### need help in counting the elements how to make it count the even digits. and am I doing anything wrong ?

Scanner input = new Scanner (System.in);

int n; int even = 0;

System.out.print(" Enter a number with four digits: ");
n = input.nextInt();

if (n < 1000 ) {
System.out.print("number is less than 4 digits");}
else if (n > 9999){
System.out.print(" number is more than 4 digits");
}
else {
for(int i = 0; i < n ; i++) {
if((n%10)%2==0)
even++;

System.out.print("There’s" + even + "even number (s).");
}
}

}
}
Last edited on What language is that? Looks like Java to me. yes it's java am I doing anything wrong ?

Might be more accurately written as "am I doing anything right?"

I'm pretty sure that if he/she puts 1000 in the result will be 1000.

But I've forgotten how to compile java. the problem is that I need to get how many even digits there is in a 4 digit integer and am I doing anything wrong ?
like posting java code in c++ forums? :P You may get some help anyway, but you would do better in ... a java forum...

also that for loop is checking N%10 every time.
I think you want to change that 10.
something like
for(stuff)
{
even+= 1 - n%2; //even, n%2 is 0, adds 1. odd, n%2 is 1, adds 0.
n = n%2; //if you need n later, this should be a copy of n...
}

or alternately try /1, 10,100 (start at 1, multiply by 10 each time).

I am not a chemist. Yet I constantly have to deal with elements and solutions. Does that seem right to you?
Last edited on ``1234567891011`` ``````#include using namespace std; int countEven( int n ) { return ( n >= 10 ? countEven( n / 10 ) : 0 ) + 1 - n % 2; } int main() { int n = 0; while ( n < 1000 || n > 9999 ) { cout << "Input n: "; cin >> n; } cout << "Number of even digits = " << countEven( n ); }`````` another topic can you give me head start to this one

Write a program that count until we got 5 numbers multiple of 3 (the program should ask the user to enter integer until we got 5 integers that are multiple of 3):

also in java also in java

You do know that this site is www.cplusplus.com, don't you? not really i thought it's just a name I"m sorry I'll disappear Oh, come back, Amin96!

C++ is a miles better language than Java! Don't go over to the dark side! i'll be back but it's for my university `if((n%10)%2==0) ` is checking if something is divisible by 20. Maybe like this:
 ``1234567891011121314151617181920212223242526272829303132`` ``````import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner input = new Scanner (System.in); int num, num_even = 0; System.out.print(" Enter a number with four digits: "); num = input.nextInt(); if (num < 1000 ) { System.out.print("number is less than 4 digits"); return; } else if (num > 9999) { System.out.print(" number is more than 4 digits"); return; } while (num > 0) { int last_digit = num % 10; if (last_digit % 2 == 0) num_even++; num = num / 10; } System.out.printf("There’s %d even number(s).", num_even); } }`````` Write a program that count until we got 5 numbers multiple of 3 (the program should ask the user to enter integer until we got 5 integers that are multiple of 3)

Forget about Java coding for a minute and concentrate on the process, using pseudocode

1. set a counter to zero
loop:
2. Enter a number
3. If the number is divisible by 3? (i.e. n%3 is zero), then add 1 to the counter
5. If the counter is < 5 loop back for another number, otherwise ...
...
6. ... THE END

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