Yep. (Add first week and last week, then multiply by 26.) Your turn again, jajdude.
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Yep. (Add first week and last week, then multiply by 26.) Your turn again, jajdude.
If mick likes, he can find one. I got nothing right now. Can do later with more time to spare.
OK, a cryptogram:
HSI KZD JST HSCDNE SI NDTJE CH ZPP SZE KIOIPX EHTGGIV HSCDNCDA.
Quote:
Originally Posted by billl
I mean - why does that work? How can you possibly know to do that?
I don't know whether you're taking the mick, but in case you're not, imagine having to add the numbers from one to five.Quote:
Originally Posted by prendrelemick
* ** *** **** *****
The middle one is the average of the others. Or, to put it another way, you could take two from the last one, and put it on the first one. And one from the fourth one and put it on the second one. You make them all the same because they 'balance' round the middle one.
*** *** *** *** ***
So the sum is the middle number (3) times the number of instances (5).
Same principle with the suns and planets. Figure out how many there are in the middle one and multiply by the number of instances.
Except 52 is an even number of elements (with no 'middle' integer--26 is actually the last member of the first half), so one has to take the other (similar) route (summing the extremes and dividing by two to find the average).
Of course, 26 would be the middle of 51, not 52.Quote:
Originally Posted by MarkBastable
Oh. Beaten by billl.
Once read a story about Gauss, a great mathematician. Teacher wanted to keep kids busy so he/she told them to add numbers 1 to 100. He solved it real fast. I believe he, at maybe 6 years old, took out 100 and the middle number 50. Then it's 99+1 + 98 + 2 + 97 +3 ... or 49 x 100. Or was it 101 x 50, what ever worked. This one had 1 + 2 + ... +52 , or 53 x 26 (twice) as you noted, but forgot to add the stars.Quote:
Originally Posted by prendrelemick
http://www.jimloy.com/algebra/gauss.htm
zdtdxkr*h?Quote:
Originally Posted by jajdude
Thanks for that everyone. I think I understand.- If you add the extremes and divide by two, that must be the average, provided it is a regular progression. But Billl being Billl, was able to jump a step and just times by 26 to get the answer.
Whoops! Wow! Finally, a wrapping-up and an end to the avalanche of carelessness (dating back at least to the mix-up about the variable Y in jajdudes puzzle from days ago. 10652!).
EDIT: Correction, I wasn't as careless as I just thought about skipping the division, and multiplying by 26. I think...
I have a whopper all loaded up, if you guys really want to deal with it. Definitely requires math, and pencil and paper. I'm taking my time, in case some people are yet to give the cryptogram a fair shot... But raise your objections now, if you must, and I'll try to come up with something else, because you might not like this next one...
:leaving:
EDIT: I've attached the solution to jajdude's cryptogram.
Five Barrels
Goran has been assigned the task of ordering a customized trailer for his company. His company makes really big barrels of something (he isn't sure what, he mostly just uses the internet at his desk), and each barrel is 5 feet in diameter. If the trailers can't be wider than 9 feet wide, what is the minimum length the trailer must be to completely accommodate 5 barrels. (The barrels cannot be stacked or lain on their side.)
Good job bill, will look at your puzzle later. Bit drunk now.
:wave:
There's a coincedence I spent yesterday stacking 5ft round bales into my shed.
That might've made for a better set-up than "Some guy has to do something with barrels full of who knows what," but I'm guessing hay is stacked and stored on its side (if drives in the country are sufficient to reveal how that works.)
That sounds great, though, to be outside (and in a shed) dealing with hay. Grass is always greener, maybe, at least some days, but still, sounds nice. Actually, the grass would have lost its green-ness, in this case, and would most certainly be greener over at the other hypothetical spot... Another nice coincidence there, this time in my application of an idiom. Hmph.
Delivering lambs, though--I think I'd have to get pretty good at it before I stopped dreading that particular task.
Its usually wet and cold Billl.
Mick got the latest puzzle (The Five Barrels).
I've attached the solution that he PM'ed to me.
