Difference between sums of odd and even digits.

Given a number n, check whether the difference between the sum of digits at odd positions and the sum of digits at even positions is zero. Positions are indexed starting from 0 (rightmost digit).

Key Insight

A number is divisible by 11 if and only if the alternating sum of its digits (difference between sum of digits at odd and even positions) is 0 or divisible by 11. So a simple n % 11 == 0 check can determine if the difference is zero.

Worked Example

For n = 1212112 −

Digits:     1  2  1  2  1  1  2
Position:   6  5  4  3  2  1  0

Sum of odd positions  (5,3,1) = 2 + 2 + 1 = 5
Sum of even positions (6,4,2,0) = 1 + 1 + 1 + 2 = 5

Difference = 5 - 5 = 0 ? Yes

Java Implementation

The following program checks if the difference is zero using the divisibility by 11 trick ?

class JavaTester {
    public static int difference(int n) {
        return (n % 11);
    }

    public static void main(String args[]) {
        int n = 1212112;
        System.out.println("Number: " + n);
        System.out.println(difference(n) == 0 ? "Yes" : "No");

        n = 12121121;
        System.out.println("Number: " + n);
        System.out.println(difference(n) == 0 ? "Yes" : "No");
    }
}

The output of the above code is ?

Number: 1212112
Yes
Number: 12121121
No

Conclusion

The difference between sums of odd-positioned and even-positioned digits is zero when the number is divisible by 11. This provides an efficient O(1) solution using the modulo operator instead of extracting individual digits.