Another mathematical problem or two, because I don't have time to solve that other one sadly.
- A hunter leaves his camp and goes 1km south. He comes across a bear. He runs 1km east, and, having escaped the bear, walks 1km north and gets back to his camp. What colour is the bear?
- A viking father has 3 sons, and 11 axes. He wants to give his first son 1/2 the axes, his second son 1/4 of the axes, and his third son 1/6 of the axes. How many axes to each of the sons get? _________________
Location: Somewhere between insanity and logical resoning.
Posted: Sun Dec 07, 2008 11:43 am Post subject:
The first one is not a problem, it is a riddle. The only place that this can be true is the north pole, so the bear is white, as it is a polar bear.
To solve the second one you must first realise that 1/2+1/4+1/6 equals less than one. In fact, it equals 11/12. So I used the number 12 instead of eleven. the first son gets six axes. The second son gets three axes. The third son gets two axes. Six plus three plus two equals eleven. Eleven axes are distributed.
Does anyone know the equasion for finding the mass and/or the volume of a spere? _________________ Chuck Norris!
Ah, but it doesn't really matter. Riddles count too. You just need to think about it.
Oh, and technically, the distances travelled could work near the south pole, too. But there aren't any bears near the south pole.
As for the sphere... If I can remember it right, it's V = (3/4)π(r^2).
But that's not a puzzle or a riddle, it's just a bit of knowledge. Although I guess you could try to work the formula out from more basic principles.
Of course, if you want the mass, multiply the volume by the density.
Can I hijack it too, then? I have a Linear Algebra paper for tomorrow. It seems rather small, but I just can't get myself to really look at it. Also have two papers in logic/proof theory for tuesday... one to be corrected, another to be written.
[joke]Anyone who wants to make them for me are welcome.[/joke]
No, they give results that won't work.
That is, nonsensical 'times'. Basically, you can exclude 3 and 4 from any potential solution(s) for that puzzle. You want to get two real times as far apart as possible.
Or alternatively, substitutions. Imagine that we used 'h' instead of the number 3. So 3 and 4 turn into each other when rotated...
I'll have to work on those, though. For now, can you get it as written?
Hmm. I think there's another problem with it the way you've stated it.
If you have the time ab:cd, then the rotated one would be dc:ba. But since for example d in the first one can be any number between 0 and 9, the second time could end up becoming something like 9x:xx. That's not a valid time either. Do we rule those out as well? _________________ Character Descriptions | deviantArt
Yup. Only regular 24-hour times, with real numbers, are valid. Anything between 00 and 23 hours, and between 00 and 59 minutes. That should help narrow down the possibilities for you.
In fact, most of the problem is figuring out what combinations are possible.
I should probably mention that I thought this up while looking up at the ceiling while my projector-clock was switched on. It was upside down, but I didn't realise it until the minutes changed on the wrong side.
Ok, no problem, my numbers should be correct, then.
And, as far as I can tell, that gives a latest time of 22:51 for time B. Anything later is apparently not possible. Unless you consider 00:00 the latest. It could be the earliest too, however. _________________ Character Descriptions | deviantArt
No, because that would be 0 hours different. The furthest apart any two times could ever be is 12 hours, because that's when you start coming back the other way.
14 hours is just 10 hours in the other direction.
Well, it really does depend on whether you can count in both directions. If you can only count one direction, the furthest apart would be 24 hours, or atleast 23 and 59 minutes depending on how you treat 00:00
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