Divisibility by 3

[briankiprono]

I'm trying to write a code for an integer n
a) n is less than 500

b) n is positive

c) n is divisible by 3

but the result for divisibility by 3 is not showing when I run it.
Here is my code.

#include <cstdlib>
#include <iostream>

using namespace std;

int main(int argc, char** argv)
{
int n,x,y;
cout <<"input the value of n"<<endl;
cin >>n;
x=500-n;
if(x>0)
cout<<"the value of n is less than 500 "<<endl;
if(n>0)
cout<<"the value of n is positive"<<endl;
y=n%3;
if(y=0)
cout<<"the value of n is divisible by 3"<<endl;
return 0;
}

Am I missing something?

[JLBorges]

if(y=0) // y=0 assigns the value 0 to y
    cout<<"the value of n is divisible by 3"<<endl;

should be

if(y==0) // y==0 tests if y is equal to 0
    cout<<"the value of n is divisible by 3"<<endl;

#include <iostream>

int main()
{
    int n ;
    std::cout << "input the value of n: " ;
    std::cin >> n;

    if( n < 500 ) std::cout << "the value of n is less than 500\n" ;

    if( n > 0 ) std::cout << "the value of n is positive\n" ;

    if( n%3 == 0 ) std::cout << "the value of n is divisible by 3\n" ;
}

[briankiprono]

Thanks, understood and it worked great.
and I always start with the using namespace std to avoid defining them over again

[JLBorges]

It's ok as long as you do not write a using namespace directive (at global scope) in a header file or before an #include

Ideally, it is best when the using namespace directive is at function local scope.

#include <iostream>

int main()
{
    using namespace std ; // this won't affect the meaning of the code
                          // in other functions that we may use later

    int n ;
    cout << "input the value of n: " ; // within this function, we can still use
    cin >> n;               // unqualified names for entities from namespace std

    if( n < 500 ) cout << "the value of n is less than 500\n" ;

    if( n > 0 ) cout << "the value of n is positive\n" ;

    if( n%3 == 0 ) cout << "the value of n is divisible by 3\n" ;
}

[jonnin]

inefficiently but interestingly if the sum of the digits are divisible by 3, so is the number. Useful for humans, but not really in code.

[briankiprono]

yeah could work but that will be another code to add individual values in a number