Given a number, the task is to check if the number is divisible by 11 or not. The input number may be large and it may not be possible to store it even if we use long long int.

Examples:

Input : n = 76945 Output : Yes Input : n = 1234567589333892 Output : Yes Input : n = 363588395960667043875487 Output : No

Since input number may be very large, we cannot use n % 11 to check if a number is divisible by 11 or not, especially in languages like C/C++. The idea is based on following fact.

A number is divisible by 11 if **difference of following two** is divisible by 11.

- Sum of digits at odd places.
- Sum of digits at even places.

**Illustration:**

For example, let us consider 76945 Sum of digits at odd places : 7 + 9 + 5 Sum of digits at even places : 6 + 4 Difference of two sums = 21 - 10 = 11 Since difference is divisible by 11, the number 7945 is divisible by 11.

**How does this work?**

Let us consider 7694, we can write it as 7694 = 7*1000 + 6*100 + 9*10 + 4 The proof is based on below observation: Remainder of 10^{i}divided by 11 is 1 if i isevenRemainder of 10^{i}divided by 11 is -1 if i isoddSo the powers of 10 only result in values either 1 or -1. Remainder of "7*1000 + 6*100 + 9*10 + 4" divided by 11 can be written as : 7*(-1) + 6*1 + 9*(-1) + 4*1 The above expression is basically difference between sum of even digits and odd digits.

Below is implementation of above fact :

// C++ program to find if a number is divisible by // 11 or not #include<bits/stdc++.h> using namespace std; // Function to find that number divisble by 11 or not int check(string str) { int n = str.length(); // Compute sum of even and odd digit // sums int oddDigSum = 0, evenDigSum = 0; for (int i=0; i<n; i++) { // When i is even, position of digit is odd if (i%2 == 0) oddDigSum += (str[i]-'0'); else evenDigSum += (str[i]-'0'); } // Check its difference is divisble by 11 or not return ((oddDigSum - evenDigSum) % 11 == 0); } // Driver code int main() { string str = "76945"; check(str)? cout << "Yes" : cout << "No "; return 0; }

Output:

Yes

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