# Check whether a large number is divisible by 53 or not

Given a large number in the form of a string N, the task is to check whether the number is divisible by 53 or not.

Examples:

Input: N = 5299947
Output: Yes

Input: N = 54
Output: No

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Extract the last digit of the given string N and remove it.
• Multiply that digit by 37.
• Subtract the product calculated in the above step from the remaining number.
• Continue until we reduce the given string to a 3 or four digit number.
• Convert the remaining string to its corresponding integer form and check if it is divisible by 53 or not.

Below is the implementation of the above approach:

## C++

 `// C++ program to check ` `// whether a number ` `// is divisible by 53 or not ` `#include ` `using` `namespace` `std; ` ` `  `// Function to check if the ` `// number is divisible by 53 or not ` `bool` `isDivisible(string s) ` `{ ` `    ``int` `flag = 0; ` `    ``while` `(s.size() > 4) { ` ` `  `        ``int` `l = s.size() - 1; ` `        ``int` `x = (s[l] - ``'0'``) * 37; ` ` `  `        ``reverse(s.begin(), s.end()); ` `        ``s.erase(0, 1); ` ` `  `        ``int` `i = 0, carry = 0; ` `        ``while` `(x) { ` `            ``int` `d = (s[i] - ``'0'``) ` `                    ``- (x % 10) ` `                    ``- carry; ` `            ``if` `(d < 0) { ` `                ``d += 10; ` `                ``carry = 1; ` `            ``} ` `            ``else` `                ``carry = 0; ` ` `  `            ``s[i] = (``char``)(d + ``'0'``); ` `            ``x /= 10; ` `            ``i++; ` `        ``} ` ` `  `        ``while` `(carry && i < l) { ` `            ``int` `d = (s[i] - ``'0'``) - carry; ` `            ``if` `(d < 0) { ` `                ``d += 10; ` `                ``carry = 1; ` `            ``} ` `            ``else` `                ``carry = 0; ` ` `  `            ``s[i] = (``char``)(d + ``'0'``); ` ` `  `            ``i++; ` `        ``} ` ` `  `        ``reverse(s.begin(), s.end()); ` `    ``} ` ` `  `    ``int` `num = 0; ` `    ``for` `(``int` `i = 0; i < s.size(); i++) { ` `        ``num = num * 10 + (s[i] - ``'0'``); ` `    ``} ` ` `  `    ``if` `(num % 53 == 0) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string N = ``"18432462191076"``; ` ` `  `    ``if` `(isDivisible(N)) ` `        ``cout << ``"Yes"` `<< endl; ` `    ``else` `        ``cout << ``"No"` `<< endl; ` ` `  `    ``return` `0; ` `} `

Output:

```Yes
```

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