6 replies to this topic

[bonanova]

In a room are 100 light bulbs, numbered 1 to 100.
Outside the room are 100 switches, numbered 1 to 100. Each switch controls its same-numbered bulb.
Initially, all the switches are off.

100 people, numbered 1 to 100, are asked to flip some or all of the switches.
"Flip" means "change its state" - if it's on, turn it off; if it's off, turn it on.

Person 1 must flip every switch: 1, 2, 3, ..., 97, 98, 99, 100.
Person 2 must flip every 2nd switch: 2, 4, 6, ..., 96, 98, 100.
Person 3 must flip every 3rd switch: 3, 6, 9, ..., 93, 96, 99.
------
Person 15 must flip every 15th switch: 15, 30, 45, 60, 75, 90.
------
Person 99 must flip every 99th switch: 99.
Person 100 must flip every 100th switch: 100.

When each person has flipped his/her assigned switches, s/he leaves.
Finally, you enter the room to check the bulbs.

Assuming the 100 people did as they were instructed:

[1] are any bulbs lit?
[2] if not, why not?
[3] if so, which one[s]?

Spoiler for Hints

There is an easy and a hard way to solve.
What did person 1 do that no one else did?
Pick a bulb. Which person[s] flipped it?

[KamZhiYhi]

I can say for sure that light bulb number 1 is on. But im not sure, or just not bothered to try to work out the rest

[unreality]

In a room are 100 light bulbs, numbered 1 to 100.
Outside the room are 100 switches, numbered 1 to 100. Each switch controls its same-numbered bulb.
Initially, all the switches are off.

100 people, numbered 1 to 100, are asked to flip some or all of the switches.
"Flip" means "change its state" - if it's on, turn it off; if it's off, turn it on.

Person 1 must flip every switch: 1, 2, 3, ..., 97, 98, 99, 100.
Person 2 must flip every 2nd switch: 2, 4, 6, ..., 96, 98, 100.
Person 3 must flip every 3rd switch: 3, 6, 9, ..., 93, 96, 99.
------
Person 15 must flip every 15th switch: 15, 30, 45, 60, 75, 90.
------
Person 99 must flip every 99th switch: 99.
Person 100 must flip every 100th switch: 100.

When each person has flipped his/her assigned switches, s/he leaves.
Finally, you enter the room to check the bulbs.

Assuming the 100 people did as they were instructed:

[1] are any bulbs lit?
[2] if not, why not?
[3] if so, which one[s]?

Spoiler for ...

okay here we go:

[1] yes of course, bulb #1
[2] there IS a bulb flipped on so this question doesnt matter
[3] the actual question... which bulbs are on?

We can do this by solving factors:
Even number of factors: bulb is OFF
Odd number of factors: bulb is ON

FACTORS
*********
1: 1 ODD=ON
2: 1,2 EVEN=OFF
3: 1,3 EVEN=OFF
4: 1,2,4 ODD=ON
5: 1,5 EVEN=OFF
6: 1,2,3,6 EVEN=OFF
7: 1,7 EVEN=OFF
8: 1,2,4,8 EVEN=OFF
9: 1,3,9 ODD=ON
16: 1,2,4,8,16 ODD=ON
25: 1,5,25 ODD = ON
36: 1,2,3,4,6,9,12,18,36 ODD=ON


from this we can devise a few truths:
1) everything is divisible by 1; the "One-Man" flips every switch once. And the man for that number switch flips it- meaning you always have 2 (EVEN=OFF) flips, what matters is what EXTRA factors you have.
2) prime numbers are going to be them and themselves which means they are OFF. 2,3 and 5 are all example prime numbers- divide by themselves and one. All prime numbers are turned OFF. Prime numbers are never squares.
3) it becomes clear that we must find an algorithm to determine the numbers with an ODD number of factors

as youve probably seen from the way my list skipped, i'm going to go ahead and guess ALL WHOLE NUMBER SQURES

1
4
9
16
25
36
49
64
81
100

[Michlips]

Bonanova:

What the HECK is the answer?!?!

[Martini]

Welcome to the boards, Michlips. It's customary for the OP to give people time to figure the answer out and for others to discuss their rationale for their answer and sometimes their rationale for why others are wrong. It makes things a little more interesting. After all, what's the rush?

[bonanova]

You can pose this puzzle with 10 bulbs instead of with 100.
But then it's too easy to grind it out; boring.

You can also pose it with 1000 bulbs.
But life is short, and people with any sense walk away, and miss the fun.
Michlips appears to be in that [sensible <!-- s;) --><!-- s;) --> ] group.

The whole idea is to find the short cut; and grinding it out partially can light the way.
ergo, 100 bulbs is only moderately complex and sucks you in to the search.

KamZhiYhi took the first step and realized bulb #1 would be on - partial credit.
unreality went farther, saw the pattern, and found the answer.

Here's another way to see the solution.

Spoiler for ...

Observation 1. Switches flipped an odd number of times are left ON.
Observation 2. Switch N is flipped once for each of its factors: [1,N], etc.
Observation 3. Factors come in pairs, but for squares one factor is repeated.

Solution: Squares are left ON, since they have an odd number of distinct factors.

Factors of 8: [1,8], [2,4] - Persons 1,2,4 and 8 flipped switch 8. Even number of flips. Light is OFF.
Factors of 9: [1,9], [3,3] - Persons 1,3 and 9 flipped switch 9. Odd number of flips. Light is ON.

[eedray]

pretty cool puzzle