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

Given two integer ‘N’ and ‘K’, the task is to find the sum of digits of ‘N’ at it’s 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

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

Approach:

• Find the sum of the digits of ‘N’ at odd places (right to left).
• Then check the divisibility of the sum by taking it’s 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 ` `?> `

Output:

```YES
```

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