}
F  Yes > G[/Print n is palindrome/] > I[Stop];
F  No > H[/Print n is not a palindrome/] > I((Stop));
Pseudocode
BEGIN
READ number to check as n
SET s = 0
SET t = n
WHILE n != 0 DO
r = n % 10
s = s * 10 + r
n = n / 10
ENDWHILE
IF n == t
DISPLAY n is palindrome
ELSE
DISPLAY n is not a palindrome
ENDIF
END
Implementation

In C:
#include <stdio.h> /* Function to check if n is Palindrome*/ int is_palindrome(int n) { int s = 0; int t = n; int r = 0; while (n != 0 ) { r = n % 10; s = s * 10 + r; n = n / 10; } // Check if rev_n and n are same or not. if (s == t) return 1; else return 0; } int main() { int tocheck = 1221; int op = is_palindrome(to_check); if (op == 1) printf("%d is a palindrome", tocheck); else printf("%d is not a palindrome", tocheck); return 0; }

In Python:
def is_palindrome(n: int) > bool: s = 0 t = n r = 0 while n != 0: r = n % 10 s = s * 10 + r n = n / 10 if t == s: return True else: return False number = 1001 palim = is_palindrom(number) if palim: print(f"{number} is a palindrom") else: print(f"{number} is not a palindrom")