a much better logical puzzle

[restoshammyman]

hello everyone.
i have a riddle that will stump you all.
i will never ever ever post the answer, i will however tell you if someone got it right.
so here it goes:

there is a room with 100 light bulbs on the ceiling, and 100 switches on the wall.
each light bulb and each switch is numbered(switch 1 turns on the first bulb, No. 2 turns 2 on, and so on).
there are also 100 goblins.
each goblin is also numbered from 1 to 100 and the number is written on they'r shirts(goblins are retarded, what can you do).
each goblin get he's turn to go and play with the switches(only one turn per goblin).
when a goblin plays with the switches, he only switches a switch if its number can be divided by the number on the goblins shirt and only switch it once(a number CAN be divided by one and itself).

and the question is.
what lights will be on after all the goblin are done.

 

[kommando367]

#1 will be off, I know this for sure. And from what else I can tell,

Spoiler: spoilin'

I have a headache and quit.

.

 

[Tzfanya]

To clarify: does each goblin switch each switch that is a multiple of their shirt number?

So goblin one flicks all the switches once,
goblin two flicks all the even switches once, etc?

 

[About To Crash]

A Goblin will only play with one switch on his turn, or as many as he can?
For instance, would Goblin 1 play with just 1 of the switches, or all of them?

 

[Kpt._Rob]

If I may adapt something Lewis Black said to our purposes here. There is not enough liquor in the universe to get one through the tedious mathematical process that would be necessary to solve this.

 

[Gitsnik]

None of them.

I think this too. But:

If I may adapt something Lewis Black said to our purposes here. There is not enough liquor in the universe to get one through the tedious mathematical process that would be necessary to solve this.

I don't know if my maths is correct:

#!/usr/bin/perl
use strict;
use warnings;
my @switch = ();
foreach (1..100) {
        $switch[$_] = 0;
}
foreach my $goblin (1..100) {
        foreach my $light (1..100) {
                for(my $counter = 2; $counter <= int($light / $goblin); $counter++) {
                        my $val = $goblin / $light;
                        print qq|$val\n|;
                        if($val !~ /\./) {
                                $switch[$light] = ($switch[$light] == 0 ? 1 : 0);
                                $counter = 2000000;
                        }
                }
        }
}
foreach (1..100) {
        print "$_ is on\n" if $switch[$_] == 1;
}

Edit:

Oh I don't think it was, that should have been my $val = $light / $goblin;

If that is the case then we have:

Spoiler

2 is on
3 is on
5 is on
6 is on
7 is on
8 is on
10 is on
11 is on
12 is on
13 is on
14 is on
15 is on
17 is on
18 is on
19 is on
20 is on
21 is on
22 is on
23 is on
24 is on
26 is on
27 is on
28 is on
29 is on
30 is on
31 is on
32 is on
33 is on
34 is on
35 is on
37 is on
38 is on
39 is on
40 is on
41 is on
42 is on
43 is on
44 is on
45 is on
46 is on
47 is on
48 is on
50 is on
51 is on
52 is on
53 is on
54 is on
55 is on
56 is on
57 is on
58 is on
59 is on
60 is on
61 is on
62 is on
63 is on
65 is on
66 is on
67 is on
68 is on
69 is on
70 is on
71 is on
72 is on
73 is on
74 is on
75 is on
76 is on
77 is on
78 is on
79 is on
80 is on
82 is on
83 is on
84 is on
85 is on
86 is on
87 is on
88 is on
89 is on
90 is on
91 is on
92 is on
93 is on
94 is on
95 is on
96 is on
97 is on
98 is on
99 is on

Again, not sure if that's correct.

 

[restoshammyman]

To clarify: does each goblin switch each switch that is a multiple of their shirt number?

So goblin one flicks all the switches once,
goblin two flicks all the even switches once, etc?

yes.
English isn't my first language, and i had trouble finding the right words to say just that.

 

[restoshammyman]

A Goblin will only play with one switch on his turn, or as many as he can?
For instance, would Goblin 1 play with just 1 of the switches, or all of them?

he plays with all the ones he should on hes turn.
not just one

 

[restoshammyman]

none of them.

wrong

 

[restoshammyman]

None of them.

I think this too. But:

If I may adapt something Lewis Black said to our purposes here. There is not enough liquor in the universe to get one through the tedious mathematical process that would be necessary to solve this.

I don't know if my maths is correct:

#!/usr/bin/perl
use strict;
use warnings;
my @switch = ();
foreach (1..100) {
        $switch[$_] = 0;
}
foreach my $goblin (1..100) {
        foreach my $light (1..100) {
                for(my $counter = 2; $counter <= int($light / $goblin); $counter++) {
                        my $val = $goblin / $light;
                        print qq|$val\n|;
                        if($val !~ /\./) {
                                $switch[$light] = ($switch[$light] == 0 ? 1 : 0);
                                $counter = 2000000;
                        }
                }
        }
}
foreach (1..100) {
        print "$_ is on\n" if $switch[$_] == 1;
}

Edit:

Oh I don't think it was, that should have been my $val = $light / $goblin;

If that is the case then we have:

