A friend posted this question on my college mailing list.

How do we find if a Number is a Palindrome?

Note: No conversion to string. And also, no traditional method of dividing the number to construct the reversed number and then compare.

Here is how you go about it.

- Lets say the number is 1987891 (odd number of digits)
- Reduce it to 198891. If the number were 19877891 (even number of digits) don’t change it (basically change it to a number with even number of digits, by excluding the middle digit)
- Now Split it into 198 and 891
- If ((891-198)%9 == 0 and (891*1000+198)%11 == 0 and (198*1000+891)%11 == 0) then its a palindrome

Sample C Code

```
#include <stdio.h>
int numdigits(int n) {
int count = 0;
while(n != 0) {
n /= 10;
++count;
}
return count;
}
int main(void) {
int n;
int l;
int a,x;
int n1, n2, n3;
scanf("%d",&n);
l=numdigits(n); /* Find the number of digits */
a=1;
for(x=0;x<l/2;x++)
a*=10;
n1=n%a;
n2=n/a;
if(l%2)
n2/=10;
if (((n1-n2)%9 == 0) && ((n2*a+n1)%11 == 0) && ((n1*a+n2)%11 == 0)) {
printf("%d\t",n);
return 0;
}
return 1;
}
```

To compile it

gcc -o palindrome palindrome.c

You can verify that above code works using the below script.

```
#!/bin/bash
for (( i = 0; i < 100000; i++ )); do
echo $i | ./palindrome
done
```

The script assumes that the palindrome binary is in the present working directory.

Jotted this down for school students scouring the web for their homework answers: 😉