Thread: How to check if a multiple of 3?

If the number is divisible by three then it is a multiple of 3. Try testing to see if when the number is divided by 3 there is a remainder using the modulus operator.

Code:

#include <iostream>

using std::cout;
using std::endl;

int main(void)
{
	int foobar = 10;

	if ((foobar % 3) == 0)
		cout << "foobar is a multiple of 3" << endl;
	else
		cout << "foobar is not a multiple of 3" << endl;

	return 0;
}

Sorry, but I just have to say this. This thread seems may too easy

Code:

int foobar;
if(!(foobar % 3))
	cout << "foobar is a multiple of 3" << endl;
else
	cout << "foobar is not a multiple of 3" << endl;

Is that above solution more complicated than it has to be? Yes. Is that just the way I am? Yes.

Another way to do this. If a number is divisible by three, the sum of its digits is divisible by three. ex: 121233666995123172990021 is divisible by three, and I know this rather easily by just adding up the digits. You could read in a number as a char array, convert it to an int array, then add up the numbers. You may be thinking "but, then, you'd still have to use the modulus operator!". Not true. If you really wanted a difficult, non-elegant solution, you could do this:

Code:

while(foobar >= 0)
{
	foobar -= 3;
}
if(!foobar)	// if(foobar == 0)
	cout << "foobar is a multiple of 3!" << endl;
else
	cout << "foobar is not a multiple of 3." << endl;

Can you tell I'm in favor of the contest idea to obfuscate a simple problem as much as possible? Oh my, I am bored. I need to find something better to do. *walks away*

Code:

#include<iostream>
int main
(){int


foobar;     std::cout<<

"foobar is "


<<
(foobar%3?
   "not ":"")
            <<"a multiple of 3."    
      
       <<       std::endl; return
            
       0;
}

pwned

For additional speed, you could use a look-up table, i.e. precompute the values.

Code:

int main()
{
  bool* Div3Array = new bool[std::num_limits<int>::max()];
  for (int i=0;i<std::num_limits<int>::max();++i)
    Div3Array[i] = i%3==0; //or another algorithm
  
  //Now we can use the array to determine if an arbitrary
  //number is divisible with 3
  int num;
  cin >> num;
  cout << Div3Array[i];
}

This solution might be a little memory-intensive, though.