That sounds tricky!

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Let $N_1$ be a subset of $N$ .The only difference between $N$ and $N_1$ is that every element of $N_1$ is not divisible by either $2$ or $3$ .If Lawrence decides to write down the numbers of $N_1$ in increasing order what would be the number in the $1999$ -th position?

The answer is 5995.

1 solution

Prasun Biswas
Feb 3, 2014

I used Turbo C++ to solve this and here is my code :

#include<iostream.h>
#include<conio.h>
void main()
    {
    long p,i=1;
    clrscr();
    for (p=1;p>0;p++)
        {
        if(p%2==0||p%3==0)
            continue;
        else
            {
            if(i==1999)
                {
                cout<<"Answer= "<<p;
                break;
                }
            }
        i++;
        }
    getch();
    }

That sounds tricky!

The answer is 5995.

1 solution

I used Turbo C++ to solve this and here is my code :

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