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Use the keyboard to enter a positive number. Determine the number of even digits and odd digits in the entered number.
If a number is divisible by 2 without a remainder (i.e. the remainder is zero), then it is even.
And we'll increase the even digit counter (static variable even) by 1.
Otherwise, the number is odd, so we'll increase the odd digit counter (static variable odd).
Display the following message: "Even: a Odd: b", where a is the number of even digits and b is the number of odd digits.

Example for 4445:
Even: 3 Odd: 1

Requirements:

The program must read data from the keyboard.
The main method should calculate how many even digits are in the entered number and then write this number to the variable even.
The main method should calculate how many odd digits are in the entered number and then write this number to the variable odd.
The program should display text on the screen.
The displayed text must match the task conditions.

package com.codegym.task.task06.task0606; import java.io.*; /* Even and odd digits */ public class Solution { public static int even = 0; public static int odd = 0; public static void main(String[] args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); while(true) { int a = Integer.parseInt(reader.readLine()); if ((a %10)%2 == 0 && a>0) { even++; } else { odd++; } break; } System.out.println("Even: " + even + " Odd: " + odd); } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

12 February 2019, 04:00

You need to count the number of even digits and the number of odd digits in the one number the user enters. First you want to get the user input outside of the loop, because you only want one user input. Next you want to write your loop code so that does this: A loop that continues while the user input is greater than 0{ check if the user input mod 2 is equal to 0 If it is then add one to even Else add one to odd Set the user input to the value of user input divided by ten } The above is code spelled out in English. Try to rewrite it in Java. If you get it, it will pass verification. Good luck!

hidden #10424566

Level 7

13 February 2019, 14:53

Thanks a lot!!