Find the sum of digits of a number at even and odd places | Java implementation of the approach ; Function to return the reverse of a number ; Function to find the sum of the odd and even positioned digits in a number ; If c is even number then it means digit extracted is at even place ; Driver code
import java . util . * ; class GFG { static int reverse ( int n ) { int rev = 0 ; while ( n != 0 ) { rev = ( rev * 10 ) + ( n % 10 ) ; n /= 10 ; } return rev ; } static void getSum ( int n ) { n = reverse ( n ) ; int sumOdd = 0 , sumEven = 0 , c = 1 ; while ( n != 0 ) { if ( c % 2 == 0 ) sumEven += n % 10 ; else sumOdd += n % 10 ; n /= 10 ; c ++ ; } System . out . println ( " Sum ▁ odd ▁ = ▁ " + sumOdd ) ; System . out . println ( " Sum ▁ even ▁ = ▁ " + sumEven ) ; } public static void main ( String args [ ] ) { int n = 457892 ; getSum ( n ) ; } }