
Thread: Gom's Weekly Test  This thread is pages long: 1 2 3 4 · «PREV / NEXT» 

Gom_Jabbar
Promising
Famous Hero
Revealer of Truth

posted August 04, 2006 10:42 PM 


Correct Angelito, that's another 2 points for you.
Test No.5
Why can't John Stuart, who lives in Canada and has sweedish origins, be burried in the USA?  2p
Note: sorry for the delay in updating this thread but I was preety busy; almost turned into Gom's Monthly Test
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angelito
Hero of Order
proud father of a princess

posted August 07, 2006 02:29 PM 


Because he is still alive?
____________
Better judged by 12 than carried by 6.


TitaniumAlloy
Honorable
Legendary Hero
Professional

posted August 07, 2006 02:34 PM 


Yeah Angel got it first
It's against the law to bury people alive in the USA.
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John says to live above hell.


Gom_Jabbar
Promising
Famous Hero
Revealer of Truth

posted August 07, 2006 10:57 PM 


Quite correct angelito +2p
Test No.6
"A long time ago an old man feeling he will die soon, called his sons and said to them:
 My dears, all my fortune is 300 gold coins, wich I'm not going to share evenly among you; I'll give more to the smartest of you. All my life I bought and sold horses, but I never kept more then 300 gold coins; all the extra money I earned I used to rase you. I always had for paying for horses 9 bags, in wich those 300 gold coins were split so that without opening the bags I could pay any sum from 1 to 300 gold coins."
Question:
How many gold coins were in each bag?  3p
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friendofgunnar
Honorable
Legendary Hero
able to speed up time

posted August 08, 2006 09:46 PM 


1,2,4,8,16,32,64,128, and 45.


Gom_Jabbar
Promising
Famous Hero
Revealer of Truth

posted August 08, 2006 10:36 PM 


Hmmmm.... gotcha Foggie  3p
Test No.7
You have 12 balls, and one has a diffrent weight (lighter or heavier). You can do only 3 measurements at the balance.
Question:
How can you find what is the different ball is?  3p
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ratmonky
Famous Hero
Abu Hur Ibn Rashka

posted August 09, 2006 07:33 AM 


Quote: Hmmmm.... gotcha Foggie  3p
Test No.7
You have 12 balls, and one has a diffrent weight (lighter or heavier). You can do only 3 measurements at the balance.
Question:
How can you find what is the different ball is?  3p
Let's suppose the different ball is lighter. Divide the balls into three groups of four balls and weigh 2 groups.
1.If they have the same weight then the different ball is in the third group.
2.Divide this group into 2 groups of 2 balls and weigh again.
3.Repeat again until the ligher ball is found.
1.If after the first measurement one of the groups is lighter than the other, follow the steps 2 and 3
____________
Dies illa, dies irae,
Calamitatis et miseriae.
Requiem aeternum
Dona eis, dona eis Domine.


friendofgunnar
Honorable
Legendary Hero
able to speed up time

posted August 09, 2006 10:06 AM 


Quote: Let's suppose the different ball is lighter.
You can't assume that. It could be heavier or lighter.
Here's my solution.
Divide the balls into four groups
0000 0000 00 00
Weigh the first four against the second four. If they are uneven then take Path 1. If they are even continue to Path 2:
Path 1: Four of the balls are heavier than the other four.
Put two of the balls from the "heavier" side on the right side of the scale and put one ball from the "heavier" side plus one ball from the "lighter side" on the left side of the scale.
If the scale tilts to the left, the original ball from the "heavy" side is the oddball.(the one that got combined with the ball from the light side)
If the scale tilts to the right, either the ball from the light side is light, or one of the two balls from the heavy side is heavy. In this case weigh the two potentially heavy balls against each other. If one is heavier than the other you have found the oddball. If they are the same, then the original "light" ball from the light side is the oddball.
If the scale is even, one of the three remaining balls from the "light" side is the oddball and it is also light (important). In this go to path 1A below
Path 1A. You have three balls from the original light side and you that know one of them is light. Weigh one against the other. If one is lighter than the other then you have found the oddball. If they are even, the third ball (the one you didn't weigh) is the oddball.
You've now captured all the possibilities if your original weigh (four versus four) was uneven. If it was even you have to follow path 2.
Path 2. (you have two groups of two balls now). Weigh one ball from one group against one ball from the other group. If they are uneven follow path 2A. If they are even follow path 2B.
2A. One ball is heavier then the other. Weigh the heavier ball against a normal ball (from the original group of four versus four). If it is heavy, you have found your oddball. If it is the same, then the one you left behind is the oddball.
2B. If your original one versus one was even, take a regular ball and weigh it against one of the remaining balls. If it is uneven you have found your oddball. If it is even then the last remaining ball (the one you never weighed at all) is the oddball.
Wshew! That took me like an hour.


