file.java

import java.util.Scanner;
public class file{
    static int reverse(int num){
        int temp = num,rev= 0;
        for(;temp!=0;temp/=10)
            rev = rev*10+temp%10;
        return rev;
    }
    static boolean isPallendrome(int num){
        if(num==reverse(num))
            return true;
        return false;
    }
    public static void main(String args[]){
        Scanner input = new Scanner(System.in);
        int t = input.nextInt();
        while(t-->0){
            int n = input.nextInt(),temp=0;
            for(int i=990;i>99;i-=11){
                for(int j=999;j>99;j--){
                    int req = i*j;
                    if(req<n && isPallendrome(req)){
                        if(req>temp)
                            temp = req;
                        break;
                    }
                }
            }
        System.out.println(temp);
        }
    }
}

/*  Explanation:
        suppose abccba be the pallendrome number(Remember input constrains are for six digit numbers)
        a*100000+b*10000+c*1000+c*100+b*10+a
        a*100001+b*10010+c*1100
        11*(a*9091+b*910+c*100).Hence a pallendrome number must be 11 multiple
        Question is largest pallendrome number i.e product of two 3-digit numbers
        Hence either of the two numbers is a 11 multiple. Hence one loop from 990 to 99 with 11 units gap.
        Other number from 999 to 99 with 1 gap.
*/