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100 doors

(3 votes, average: 2.33 out of 5)

November 30, 2007 by Ankur 3 Comments

There are 100 doors, all closed. In a nearby cage are 100 monkeys. The first monkey is let out, and runs along the doors opening every one. The second monkey is then let out, and runs along the doors closing the 2nd, 4th, 6th,… all the even-numbered doors. The third monkey is let out. He attends only to the 3rd, 6th, 9th,… doors (every third door, in other words), closing any that is open and opening any that is closed, and so on. After all 100 monkeys have done their work in this way, what state are the doors in after the last pass, which doors are left open and which are closed ?

See Solution : 100 doorsHide Solution

Consider door number 56, monkeys will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1 1st monkey will open the door, pass 2 2nd one will close it, pass 4 open, pass 7 close, pass 8 open, pass 14 close, pass 28 open, pass 56 close. For every pair of divisors the door will just end up back in its initial state. But there are cases in which the pair of divisor has same number for example door number 16. 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is repeated because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will be open at the end.

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Comments

  1. Gaurav khurana says

    February 1, 2017 at 5:34 pm

    Here is the javacode to test this

    package javaPractise;

    public class DoorOpen {

    public static void main(String[] args) {
    // TODO Auto-generated method stub

    int length=101;
    int arr[]=new int[length]; // by default all doors are closed . Lets assume

    System.out.println(“0 is closed”);
    System.out.println(“1 is open”);

    for (int i=1;i <length; i++)
    {
    // System.out.println();

    for(int j=i; j<length && j<=100; j=j+i)
    {
    // Lets revert the condition on each pass , if o(closed) then 1(open) and vice versa
    if(arr[j]==0)
    {

    arr[j]=1; // door is open now
    }
    else if(arr[j]==1)
    {
    arr[j]=0; // Door is closed
    }
    }

    }

    System.out.println(" The below doors are open");
    for (int i=1;i <length; i++)
    {
    if(arr[i]==1)
    {
    System.out.println(i);
    }
    }
    }

    }

    Reply
  2. Mary says

    June 27, 2015 at 11:08 am

    1,7,11,13,17,19,23,29,31,37,41,43,47,53,59,,61,67,71,73,79,83,89,97 are open.remaining are closed

    Reply
  3. vinay says

    December 28, 2012 at 12:51 pm

    1,4,9,16,25,36,49,64,81,100 are open.remaining are closed. only these numbers(perfect squares) having odd no.of factors.

    Reply

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