Ok, enough of all this logic and maths. Its time for some intuition, a little knowledge and possibly a bit of wiki-ing.
A King fisher is flying over the countryside, when it sees a small pond. Next to the pond is a convenient stone cairn. It lands on the cairn and peers into the water. In the pond is a freshwater fish, that is unfortunetly too large for the bird to catch. My question is, how much stone was needed to build the cairn?
Enough stone for the bird to spot a rock perch?
("Perch" has two meanings in this undoubtedly incorrect answer, with one of them being a species of freshwater fish, including something called a "rock perch" which is not actually a perch at all, but is instead a type of small bass.)
Thanks billl, my first clue was going to be that the fish is a perch, and not a red herring. so you're in the right area.
24.75 cubic feet of stone?
http://en.wikipedia.org/wiki/Perch_(unit)#Volume
That is correct. The cairn was a perch, and a perch is just under 25 cubic feet of stone. It's not all that archaic either, a few years ago I did a walling job for a large estate and was paid by the perch.
The dictionary on my computer mentioned it as a unit to measure length, but didn't have the volume definition so I ended up thinking along different lines. So close...
I'll try and come up with something soon, but anyone else with a good puzzle/problem is welcome to jump in...
Ike The Ice Cream Man
Ike the ice cream man has a problem. A bunch of sorority girls just got off at the shuttle bus stop outside his shop, and they are about to come streaming inside his door. Ike can't resist these girls, they get anything they want from him, and he has foolishly left a 750ml bottle of Schnapps that he recently purchased sitting at the side of the counter, in plain view.
The girls love only one thing more than Schnapps, and that's ice cream, and so Ike starts scooping it into cups at the rate of one cup every 10 seconds (the same rate at which the girls are now entering his ice cream shop).
If it takes a girl 60 seconds to finish one cup of ice cream, and 30 seconds to pour and drink a 50ml shot of Schnapps, How long (from the moment he begins scooping) will it be before the sorority girls finish off Ike's Schnapps?
Can I ask, billl - do the girls pour their own schnapps or does Ike stop scooping ice-cream and pour it for them?
Does he keep them supplied with endless cups of ice cream, or does he stop after one each?
They pour their own Schnapps, and he supplies as much ice cream as he can (he's still scooping when the bottle's finished).
I can't see a reason why it isn't 7.5 mins.
I could have maybe done a better job explaining, the situation is this:
Girls are pouring in at a rate of 1 every 10 seconds, and Ike serves each one of them ice cream at the rate of one cup per 10 seconds (which each of the girls are able to finish in 60 seconds). Those who don't have ice cream will do shots of Schnapps, at a rate of 50ml per 30 seconds, and girls without any ice cream to enjoy will share the bottle. (But any girl with an opportunity to have a cup of ice cream will choose ice cream instead of the Schnapps.)
The girls remain in the shop, eating ice cream as quickly as Ike can serve it to them, and so he is busy with a store filled with sorority girls until the following Friday, when a bunch of mandatory keg parties happen.
So he's okay for the first six girls, but when the seventh girl comes in, he has a problem, because the first girl has finished her ice cream, so he has two girls to give ice cream to, and only ten seconds before another girl's finished and an eighth arrives - and every ten seconds that pass, he has another girl he can't keep occupied with ice cream.
The seventh girl in does a shot of Schnapps in thirty seconds, by which time there are ample girls available to drink further Schnapps.
So from the moment he starts scooping it's 1 minute (while he's holding the first six girls off with ice cream) + 750/50*30secs =
8.5 minutes
The only thing that seems odd to me is that you say it takes thirty seconds "to pour and drink a 50ml shot of Schnapps". I'd'a thought that the next one could be poured as the previous one was being drunk, so what we really care about is how long it takes to pour.
