MrRubix's Riddle Thread

[FictionJunction]

Re: MrRubix's Riddle Thread

maybe I'm going too far into this but all 60 minutes are represented on the clock. That said, when we use all minute markers to measure the position of the hour hand as it goes from, let's say, 12 to 1, it would take 12 minutes for it to reach each one of those markers. So, he'd be able to tell times such as 12:00, 12:12, 12:24, 12:36, 12:48, and 1:00. 12:15, 12:30, and 12:45 would actually be harder to tell if the clock doesn't move its hands at a continuous speed and moves from marker to marker when it is designated to do so.

I'm bored at work.

[s0ulst0n3]

Re: MrRubix's Riddle Thread

22 times. Every hour you can't notice it 2 times, and you don't count a 13th hour. The 13th hour would include the 2 extra needed for the 24. I may have to explain at which point. If I do, quote me and ask me to. I don't want to bother explaining if it's wrong.

[Cyanite]

Re: MrRubix's Riddle Thread

You guys need to stop focusing on overlapping hands and focus more on positions that occur twice within a 12 hour period. The 12:00 position is one example.

Besides, don't overlapping hands only have one option? I think they overlap somewhere within 3:15 and 3:20, but that occurs twice in a 24 hour period, meaning the only time confusion would be between AM/PM. Unless that creates an acceptable answer...

[AznGreen]

Re: MrRubix's Riddle Thread

Come on guys, you can do it :P

i beg to differ XD. seeing how over 10 attempts at this isn't getting anywhere.

[s0ulst0n3]

Re: MrRubix's Riddle Thread

You guys need to stop focusing on overlapping hands and focus more on positions that occur twice within a 12 hour period. The 12:00 position is one example.

Wrong. NO TIME overlaps once within a 12 hour period. There are 720 minutes in a 12 hour frame. When you start at 12:00, that counts as the first minute, meaning 10 minutes into the first hour would be 12:11. The 720th minute of the 12 hours ends at 11:59, where it begins the next hour at 12:00.

Your theory is flawed. Or maybe mine, I dunno I'm tired.

[MrRubix_MK5]

Re: MrRubix's Riddle Thread

Guys, if the hands overlap, you know what time it is. How are you possibly going to get a different value for your hour and minute hands if they point to the same spot?

The kind of times you are looking for here are ones where the hands could literally represent two different times, even when you account for the fact that the hour hand moves along with the minute hand. What makes the riddle tricky is how you choose to deal with the angles involved, and how to reconcile the fact that different hours begin at different spots on the clock.

[PuRoDoMiNiCaNo]

Re: MrRubix's Riddle Thread

Well... messing around on with the clock on my PC...7:33 and 6:37 can be confused,but it doesn't flow, it clicks =(. Leaves lotsa room for mistakes. w/e angle i just got..ill try to apply it to other times to see if that works..be back in a bit

[emerald000]

Re: MrRubix's Riddle Thread

132 times? I got the reasoning behind it, so tell me it is correct before saying how I got it.

[MahouMinachan]

Re: MrRubix's Riddle Thread

so far the farthest I've gotten in this problem is

If 6m2=.5h1
then .5h2=6m1

[MrRubix_MK5]

Re: MrRubix's Riddle Thread

Emerald's got it. Very elegant solution too with hardly any math

[Fractal_Monkey]

Re: MrRubix's Riddle Thread

Damnit. As soon as I saw 132 I realised what I'd done wrong. The interchangeable times are the 142 I got in my previous post, minus the 10 where the hands overlap, which I forgot to remove.

[s0ulst0n3]

Re: MrRubix's Riddle Thread

Damnit. As soon as I saw 132 I realised what I'd done wrong. The interchangeable times are the 142 I got in my previous post, minus the 10 where the hands overlap, which I forgot to remove.

144-12, actually.

Pfft...I was pretending to be dumb, I KNEW IT ALL ALONG! I just wanted to see who else would figure it out. xD

[JenovaSephiroth]

Re: MrRubix's Riddle Thread

I found the solution but it requires a lot of graphs in MSPaint to show what goes on that I'm too lazy to make. =(

I'll try to show you anyway, though, once I have the pictures ready. ;D

[emerald000]

Re: MrRubix's Riddle Thread

I guess I will post mine. It is 132 times, for those wondering.

In order to put it as simpler as possible, we'll start with the number of times in the 12x range. We need to find times when we swap the hands, it is a correct time.

The first and easiest one is the 12:00. The only problem is that we know which one it is, because the hands are overlapped. So we don't count this one.

Then, let's analyze the 12:05 to 12:10 range. Since the minutes hand is between the 1 and the 2, the inverted time will have an time of 1x, and the inverted minutes will be around 6-7. Note that if you take all the possibilities of the normal time (the infinite number of them), you will see that the inverted hour hand will take every position between the 1 and the 2. That way, it is sure to be in the good place once in its travel.

You can apply the same reasoning to every other 5 minutes on the clock, except one, which will be when the hands are overlapped. You can also apply that reasoning to the other 12 hours of the clock.

So, that gives: 12 hours x 11 times per hour = 132 times. (Note that the "11 times per hour" comes from the twelve 5-minutes in an hour minus the 5-minute in which the hands are overlapped)

Bow down to my superior intelligence.

PS: I would like to see your solution Rubix.

[ieatyourlvllol]

Re: MrRubix's Riddle Thread

@Fractal: Yeah, you got the cyclical interval correct but forgot to subtract 11 from the coincidental positions [which fall at h:h(60/11)].

I was going to post an answer earlier, but you were close enough that it would have spoiled it for you. I think Emerald beat me to solving it, though.

A simple way to think of it is by separating the time frames of both hands into independent objects. Doing so yields twelve hours with twelve corresponding values (the count of which is tied to the clock parameters) whose angles/distances will be the same, resulting in 144 possible values. Remove 12:00 once (since 12 hours is singularly inclusive), then omit the 11 unique overlapping times to end up with 132.

EDIT: Ninja'd by Emerald