Member Avatar for arjen

print a list of integers from 1 to N that are both divisible by 2 and 3.and after producing the list, count the numbers of integers found.

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Numbers divisible by both 2 and 3 are exactly the multiples of 6. That means you don't need to test every number with two modulus operations — you can either (A) list multiples of 6 by stepping by 6, or (B) compute the count directly as floor(N / 6). Both are simpler and less error-prone than checking i % 2 == 0 && i % 3 == 0 for every i.

A concise, robust approach: validate the input, then either iterate i = 6; i <= N; i += 6 to print every match, or compute count = N / 6 (integer division) if you only need the quantity. Use a 64-bit integer for large N and handle N < 6 as a no-results case.

#include <iostream>

int main() {
    long long N;
    if (!(std::cin >> N)) return 0;
    if (N < 6) { std::cout << "Count: 0\n"; return 0; }

    long long count = 0;
    for (long long i = 6; i <= N; i += 6) {
        std::cout << i << '\n';
        ++count;
    }
    std::cout << "Count: " << count << '\n';
}

Practical notes: if you prefer the modulus method, be careful to compare to zero (e.g., i % 3 == 0), initialize your counter before use, and advance the loop variable each iteration — mistakes along those lines were the main issues in the earlier posts. correctly pointed out the variable mix-up, and 's suggestion to loop is sound; using the multiple-of-6 trick keeps the loop simple and fast.

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Member Avatar for LevyDee

You have to try first. Forum rules =)

But to get you started, you'll want to use a for loop. Inside your for loop you will do some math to figure if your number is divisible by either 2 or 3, and then have a counter variable for each number that passes the requirments in that iteration of the loop.

I just basically told you exactly what to do, so if you cant figure it out, ask your instructor for more help.

Member Avatar for prvnkmr449 
LOOP num from 1 to N
if(num%2==0 && num%3)
    {
     print num
     counter++
     }
end loop

I hope it will help you
Best Of Luck

Member Avatar for arjen

how about this...

#include <iostream>

using namespace std;

int main()
{
    int n;
    int x = 1;
     cout << "enter the value for n : " << endl;
     cin >> n;

     if(x <= n)
     {
             if(n%2==0 && n%3)
             {
                       cout << n << endl;
                       x++;
             }
     }


     return 0;
     }
Edited by Dani because: Formatting fixed
Member Avatar for arjen 
#include <iostream>

using namespace std;

int main()
{
    int n;
    int x = 1;
     cout << "enter the value for n : " << endl;
     cin >> n;
     
     if(x <= n)
     {
             if(n%2==0 && n%3)
             {
                       cout << n << endl;
                       x++;
             }
     }
                       
 
     return 0;
     }
Member Avatar for prvnkmr449

Try to understand what happen in ur code

#include <iostream>

using namespace std;

int main()
{
    int n;
    int x = 1;
     cout << "enter the value for n : " << endl;
     cin >> n; // n is a specified number of input 
     
     if(x <= n)
     {
             if(n%2==0 && n%3) // u always check n logical error, it must be x
             {
                       cout << n << endl; // same problem here also x in place of n
                       x++;
             }
     }
                       
 
     return 0;
     }

I hope u understand
Best Of Luck

Member Avatar for LevyDee

Remember you also need to loop the code, what your doing now is checking the number 1, once and then the program ends.

while(x <= n)
{
    if(x % 2 == 0 && x % 3 == 0)
       {
            cout << x << endl;
            count++;
       }

    x++;
}

This is basically what your looking for.

Edited by LevyDee because: n/a
Member Avatar for prvnkmr449

Sorry i don't understand what u try to say

while(x <= n) loop up to N
{
    if(x % 2 == 0 && x % 3 == 0) // it will check ur x (n >= x >= 1) if x is 
       {    // divisible by both 2 and 3 then it will print and increment the counter
            cout << x << endl;
            count++;
       }

    x++; //increase the value of by 1
}
Member Avatar for arjen

so what that count for count++ which mean count = count + 1 am i right or wrong?? so in that case count i should declare the count as int count = 0??

Member Avatar for prvnkmr449

YES, U Have declare the count as

int count=0;
Member Avatar for LevyDee

Sorry i don't understand what u try to say

while(x <= n) loop up to N
{
    if(x % 2 == 0 && x % 3 == 0) // it will check ur x (n >= x >= 1) if x is 
       {    // divisible by both 2 and 3 then it will print and increment the counter
            cout << x << endl;
            count++;
       }

    x++; //increase the value of by 1
}

lol

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