Array values times 10

Array values times 10
Given an array of ints, compute recursively if the array contains somewhere a value followed in the array by that value times 10.
We'll use the convention of considering only the part of the array that begins at the given index. In this way, a recursive call can pass index+1 to move down the array. The initial call will pass in index as 0.

Input
On the first line you will be given the comma separated array. On the second line you will receive the index.

Output
On the only output line you should print the result.

Sample tests
Input
1,2,20
0
Output
true
Input
3,30
0
Output
true

import java.util.Scanner;
public class ArraysContaining11 {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String[] input = scanner.nextLine().split(",");
        int index = Integer.parseInt(scanner.nextLine());
        int[] numbers = new int[input.length];

        for (int i = 0; i < input.length; i++) {
            numbers[i] = Integer.parseInt(input[i]);
        }
        System.out.println(isEleven(numbers, index));
    }

    public static  boolean isEleven (int[] numbers, int index){
        if(index<0 || index>numbers.length-2){
            return false;
        }
        if(numbers[index] * 10 == numbers[index + 1]){
            return true;
        }
        return isEleven(numbers, index+1);
    }
}