angelito
Hero of Order
proud father of a princess

posted August 09, 2006 10:21 AM 


It doesn't matter if the "wrong" ball is heavier or lighter.
ratmonkeys solution seems correct for me, because all u need to see is the UNbalance of 1 of the groups, no matter if it goes up or down on one side.
____________
Better judged by 12 than carried by 6.


friendofgunnar
Honorable
Legendary Hero
able to speed up time

posted August 09, 2006 10:41 AM 


Quote: 2.Divide this group into 2 groups of 2 balls and weigh again.
When Ratmonkey does this weighing it is completely useless. It will be uneven but you won't gain any information whatsoever because you will not know whether the right side is light or whether the left side is heavy. (or vice versa)
Also, when Ratmonkey does the original 4 versus 4 if it is uneven he won't gain any information other than the fact that a heavy ball might on one side or the light ball might be on the other. In other words he won't know which group of 4 contains the oddball, so he can't continue his reductions.
Keep in mind that Ratmonkey wasn't answering the riddle. His method would work if the oddball was lighter. However in Gom's riddle the oddball could be lighter or heavier.


angelito
Hero of Order
proud father of a princess

posted August 09, 2006 11:32 AM 


You are right. My bad.
____________
Better judged by 12 than carried by 6.


Lady_Milena
Honorable
Known Hero
Grannie Sweet Cheeks

posted August 09, 2006 03:48 PM 


Here is my solution:
Divide those 12 balls in 3 groups: 5, 5, 2
Put 5 and 5 on the scales. If they are even, measure the rest two to find out which one is lighter.
If one of 5/5 is heavier, take it and divide it in 2/2/1. Measure 2 and 2, see if they are even. If that is so, the one ball is the one we're looking for.
If they are not even, our ball is one of the two. That's third measurement too.
P.S. The same idea can be used for 14 balls.
Divide in 7/7/2.
If 7/7 even out, ball is one of the two: measure them
If one group is lighter, pick it, divide in 3/3/1. If 3/3 even out, measure 1/1 and leave one aside. If they even out the third ball is the one we're looking for.
~Edit P.S. I was under the wrong impression that we KNEW if the ball was either lighter or heavier. It didn't come to my mind we actually had to find out IF it was lighter or heavier.
My bad.
____________
God does not need exist to save us...


russ
Promising
Supreme Hero
blah, blah, blah

posted August 09, 2006 05:20 PM 


Hehe, FoG, GJ! I had the same idea, but it was 5pm when I was thinking about 1A (i.e. uneven), so I didn't finish it and went home I even realized that you should weigh light+heavy vrs 2 heavies since any other combination would not let you come to a conclusion after the last weigh, but I didn't have time to figure out how weighting those would help me


friendofgunnar
Honorable
Legendary Hero
able to speed up time

posted August 09, 2006 06:40 PM 


GAH!
Do you know how sometimes when you are thinking on a problem and then go to sleep and when you wake up the next day you know the answer? like your brain was working on it the entire time you were sleeping?
I woke up this morning and realized that my answer was incorrect.
It needs a slight adjustment. You have to weigh three heavies against one heavy and two lights. That should make it work.
I'll adjust my answer soon...
BTW Milena, your solution has the same problem that Ratmonkey's has. In your final 2 x 2 weigh you won't gain any information if it is unbalanced. This is because you don't know whether the oddball you are looking for is heavier or lighter. All you will know is that the two on the right have the heavy ball or the two on the left have the light ball (or vice versa)


russ
Promising
Supreme Hero
blah, blah, blah

posted August 09, 2006 07:00 PM 


Quote: It needs a slight adjustment. You have to weigh three heavies against one heavy and two lights. That should make it work. ;
An adjustment? It looks correct to me...


