GeeksforGeeks App
Open App
Browser
Continue

Check whether sum of digits at odd places of a number is divisible by K

Given two integers ‘N’ and ‘K’, the task is to find the sum of digits of ‘N’ at its odd places (right to left) and check whether the sum is divisible by ‘K’. If it is divisible, output YES, otherwise output NO.

Examples:

Input: N = 4325, K = 4
Output: YES
Since, 3 + 5 = 8, which is divisible by 4.

Input: N = 1209, K = 3
Output: NO

Approach:

• Find the sum of the digits of ‘N’ at odd places (right to left).
• Then check the divisibility of the sum by taking its modulo with ‘K’.
• If it is divisible, then output ‘YES’, otherwise output ‘NO’.

Below is the implementation of the above approach:

C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// function that checks the``// divisibility of the sum``// of the digits at odd places``// of the given number``bool` `SumDivisible(``int` `n, ``int` `k)``{``    ``int` `sum = 0, position = 1;``    ``while` `(n > 0) {` `        ``// if position is odd``        ``if` `(position % 2 == 1)``            ``sum += n % 10;``        ``n = n / 10;``        ``position++;``    ``}` `    ``if` `(sum % k == 0)``        ``return` `true``;``    ``return` `false``;``}` `// Driver code``int` `main()``{``    ``int` `n = 592452;``    ``int` `k = 3;` `    ``if` `(SumDivisible(n, k))``        ``cout << ``"YES"``;``    ``else``        ``cout << ``"NO"``;``    ``return` `0;``}`

Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `solution``{` `// function that checks the``// divisibility of the sum``// of the digits at odd places``// of the given number``static` `boolean` `SumDivisible(``int` `n, ``int` `k)``{``    ``int` `sum = ``0``, position = ``1``;``    ``while` `(n > ``0``) {` `        ``// if position is odd``        ``if` `(position % ``2` `== ``1``)``            ``sum += n % ``10``;``        ``n = n / ``10``;``        ``position++;``    ``}` `    ``if` `(sum % k == ``0``)``        ``return` `true``;``    ``return` `false``;``}` `// Driver code``public` `static` `void` `main(String arr[])``{``    ``int` `n = ``592452``;``    ``int` `k = ``3``;` `    ``if` `(SumDivisible(n, k))``        ``System.out.println(``"YES"``);``    ``else``        ``System.out.println(``"NO"``);` `}``}``//This code is contributed by Surendra_Gangwar`

Python 3

 `# Python 3 implementation of the approach` `# function that checks the divisibility``# of the sum of the digits at odd places``# of the given number``def` `SumDivisible(n, k):` `    ``sum` `=` `0``    ``position ``=` `1``    ``while` `(n > ``0``) :` `        ``# if position is odd``        ``if` `(position ``%` `2` `=``=` `1``):``            ``sum` `+``=` `n ``%` `10``        ``n ``=` `n ``/``/` `10``        ``position ``+``=` `1``    ` `    ``if` `(``sum` `%` `k ``=``=` `0``):``        ``return` `True``    ``return` `False` `# Driver code``if` `__name__ ``=``=``"__main__"``:``    ``n ``=` `592452``    ``k ``=` `3` `    ``if` `(SumDivisible(n, k)):``        ``print``(``"YES"``)``    ``else``:``        ``print``(``"NO"``)` `# This code is contributed``# by ChitraNayal`

C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``// function that checks the``// divisibility of the sum``// of the digits at odd places``// of the given number``static` `bool` `SumDivisible(``int` `n, ``int` `k)``{``    ``int` `sum = 0, position = 1;``    ``while` `(n > 0)``    ``{` `        ``// if position is odd``        ``if` `(position % 2 == 1)``            ``sum += n % 10;``        ``n = n / 10;``        ``position++;``    ``}` `    ``if` `(sum % k == 0)``        ``return` `true``;``    ``return` `false``;``}` `// Driver code``static` `public` `void` `Main ()``{``    ``int` `n = 592452;``    ``int` `k = 3;` `    ``if` `(SumDivisible(n, k))``        ``Console.WriteLine(``"YES"``);``    ``else``        ``Console.WriteLine(``"NO"``);``}``}` `// This code is contributed by Sachin`

PHP

 ` 0)``    ``{` `        ``// if position is odd``        ``if` `(``\$position` `% 2 == 1)``            ``\$sum` `+= ``\$n` `% 10;``        ``\$n` `= (int)``\$n` `/ 10;``        ``\$position``++;``    ``}` `    ``if` `(``\$sum` `% ``\$k` `== 0)``        ``return` `true;``    ``return` `false;``}` `// Driver code``\$n` `= 592452;``\$k` `= 3;` `if` `(SumDivisible(``\$n``, ``\$k``))``    ``echo` `"YES"``;``else``    ``echo` `"NO"``;` `// This code is contributed``// by Sach_Code``?>`

Javascript

 ``

Output:

`YES`

Time Complexity: O(log10n), since each time the value of n is reduced to n/10.

