c++ that print the list of integers

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.

 

whatz this dude....it is so simple...u could have tried it...

Code:

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if (n%2==0 && n%3==0)
consider the number;
increase the counter;

else
do nothing;
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1.first of all take number from the user
2.then run a loop from 1 to user number
3.then inside loop write the code of if statement that if the remainder of the running loop by 2 and 3 are zero then take the number and add in sum integer otherwise step forward
4.finally print the sum

 

you have to display "counter " out of the loop to display number of integers found

 

int sum,i,n;
cin>>n;
sum=0;
for(i=1;i<=n;i++)
if(i%2==0 && i%3==0)
{
cout<<i;
sum=sum+i;
}
end of for loop
cout<<"sum="sum;
end of program run this one then reply

 

Code:

#include "stdafx.h"
#include "iostream"
#include "stdlib.h"
#include "conio.h"
using namespace std;
void main()
{
	int a;
	static int m=0;
	cout<<"input <n> (integer): ";
	cin>>a;
	cout<<"\n";
	for(int i=1;i<(a+1);i++){
		if(i%3==0){
			if(i%2==0){
				cout<<i<<" ";
				m=m+1;
			}
		}
	}
	cout<<"\n number of them= "<<m;
	cout<<"\n program by AHA";
	getch();
}

good luck

 

Code:

#include <iostream>

using namespace std;

int main()
{
    int n;
    cout<<"Please Enter the maximum Limit(N)\n";
    cin>>n;

    int count=0;

    for(int i=1;i<=n;++i)
    {
        if(i%2==0&&i%3==0)
        {
            count++;
            cout<<i<<" ";
        }
    }
    
   cout<<"\n\nThe number of integers found is "<<count;

return 0;
}