Spoiler

2 is on
3 is on
5 is on
6 is on
7 is on
8 is on
10 is on
11 is on
12 is on
13 is on
14 is on
15 is on
17 is on
18 is on
19 is on
20 is on
21 is on
22 is on
23 is on
24 is on
26 is on
27 is on
28 is on
29 is on
30 is on
31 is on
32 is on
33 is on
34 is on
35 is on
37 is on
38 is on
39 is on
40 is on
41 is on
42 is on
43 is on
44 is on
45 is on
46 is on
47 is on
48 is on
50 is on
51 is on
52 is on
53 is on
54 is on
55 is on
56 is on
57 is on
58 is on
59 is on
60 is on
61 is on
62 is on
63 is on
65 is on
66 is on
67 is on
68 is on
69 is on
70 is on
71 is on
72 is on
73 is on
74 is on
75 is on
76 is on
77 is on
78 is on
79 is on
80 is on
82 is on
83 is on
84 is on
85 is on
86 is on
87 is on
88 is on
89 is on
90 is on
91 is on
92 is on
93 is on
94 is on
95 is on
96 is on
97 is on
98 is on
99 is on

Again, not sure if that's correct.

so freaking wrong. check your program. to bad i only know pascal and javascript

 

[waggmd]

I think the only one's left on are the lights which are perfect squares.

 

[Gitsnik]

so freaking wrong. check your program. to bad i only know pascal and javascript

It's the logic not the program

Teach me to write code when I should be working.

Edit:

Well there you go, change that 2 for a 1 in the loop:

Spoiler

1 is on
4 is on
9 is on
16 is on
25 is on
36 is on
49 is on
64 is on
81 is on
100 is on

Personally, I don't even know if that is right or not (still) - but there it is.

 

[restoshammyman]

I think the only one's left on are the are lights which are perfect squares.

wow that was fast.
we got a guy with a working brain over here.

 

[restoshammyman]

None of them.

I think this too. But:

If I may adapt something Lewis Black said to our purposes here. There is not enough liquor in the universe to get one through the tedious mathematical process that would be necessary to solve this.

I don't know if my maths is correct:

#!/usr/bin/perl
use strict;
use warnings;
my @switch = ();
foreach (1..100) {
        $switch[$_] = 0;
}
foreach my $goblin (1..100) {
        foreach my $light (1..100) {
                for(my $counter = 2; $counter <= int($light / $goblin); $counter++) {
                        my $val = $goblin / $light;
                        print qq|$val\n|;
                        if($val !~ /\./) {
                                $switch[$light] = ($switch[$light] == 0 ? 1 : 0);
                                $counter = 2000000;
                        }
                }
        }
}
foreach (1..100) {
        print "$_ is on\n" if $switch[$_] == 1;
}

Edit:

Oh I don't think it was, that should have been my $val = $light / $goblin;

If that is the case then we have:

Spoiler

2 is on
3 is on
5 is on
6 is on
7 is on
8 is on
10 is on
11 is on
12 is on
13 is on
14 is on
15 is on
17 is on
18 is on
19 is on
20 is on
21 is on
22 is on
23 is on
24 is on
26 is on
27 is on
28 is on
29 is on
30 is on
31 is on
32 is on
33 is on
34 is on
35 is on
37 is on
38 is on
39 is on
40 is on
41 is on
42 is on
43 is on
44 is on
45 is on
46 is on
47 is on
48 is on
50 is on
51 is on
52 is on
53 is on
54 is on
55 is on
56 is on
57 is on
58 is on
59 is on
60 is on
61 is on
62 is on
63 is on
65 is on
66 is on
67 is on
68 is on
69 is on
70 is on
71 is on
72 is on
73 is on
74 is on
75 is on
76 is on
77 is on
78 is on
79 is on
80 is on
82 is on
83 is on
84 is on
85 is on
86 is on
87 is on
88 is on
89 is on
90 is on
91 is on
92 is on
93 is on
94 is on
95 is on
96 is on
97 is on
98 is on
99 is on

Again, not sure if that's correct.

so freaking wrong. check your program. to bad i only know pascal and javascript

It's the logic not the program

Teach me to write code when I should be working.

i looked at your results.
they are backwards. your printing the lights that are off.

 

[waggmd]

I think the only one's left on are the are lights which are perfect squares.

wow that was fast.
we got a guy with a working brain over here.

Actually I just helped my sister with this problem not too long ago. It was part of her Grade 8 math assignment.

 

[Gitsnik]

i looked at your results.
they are backwards. your printing the lights that are off.

My divisor was off by one I think. See the edit to the post you quoted.

 

[dragontiers]

Um, quick question. Which lights, if any, were on before the goblins started throwing switches? That is a major factor in the equations.

 

[About To Crash]

I think that any number that has an odd number of different roots will be on. That's because it will be flicked an odd number of times, and end up in the on position.
Any number with an even number of different roots will be off, because it will be flicked an even number of times.

1, 4, 9, 24, etc.

I think.

And I'm assuming (Though this wasn't noted) that all the lights start off. If they do, groovy. If not, balls...

EDIT: Nope, nevermind, I'm way off.

 

[restoshammyman]

so freaking wrong. check your program. to bad i only know pascal and javascript

It's the logic not the program

Teach me to write code when I should be working.

Edit:

Well there you go, change that 2 for a 1 in the loop:

Spoiler

1 is on
4 is on
9 is on
16 is on
25 is on
36 is on
49 is on
64 is on
81 is on
100 is on

Personally, I don't even know if that is right or not (still) - but there it is.

100 is not on.
do you notice a pattern in the other ones?

 

[restoshammyman]

Um, quick question. Which lights, if any, were on before the goblins started throwing switches? That is a major factor in the equations.

they all start off.
i spent an hour idiot proofing and i missed that.