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Thread: Gom's Weekly Test | This thread is pages long: 1 2 3 4 · «PREV / NEXT» |
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angelito
Honorable
Undefeatable Hero
proud father of a princess
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posted July 10, 2006 11:01 PM |
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There is a logical problem Russ.
Different letters in maths always have different values...
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Gom_Jabbar
Promising
Famous Hero
Revealer of Truth
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posted July 10, 2006 11:25 PM |
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Quote: It does not break any laws of logic either.
You are perfectly right. There are many ships that can carry 4752855 passangers.
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Gnoll_Mage
Responsible
Supreme Hero
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posted July 12, 2006 08:07 PM |
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Quote:
quote:
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You have 2 sticks of incense. Each burns for exactly one hour. However, the sticks are not uniformally distributed (half a stick does NOT equal half an hour). Using these two sticks of incense, how can you tell when exactly 45 mins has past?
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You can't. Plus don't you have a watch, dear ? lmao
Seriously I have no idea. Can I have the solution ?
Can I have it too please? I can't work that one out, it's really frustrating
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Gom_Jabbar
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Revealer of Truth
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posted July 12, 2006 10:26 PM |
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Quote:
Can I have it too please? I can't work that one out, it's really frustrating
Ok... I'll tell you the answer(since this isn't my riddle anyway ):
Step one: you light one stick at one end, and the other at both ends
Step two: when the second stick is out(30 minutes) you light the other end of the first stick; the first stick will now have only 15 minutes untill is out.
So: 30+15=45 minutes
And now everyone get to work on my riddle
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Gom_Jabbar
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Revealer of Truth
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posted July 14, 2006 10:42 AM |
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Helooooo... is there noone here that can solve the ridle?
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angelito
Honorable
Undefeatable Hero
proud father of a princess
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posted July 14, 2006 02:56 PM |
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Quote: Helooooo... is there noone here that can solve the ridle?
Would "Titanic" be a good hint?
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Better judged by 12 than carried by 6.
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Xarfax111
Bad-mannered
Supreme Hero
The last hero standing
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posted July 15, 2006 12:31 AM |
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Quote: Test No.2 :
"A ship had X ship funnels, Y propelers and T passangers. It lifted its anchor on the day N of the month P of the year 1900+Z. If you multiply this numbers (X*Y*T*N*P*Z) and add the square root of the captain's age you end up with the number 4752862.
Im not good at riddles..but lets see if we can near this:
Lets find out about the maximum value things can have...start with
the easy ones:
N max 31
P max 12
Z max 99
The Captains age can be:
25 (5x5)
36 (6x6)
49 (7x7)
64 (8x8)
81 (9x9) which is unlikely
so the numbers to add at the end can be 5, 6, 7, 8 (or 9).
Lets assume that all numbers are at value 10:
X*Y*T*N*P*Z = 10 high 6 = 1.000.000
With 4.752.862 at the end the average values cant be that high, so at least Z must be lower then approx. 60.
Lets get back to the captains age:
Can a multiplication of several numbers end with a 3? 4? 5? 6? 7?
...erm i will think about it next time
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Xarfax111
Bad-mannered
Supreme Hero
The last hero standing
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posted July 15, 2006 03:19 AM |
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Quote: Would "Titanic" be a good hint?
Titanic leaving Belfast for sea trials, April 2, 1912.
Owners: White Star Line
Builders: Harland and Wolff yards in Belfast, Ireland
Port of registry: Liverpool, United Kingdom
Laid down: March 31, 1909
Launched: May 31, 1911
Christened: Not christened
Maiden voyage: April 10, 1912
Fate: Hit an iceberg at 11:40 P.M on April 14, 1912. Sank on April 15, 1912; wreck discovered in 1985.
General Characteristics
Gross Tonnage: 46,328 GRT
Displacement: 52,310 Long Tons
Length: 882 ft. 9 in. (269 m)
Beam: 92.5 feet (28 m)
Power: 29 boilers. Two four cylinder triple expansion reciprocating engines each producing 16000 hp (12 MW) for outer two propellers. One low pressure steam turbine producing 18000 hp (13.5 MW) for the center propeller. Total 50,000 hp (37 MW)
Propulsion: Two bronze triple blade side propellers. One bronze quadruple blade central propeller.
Speed: 23 knots (42.5 km/h) (26.4 mi/h)
Number of Passengers (Maiden Voyage): 1912 - Total 2,223 First Class:329 Second Class:285 Third Class:710
Crew:899
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Gom_Jabbar
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Revealer of Truth
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posted July 17, 2006 07:33 PM |
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Nice Titanic thing Xarfax, but no, Titanic is not a good hint.
Were you guys too lazy or was the riddle too hard? I'm speechless. I'll edit the post later when I get home and give you guys the solution, and a brand new riddle.
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Gom_Jabbar
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Revealer of Truth
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posted July 17, 2006 09:44 PM |
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Ohhh my god.... Mea Culpa:
Since not many tried to answer my riddle I only saw now that a slight mistake slipped in. So
Quote: the square root of the captain's age
...is actually the cubic root of the captain's age. For those that do not know what a cubic root is I'll demonstrate:
100*100=10000 ---> square root of 10000=100
100*100*100=1000000 ---> cubic root of 1000000=100
Because of this mistake I'll leave the riddle unanswered untill next Monday.
