Finding integers that are divisible by 6 or 7 but not both

Question

I am trying to write a program that displays the integers between 1 and 100 that are divisible by either 6 or 7 but not both.

Here is my code:

import acm.program.*;

public class Problem4 extends ConsoleProgram
{
    public void run()
    {
        for (int i = 1; i <= 100; i++)
        {
            boolean num = ((i % 6 == 0) || (i % 7 == 0));

            if (num == true)
            println(i + " is divisible");
        }
    }
}

The above code shows the following answers: 6,7,12,14,18,21,24,28,30,35,36,42,48,49,54,56,60,63,66,70,72,77,78,84,90,91,96,98

Now the bold numbers 42 and 84 are both divisbile by 6 and 7. Now If I change the || to && in the above code, the result shows only 42 and 84.

What change should I do to remove these 2 numbers from the final result?

Answer 1

you have to make your condition look like:

boolean num = (i % 6 == 0 || i % 7 == 0) && !(i % 6 == 0 && i % 7 == 0);

that's basically converting "but not both" to Java code :)

Answer 2

XOR is the way to go.

import acm.program.*;

public class Problem4 extends ConsoleProgram {
  public void run() {
    for (int i = 1; i <= 100; i++) {
      if ( (i % 6 == 0) ^ (i % 7 == 0) ) {
        println(i + " is divisible");
      }
    }
  }
}

Answer 3

You need an extra check for "but not both". I think it should be:

boolean num = ((i % 6 == 0) || (i % 7 == 0)) && (i % 42 != 0);

Answer 4

you can also try

boolean num = ((i % 6 == 0) != (i % 7 == 0));

Answer 5

Think about what it means to be divisible by 6 and 7... the answer to life the universe and everything.

Answer 6

import acm.program.*;

public class Problem4 extends ConsoleProgram
{
    public void run()
    {
        for (int i = 1; i <= 100; i++)
        {
            boolean num = ((i % 6 == 0) || (i % 7 == 0));
            boolean both = ((i % 6 == 0) && (i % 7 == 0));

            if ((num == true) &&  (both == false))
            println(i + " is divisible");
        }
    }
}

Answer 7

Here is a snippet that should work as well- in C++ but change to boolean...

int value;
if ((value % 6 == 0 && value % 7 != 0) || (value % 6 != 0 && value % 7 == 0))
    cout << "Is " << value << " divisible by 6 or 7, but not both? true" << endl;
  else
    cout << "Is " << value << " divisible by 6 or 7, but not both? false" << endl;

Answer 8

A little simplified version

for(int i=1; i<=100; i++) {

            // Either Divisible by 6 or 7 but not both


            if((i%6==0 && i%7!=0) ||( i%7==0 && i%6!=0)) {


                println(i);