primePalindrome: easy solution with shortcut. 
IntuitionPalindromes have a symmetrical property which can be leveraged in combination with primality checks.Approach
Start by checking if n is 1 or 2, return 2
# Intuition
Palindromes have a symmetrical property which can be leveraged in combination with primality checks.

# Approach
1. Start by checking if n is 1 or 2, return 2 if true.
2. Ensure n is an odd number by incrementing it if it's even.
3. Use a loop to iterate through numbers, checking for the following:
- Skip numbers between 10,000,000 and 100,000,000 as they cannot be prime palindromes.
- Reverse the number to check if it is a palindrome.
- Verify if the palindrome number is prime by checking divisibility up to its square root.
4. Return the number if it's a prime palindrome, otherwise, increment n and repeat the process.

# Complexity
- Time complexity:
O(n)

- Space complexity:
O(1)

# Code
```c []
int primePalindrome(int n) {
    if(n==1 || n==2) return 2;
    if(n%2==0) n=n+1;
    while(true){
        int temp=n;
        if(10000000 < temp && temp < 100000000){
            n=100000000;
        }
        int num=0;
        while(temp!=0){
            num*=10;
            num+=temp%10;
            temp/=10;
        }
        if(n==num){
            bool divisible=false;
            for(int i=2;i<=sqrt(num);i++){
                if(num%i==0){
                    divisible=true;
                    break;
                }
            }
            if(!divisible) return num;
        }
        n++;
    }
    return -1;
}
```
['Math', 'C', 'Number Theory']