Test No.3:
"The train was so fast on the wide plain. The view was so dull. As long as you could see there was nothing but grass. The only least dull thing around were the telegraph polles spread along the the railway at equal distances, that seemed to run infront of my window. So, having nothing else better to do, as I kept on looking, I noticed that the number of polles I was seeing in one minute represented exactly the third part of the train's speed(km/h)."
Question:
What is the distance between the polles?
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Iris
Responsible
Supreme Hero
of Typos
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posted July 18, 2006 05:03 PM |
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Gom, the answer to your 2nd test:
The numbers (mix and match at will): 1, 2, 3, 4, 11, 31, 2323
I'll do your other one when I come back from lunch.
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Gom_Jabbar
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Revealer of Truth
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posted July 18, 2006 05:52 PM |
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Quote: Gom, the answer to your 2nd test:
The numbers (mix and match at will): 1, 2, 3, 4, 11, 31, 2323
I'll do your other one when I come back from lunch.
They probably do. But I'll have to ask you to explain your logic. Also don't forget to look at my review of the test(only a few posts above).
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Iris
Responsible
Supreme Hero
of Typos
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posted July 18, 2006 06:11 PM |
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Oops, I guess I should have specified. The captain's age is 4^3 = 64.
What was my logic? Well, by order of operation, I had to figure out the captain's age first. My choices were rather limited:
2^3 = 8 (most likely not)
3^3 = 27 (kind of young to be a captain)
4^3 = 64 (that sounds good, so let's try that)
5^3 = 125 (ehh...)
So now, we have 4752862 - 4 = 4752858
So now, the rest of the numbers are factors of 4752858. Well then, start dividing. Start with 1 and the primes.
4752858/1/2/3 = 792143
This number cannot be divided evenly by the single digits anymore. So, continue with the primes.
792143/11/31 = 2323
There you have it. All the numbers.
And I just realized that I don't understand the question to your 3rd riddle. What do you mean by "third part of the train"?
And what review?
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Gom_Jabbar
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posted July 18, 2006 10:17 PM |
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Quote: What do you mean by "third part of the train"?
It's "the third part of the train's speed", meaning that if the train's speed is 600 km/h the number is 600/3.
I just think that I'll give you the points(3 out of 4) for test nr.2, even though your logic wasn't so good, because not even the Titanic had 2323 passangers. If you could explain me each number (X=.. , Y=.., and so on) I'll give you 4 points.
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friendofgunnar
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posted July 18, 2006 10:32 PM |
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Is it 50 meters between posts?
I just plugged in some random numbers and did the math.
I'm screaming along at 120 km per hour. I see 40 posts in a minute. That equals 50 meters between posts.
(hope I got the math right)
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angelito
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posted July 18, 2006 11:20 PM |
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I guess the numbers of the "ship" riddle are slightly uncorrect.
My chocies:
3 ship funnels
2 propellers
101 passengers.
day 23
month 11
year 1900+31.
cubic root of captain's age 4
3 * 2 * 101 * 23 * 11 * 31 + 4 = 4752858
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Better judged by 12 than carried by 6.
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Gom_Jabbar
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posted July 19, 2006 12:06 AM |
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Quote: Is it 50 meters between posts?
I just plugged in some random numbers and did the math.
I'm screaming along at 120 km per hour. I see 40 posts in a minute. That equals 50 meters between posts.
(hope I got the math right)
Sorry FOG but this aint a guessing game. The only information you have to use is the one I gave you, and I never mentioned 120km/h. Also, if you make a statement prove it.
Quote: 3 ship funnels
2 propellers
101 passengers.
day 23
month 11
year 1900+31.
cubic root of captain's age 4
3 * 2 * 101 * 23 * 11 * 31 + 4 = 4752858
Absolutely right Angelito. Anyway I'll only give 1 point since the other 3 went Iris's way (you all know I love her).
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dschingi
Famous Hero
the guy with the dragon golem
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posted July 19, 2006 02:02 PM |
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here is my solution for riddle nr 3:
first let me set some variables
pm = polles/minute
ph = polles/hour = 60*pm
s = train's speed in km/h
d = distance between polles
so pm = 1/3 * s
d = s / ph
ph = 60*pm = 60*1/3*s
-> d = s / (60/3*s) = 3/60 = 1/20 km = 50 m
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Gom_Jabbar
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posted July 19, 2006 02:29 PM |
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Absolutely right dschingi. I see your new(only two posts so far), and I hope to see more of you around here. Actually the distance of 50 meters is the standard distance between telegraph poles all around the world. You got the 4 points . Next test tonight.
Edit:
To make this thread more pleaseant I decided to post new riddles as soon as the old ones are solved.
Test No.4:
There is a barrel with no lid and some wine in it. "This barrel of wine is more than half full," said her. "No it's not," says him. "It's less than half full."
Question:
Without any measuring implements and without removing any wine from the barrel, how can they easily determine who is correct? -2p
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angelito
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posted July 19, 2006 11:06 PM |
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Edited by angelito at 12:54, 20 Jul 2006.
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Test no.3:
You have to "overturn" (not sure thatīs the right word) the barrel, but only as far until the wine "touches" the lip.
Now u just have to take a look into the barrel. If u can see a part of the bottom of the barrel, it is less than half full of wine, if u canīt see anything of the bottom, the barrel is more than half full.
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Better judged by 12 than carried by 6.
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