Pre-requisite: Input number n

Input Example:

    Input number: 24
    Binary representation: 00011000
    Thus it's a palindrome

    Input number 153:
    Binary representation: 10011001
    Thus it's a palindrome

    Input number: 25
    Binary representation: 00011001
    Thus it's not a palindrome

Algorithm:

1)  Take the input.
2)  Create an array of length 8 to store 8 bit binary representation 
    of input number
3)  Use bitwise operators to convert into binary from
    a)  Initialize i to 7 (8-1)
    b)  Find ith bit & store in the array
        array[i]=n & 1
    c)  Right shift n & decrement i
        n=n>>1
    d)  IF n=0
            Break
        ELSE
        Repeat step b-d
    4)  Check the array where the binary representation is stored, 
        for palindrome
        a)  Set j= 0 & k=7 (8-1, 8 is the MAX SIZE)
        b)  IF (array [j]!= array [k])
            It's not a palindrome
        c)  IF j < k
                Increment j& decrement k
                Repeat b, c
            ELSE
                It's a palindrome

Example with explanation:

Input no: 24
Converting to binary representation using bitwise operator

Initially,
N=24
Array[8]={0};

0	0	0	0	0	0	0	0 (LSB)
i=7
-----------------------------------------------------------------

first iteration,
array[7]=n&1 = 0 (refer to bitwise operators and 
their working for understanding the outcome)
0	0	0	0	0	0	0	0 (current bit)
n=n>>1=12
i=6
-----------------------------------------------------------------

second iteration,
array [6]=n&1 = 0 
0	0	0	0	0	0	0 (current bit)	0 

n=n>>1=6
i=5
-----------------------------------------------------------------

third iteration,
array [5]=n&1 = 0 
0	0	0	0	0	0 (current bit)	0 	0 

n=n>>1=3
i=4
-----------------------------------------------------------------

fourth iteration,
array [4]=n&1 = 1 
0	0	0	0	1(current bit)	0 	0 	0 

n=n>>1=1
i=3
-----------------------------------------------------------------

fifth iteration,
array [3]=n&1 = 1 
0	0	0	1 (current bit)	1	0 	0 	0 

n=n>>1=0
i=2
-----------------------------------------------------------------

sixth iteration,
n=0
so no more processing
thus the array finally becomes which is 
the binary representation
0	0	0	1 	1	0 	0 	0  (LSB)

Checking palindrome
Initially,
j=0 & k=7 (8-1)
-----------------------------------------------------------------

First iteration
Array[j] == array [k] (both 0)
j=j+1=1
k=k-1=6
j<k
-----------------------------------------------------------------

Second iteration
Array[j] == array [k] (both 0)
j=j+1=2
k=k-1=5
j<k
-----------------------------------------------------------------

Third iteration
Array[j] == array [k] (both 0)
j=j+1=3
k=k-1=4
j<k
-----------------------------------------------------------------

Fourth iteration
Array[j] == array [k] (both 1)
j=j+1=4
k=k-1=3
j>k
thus, stops processing & prints it's a palindrome

C Implementation

#include <stdio.h>
#define SIZE 8

int main()
{
    unsigned int n;
    printf("enter the no ( max range 255)\n");
    scanf("%d",&n);
    int c[SIZE]={0};
    int i=SIZE-1;
    printf("binary representation is: ");
    while(n!=0){
        c[i--]=n&1;
        n=n>>1;
        
    }
    for(int j=0;j<SIZE;j++)
    printf("%d",c[j]);
    printf("\n");
    
    for(int j=0,k=SIZE-1;j<k;j++,k--){
        if(c[j]!=c[k]){
            printf("Not palindrome\n");
            return 0;
        }
    }
    
    printf("it's palindrome\n");

    return 0;
}

Output

First run:
enter the no ( max range 255)
153
binary representation is: 10011001
it's palindrome

Second run:
enter the no ( max range 255)
24
binary representation is: 00011000
it's palindrome

Third run:
enter the no ( max range 255)
-8
binary representation is: 11111000
Not palindrome