Yes, I had intended to convey the idea that more than one girl could (would) be working on the Schnapp's at the same time. Sorry, again, I'm afraid the correct answer requires a little more jumping through mathematical hoops. (I'll look over my notes again, but I don't think there ends up being a line of people waiting at the Schnapps bottle.)Quote:
Originally Posted by MarkBastable
EDIT: to clarify, more than one girl might be drinking Schnapps at the same time, and each of them would be separately progressing at a pace of 50ml per 30 seconds. There are, as it happens, no cases of more than one girl finishing their drinks at the same time and arguing over the next pour, however (but such a thing could happen in slightly different scenarios).
So I've obviously gone wrong somewhere, but this was the way it came out for me.Quote:
Originally Posted by billl
Ike can keep six girls happy with ice cream.
At sixty seconds there'll be seven girls in the store, one of whom'll hit the Schnapps.
At seventy seconds there'll be eight girls in the store - six occupied with ice cream, one with the Schapps and one free. If the Schnapps is free too, she'll hit that - so it depends whether the 'pour' part of 'pour and drink' takes less than thirty seconds. If it takes ten, say, she can hit the Schnapps. If it takes thirty, she has to wait.
By ninety seconds there'll be ten girls in the store - six on ice cream, one starting the second Schnapps (if the thirty-second 'pour and drink' cycles can't overlap) or starting the fourth (if the 'pour' part takes, say, ten seconds), and three hanging about, waiting for some kind of gratification (or two of them could be drinking Schnapps they poured earlier).
But if it's not necessary to know how long the 'pour' bit is, this may be where I've gone wrong - because in my way of working it out, you only need seven girls in the store - six on ice cream (on the 10 and 60 cycle for serve and consume) and one on Schnapps (on the 30 cycle for drink and pour). Any more girls than that are just hanging about. It seems unlikely to me that bill would have set up the problem in this way if he only needed seven girls - so my logic might have gone awry somewhere around there.
Yes, you're right--the problem assumes (but doesn't state) that the pouring time is inconsequential. Unfortunately for me, it turns out that my solution involves more than one girl pouring herself Schnapps at the same time on two occasions (during the 50 second period before the bottle is finally emptied).Quote:
Originally Posted by MarkBastable
Anyhow, a bit of a screw-up there, sorry--but you've really saved the day on this one, Mark. You've nailed the issues down perfectly well, and since the replies haven't been exactly pouring in like sorority girls at a late-night ice cream parlour, I'm going to post the solution here in an attachment.
Too much ice cream and schnapps for me.
Got a good one, Mark?
This one was lifted straight from the 'Net, so it's easy to cheat. I'm going to have only intermittent access to LitNet over the next couple of weeks, so if everyone gets sick of the question, you might all agree to Google it.
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Three men were standing in a row, all facing the same direction, so that there was one in back who could see the two in front of him, one in the middle who could see the guy at the front, and the one in front who could not see either of the other two.
They were shown five hats - three blue and two red.
One hat was placed on each man, without them knowing which, or knowing which two were left over.
First the man in the back was asked if he could deduce what color hat he had on.
"No, I can't," he said.
The man in the middle was asked the same question.
"Nope," he said.
Then man in the front was asked - and he knew what color hat he was wearing.
What was his answer, and why?
I had a beautiful lucid moment just then and worked out it was blue. Trouble is, I can't remember why.
How about this:Quote:
Originally Posted by prendrelemick
The back guy says no, so the guys in front of him can't BOTH be wearing red hats (that would mean only blue ones were left over, so he could deduce "blue").
So the second guy knows that at least one blue must be on either his head or the front guy's head (and maybe both are wearing blue...). If the person in front is wearing blue, then the middle guy can't be sure if he is wearing a blue hat himself--it could be blue OR red. However, if the guy in front of him is wearing a red hat, then he (the guy in the middle) CAN'T be wearing a red hat, because of the situation with the guy in the back (noted above, i.e. he couldn't have seen two red hats without knowing he was wearing blue).
Therefore, since the middle guy can't deduce anything, he must be seeing a blue hat on the front guy. So the front guy deduces that he is wearing a blue hat.
Yep, good. Me and the family are off to NY and the Virgin Islands. Carry on.
Quote:
Originally Posted by billl