friendofgunnar
Honorable
Legendary Hero
able to speed up time

posted August 09, 2006 08:23 PM 


My second solution was incorrect also. In the second step you must weigh two heavies plus a regular versus two heavies and a light. (In the first solution if the second balancing was even, the oddball could be amongst the three remaing lights or the 4th heavy that was never weighed. That's what killed it. In my second solution if the second balancing was even then that didn't get you anywhere either. It was worse than the first)
Solution:
Divide the balls into four groups
0000 0000 00 00
Weigh the first four against the second four. If they are uneven then take Path 1. If they are even continue to Path 2:
Path 1: Four of the balls are heavier than the other four.
Put two of the balls from the "heavier" side plus one regular ball on the right side of the scale and put two balls from the "heavier" side plus one ball from the "lighter side" on the left side of the scale.
If the scale tilts to the left, the oddball is one of the two heavy balls. Weigh them against each other to find the oddball
If the scale tilts to the right, the oddball is one of two heavies on the right, or it is the light one on the left. Weigh the two heavies against each other in this eventuality. If one of them is heavier, you found the oddball. If they are even, the oddball is the light one from the second weighing.
If the scale is even, one of the three remaining balls from the "light" side is the oddball and it is also light (important). If this is true go to path 1A below
Path 1A. You have three balls from the original light side and you that know one of them is light. Weigh one against the other. If one is lighter than the other then you have found the oddball. If they are even, the third ball (the one you didn't weigh) is the oddball.
You've now captured all the possibilities if your original weigh (four versus four) was uneven. If it was even you have to follow path 2.
Path 2. (you have two groups of two balls now). Weigh one ball from one group against one ball from the other group. If they are uneven follow path 2A. If they are even follow path 2B.
2A. One ball is heavier then the other. Weigh the heavier ball against a normal ball (from the original group of four versus four). If it is heavy, you have found your oddball. If it is the same, then the one you left behind is the oddball.
2B. If your original one versus one was even, take a regular ball and weigh it against one of the remaining balls. If it is uneven you have found your oddball. If it is even then the last remaining ball (the one you never weighed at all) is the oddball.
I think I deserve more than three points


Ecoris
Promising
Supreme Hero

posted August 10, 2006 11:40 AM 


After reading that I think you're right .


Gom_Jabbar
Promising
Famous Hero
Revealer of Truth

posted August 10, 2006 08:20 PM 


@Lady_Milena: quite incorrect my dear!
@FriendOfGunnar: after a long deliberation from the jury I decided to give you that extra point; so it's 4p
Test No.8
A pichipontz is a farlifus bigger then a moshtrofontz;
A farlifus is a moshtrofontz smaller then a pichipontz;
A moshtrofontz is a pichipontz bigger then a farlifus.
Question:
Wich one is the smallest and wich one is the biggest?
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angelito
Hero of Order
proud father of a princess

posted August 10, 2006 09:15 PM 


Quote: A pichipontz is a farlifus bigger then a moshtrofontz;
A farlifus is a moshtrofontz smaller then a pichipontz;
A moshtrofontz is a pichipontz bigger then a farlifus.
Question:
Wich one is the smallest and wich one is the biggest?
Smallest: farlifus
Biggest: pichipontz
____________
Better judged by 12 than carried by 6.


Gom_Jabbar
Promising
Famous Hero
Revealer of Truth

posted August 10, 2006 11:39 PM 


Hehe... right on angelito +1p
I just thought to give away some easy ones
Test No.9
A high tonnage truck had to cross a 35km bridge. The bridge can only sustain a maximum weight of 120 tons, exactly the weight of the truck now. The truck goes right to the middle of the bridge and stops. In that moment a bird lands on the truck.
Question:
What happens to the bridge? It collapses? Why? 1p
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