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There is a room with 1000 light switches, numbered 1, 2, 3,
4, ... 1000, all turned off.
Outside the room there are 1000 men, numbered man1, man2,
...man 1000
In order, each man walks into the room and changes the
position of each switch that is a multiple of his number.
That is: man1 flips every switch
man2 flips switches 2, 4, 6, 8 ....1000
man3 flips switches 3, 6, 9, ..... 999
.....
Man 1000 flips switch 1000
After all 1000 men are done, how many switches are on?

Answers were Sorted based on User's Feedback



There is a room with 1000 light switches, numbered 1, 2, 3, 4, ... 1000, all turned off. Outside ..

Answer / rams

The answer is 31. All the perfect squares are on. ie 1,4,9,16,25,36....961(31*31).

Here is the logic.

The first person switches on every thing.

The multiple factors come in pairs, so the changes done by multiple factors has no effect. So all the Switches are on.

The switches are turned off by the person with same number.

The square root switches on the lights for the perfect squares.

Is This Answer Correct ?    9 Yes 0 No

There is a room with 1000 light switches, numbered 1, 2, 3, 4, ... 1000, all turned off. Outside ..

Answer / orenise

The above code does not work, Please ignore it. I will Post
the answer, when i got the correct Code.

Is This Answer Correct ?    2 Yes 0 No

There is a room with 1000 light switches, numbered 1, 2, 3, 4, ... 1000, all turned off. Outside ..

Answer / orenise

The answer is 31. here is the code...

private static void Q1() {
int firstMan = 1;
int lastMan = 1000;
int numOfSwitches = 1000;
boolean[] switches = new boolean[numOfSwitches + 1];
// Print the switches initial state.
System.out.println("switches");
int col = 0;
for (int j = 0; j < switches.length; j++) {
System.out.print("[" + switches[j] + "], ");
if (col >= 10) {
System.out.println();
col = 0;
}
col++;
}

for (int currentMan = firstMan; currentMan <= lastMan;
currentMan++) {
for (int switchNum = currentMan; switchNum <=
lastMan; switchNum++) {
int divids = switchNum % currentMan;
if (divids == 0) {
switches[switchNum] = !switches[switchNum];
}
}
}

// Print the switches final state.
System.out.println("switches");
int col2 = 0, sum = 0;
for (int j = 0; j < switches.length; j++) {
System.out.print("[" + switches[j] + "], ");
if (switches[j]) {
sum++;
}
if (col2 >= 10) {
System.out.println();
col2 = 0;
}
col2++;
}
System.out.println(" sum = " + sum);
}

Is This Answer Correct ?    3 Yes 2 No

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