Auxiliary Space: (1), since no extra space has been taken.

Method #2:Using string() method:

1. Convert the integer to a string, then traverse the string and find the sum of all odd indices by storing it in sum.
2. If the sum is divisible by k, then return True else False.

Below is the implementation:

C++

 `// C++ implementation of the``// above approach``#include ``using` `namespace` `std;` `bool` `sumDivisible(``int` `n, ``int` `k)``{``    ``int` `sum = 0;` `    ``// Converting integer to string``    ``string num = to_string(n);``    ``int` `i;``  ` `    ``// Traversing the string``    ``for` `(i = 0; i < num.size(); i++) {``        ``if` `(i % 2 != 0) {``            ``sum = sum + (num[i] - ``'0'``);``        ``}``    ``}` `    ``if` `(sum % k == 0) {``        ``return` `true``;``    ``}``    ``else` `{``        ``return` `false``;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `n = 592452;``    ``int` `k = 3;``    ``if` `(sumDivisible(n, k)) {``        ``cout << (``"YES"``);``    ``}``    ``else` `{``        ``cout << (``"NO"``);``    ``}``    ``return` `0;``}` `// This code is contributed by gauravrajput1`

Java

 `// Java implementation of the``// above approach``import` `java.io.*;``class` `GFG``{``static` `boolean` `sumDivisible(``int` `n, ``int` `k)``{``    ``int` `sum = ``0``;` `    ``// Converting integer to string``    ``String num = Integer.toString(n);``    ``int` `i;``  ` `    ``// Traversing the string``    ``for` `(i = ``0``; i < num.length(); i++) {``        ``if` `(i % ``2` `!= ``0``) {``            ``sum = sum + (num.charAt(i) - ``'0'``);``        ``}``    ``}` `    ``if` `(sum % k == ``0``) {``        ``return` `true``;``    ``}``    ``else` `{``        ``return` `false``;``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``592452``;``    ``int` `k = ``3``;``    ``if` `(sumDivisible(n, k)) {``        ``System.out.println(``"YES"``);``    ``}``    ``else` `{``        ``System.out.println(``"NO"``);``    ``}``}``}` `// This code is contributed by shivanisinghss2110`

Python3

 `# Python3 implementation of the``# above approach` `def` `sumDivisible(n, k):``    ``sum` `=` `0``    ` `    ``# Converting integer to string``    ``num ``=` `str``(n)``    ` `    ``# Traversing the string``    ``for` `i ``in` `range``(``len``(num)):``        ``if``(i ``%` `2` `!``=` `0``):``            ``sum` `=` `sum``+``int``(num[i])` `    ``if` `sum` `%` `k ``=``=` `0``:``        ``return` `True``    ``return` `False`  `# Driver code``n ``=` `592452``k ``=` `3``if` `sumDivisible(n, k) ``=``=` `True``:``    ``print``(``"YES"``)``else``:``    ``print``(``"NO"``)` `# This code is contributed by vikkycirus`

C#

 `// C# implementation of the``// above approach``using` `System;``class` `GFG``{``static` `bool` `sumDivisible(``int` `n, ``int` `k)``{``    ``int` `sum = 0;` `    ``// Converting integer to string``    ``string` `num = n.ToString();``    ``int` `i;``  ` `    ``// Traversing the string``    ``for` `(i = 0; i < num.Length; i++) {``        ``if` `(i % 2 != 0) {``            ``sum = sum + (num[i] - ``'0'``);``        ``}``    ``}` `    ``if` `(sum % k == 0) {``        ``return` `true``;``    ``}``    ``else` `{``        ``return` `false``;``    ``}``}` `// Driver code``static` `public` `void` `Main ()``{``    ``int` `n = 592452;``    ``int` `k = 3;``    ``if` `(sumDivisible(n, k)) {``        ``Console.Write(``"YES"``);``    ``}``    ``else` `{``        ``Console.Write(``"NO"``);``    ``}``}``}` `// This code is contributed by shivanisinghss2110`

Javascript

 ``

Output:

`Yes`

Time Complexity: O(d), where d is the number of digits in n.

Auxiliary Space: O(d), where d is the number of digits in n.

My Personal Notes arrow_drop_up