Seems to be the right answer but...

Question about the task

Even and odd digits

Java Syntax, Level 6, Lesson 5

Under discussion

I'm getting the right answer but not fulfilling the requirements somehow. What am I missing?

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; public static int odd; public static void main(String[] args) throws IOException { //write your code here int c; int d; int e; int f; int g; BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); int x = Integer.parseInt(br.readLine()); if(x%2==0){ c = x/10000; d = (x-10000*c)/1000; e = (x-10000*c-1000*d)/100; f = (x-10000*c-1000*d-100*e)/10; g = (x-10000*c-1000*d-100*e-10*f); } else { return; } if (c%2==0){ even++; } else { odd++; } if (d%2==0){ even++; } else { odd++; } if (e%2==0){ even++; } else { odd++; } if (f%2==0){ even++; } else { odd++; } if (g%2==0){ even++; } else { odd++; } System.out.println("Even: "+even+" Odd: "+odd); } }

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David Ogundipe

Level 19 , Lagos, Nigeria

21 August 2020, 21:06

Follow this algorithm.

int n = read int from console
int even, odd
while true
    int t = n % 10
    n /= 10
    if t % 2 == 0
        even++
    else
        odd++
    if n == 0
        break
print(even odd)

Guadalupe Gagnon

Level 37 , Tampa, United States

21 August 2020, 02:20

Line 26 checks if the input number is even and if it is not then the program ends. What about all the odd input numbers? Also, the program seems to only work when the entered number has 5 digits or less. What if I entered a number such as 58925843? Or 589258431?