05-31-2008, 08:00 PM
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#481 | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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All Star
Join Date: Feb 2005
Posts: 726
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Scenarios 17 to 24 are good, but there are a few minor problems with the others. For example:
What happens if AE > BC ? (after the first two weighings, either A is lighter, or E is heavier) Your scenario for when E is heavier has AE > BI as the last weighing. While this is correct, what happens when AE < BI? (which could happen, since A may be lighter than the rest). Basically, either of the two final weighings you've given for these scenarios will work as long as you use the same one in both. I'll have to knock off one mark for that: 9/10. Not quite a gold star! 
Yes! Correct. 10/10. And you get a gold star. Interestingly, these are different to the solution I had (the cases when ABCD=EFGH are the same though): Scenario Set 1) ABCD = EFGH IJK = ABC (=> L is heavy or light) A > L implies L is light A < L implies L is heavy A = L is impossible (otherwise we'd have 12 equally weighted coins) Scenario Set 2) ABCD = EFGH IJK > ABC (=> I is heavy, or J is heavy, or K is heavy) I > J implies I is heavy I < J implies J is heavy I = J implies K is heavy Scenario Set 3) ABCD = EFGH IJK < ABC (=> I is light, or J is light, or K is light) I > J implies J is light I < J implies I is light I = J implies K is light Scenario Set 4) ABCD < EFGH ABEF = CGIJ (=> D is light, or H is heavy) D = K implies H is heavy D < K implies D is light D > K is impossible since ABCD < EFGH Scenario Set 5) ABCD < EFGH ABEF < CGIJ (=> A is light, B is light, or G is heavy) A = B implies G is heavy A < B implies A is light A > B implies B is light Scenario Set 6) ABCD < EFGH ABEF > CGIJ (=> E is heavy, F is heavy, or C is light) E = F implies C is light E < F implies F is heavy E > F implies E is heavy Scenario Set 7) ABCD > EFGH ABEF = CGIJ (=> D is heavy, or H is light) D = K implies H is light D > K implies D is heavy D < K is impossible since ABCD > EFGH Scenario Set 8) ABCD > EFGH ABEF < CGIJ (=> E is light, F is light, or C is heavy) E = F implies C is heavy E < F implies E is light E > F implies F is light Scenario Set 9) ABCD > EFGH ABEF > CGIJ (=> A is heavy, B is heavy, or G is light) A = B implies G is light A < B implies B is heavy A > B implies A is heavy
__________________
Question everything -- especially that which you already believe to be true. |
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05-31-2008, 08:41 PM
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#482 | |||||||||||||||||||||||
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MVP
Join Date: May 2004
Location: Melbourne, AUS
Posts: 1,576
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Since the weights are still uneven after the removal of C it has been established to be of equal weight. As this is the case it could be substituted for I. So if AE > BI then AE > BC. If AE > BC then A cannot be lighter otherwise AE would be lighter than BC. If AE < BI then E cannot be heavier as it has changed sides without affecting the balance. Therefore it must be A ( I think that makes sense ).
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Inspire you of think the elephant dint Inspire you of think the elephant dint |
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06-01-2008, 08:58 AM
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#483 | |||||||||||||||||||||||
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All Star
Join Date: May 2007
Posts: 891
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BoomChickaWaWa!
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06-01-2008, 09:49 AM
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#484 | |||||||||||||||||||||||
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MVP
Join Date: Nov 2006
Location: Manchester
Posts: 1,906
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he should give you 10 bucks for solving that.....(by the way your all sad for spending so much time on that  )
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27" imac 3.4ghz i7 16gb Ram 1TB HD GTX680 Mx imac core duo 1.83ghz: OS X 10.6.8, 2gb ram, 1Tb internal |
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06-01-2008, 11:02 AM
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#485 | |||||||||||||||||||||||
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All Star
Join Date: Feb 2005
Posts: 726
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You're just jealous that you never got the answer in first! As for spending too much time on the problem, it may just have ruined my life. When I first heard the riddle, I'd not long moved to Italy to work, and my Italian was non-existant to say the least. Instead of going to Italian lessons one day, I stayed at home and worked out the answer. I never went back to the Italian lessons, and five months later I left the country. Had I never set eyes on the riddle, I could well be settled in Italy (and we all know what the Italian weather and women are like...  ).Of course, it could just be my fault for being too lazy to go to Italian lessons... (they were first thing on a Saturday)
__________________
Question everything -- especially that which you already believe to be true. |
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06-01-2008, 11:39 AM
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#486 |
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All Star
Join Date: May 2007
Posts: 891
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hey, I actually had fun solving the riddle!!
you're just jealous kel cuz I can get girls sand solve riddles at the same time
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06-01-2008, 11:46 AM
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#487 | |||||||||||||||||||||||
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MVP
Join Date: Nov 2006
Location: Manchester
Posts: 1,906
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dude, girls are riddles
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27" imac 3.4ghz i7 16gb Ram 1TB HD GTX680 Mx imac core duo 1.83ghz: OS X 10.6.8, 2gb ram, 1Tb internal |
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06-01-2008, 12:34 PM
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#488 | |||||||||||||||||||||||
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All Star
Join Date: May 2007
Posts: 891
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very true ..Im solving one later today ..
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06-01-2008, 07:59 PM
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MVP
Join Date: May 2004
Location: Melbourne, AUS
Posts: 1,576
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Alright here's the looooong winded version of the same solution;
Scenario 1) Weigh A, B, C & D against E, F, G & H If A, B, C & D are lighter than E, F, G & H this means either one of the left weights is lighter or one of the right weights is heavier. By elimination this also means I, J, K & L are all even. Remove C, D & H, move B to the right side and F & G to the left side and add one of the remaining weights ( I ), which we know is normal from step one, to the right side. If A, F & G are lighter than E, B & I then we can eliminate F, G & B as they have changed sides without affecting the balance. We can also eliminate C, D & H as they were removed without affecting the balance. This just leaves A & E. Remove F & G and move E to the left side leaving two weights on the right. In my original explanation I used B & C but let's leave it as B & I seeing they're already on the imaginary scale. As we know all the others are equal you could use any two of the other weights on the right side. If A & E are lighter than B & I then we eliminate E as it has changed sides without affecting the balance. This leaves A being lighter than the rest of the weights.
Scenario 2) Uses the same process of elimination as scenario 1 to prove that A is heavier.
Scenario 3) Weigh A, B, C & D against E, F, G & H If A, B, C & D are lighter than E, F, G & H this means either one of the left weights is lighter or one of the right weights is heavier. By elimination this also means I, J, K & L are all even. Remove C, D & H, move B to the right side and F & G to the left side and add one of the remaining weights ( I ), which we know is normal from step one, to the right side. If A, F & G are heavier than E, B & I then we can eliminate A & E as they are still on the same sides but the balance has reversed. We can also eliminate C, D & H as they were removed and it is still out of balance. This just leaves F, G & B. Remove A & F and move B to the left side leaving two weights on the right. In my original explanation I used A & I but let's leave it as E & I seeing they're already on the imaginary scale. As we know all the others are equal you could use any two of the other weights on the right side. If G & B are lighter than E & I then we eliminate F as it was removed and the scales are still out of balance. We also eliminate G as it has stayed on the same side but the balance has reversed. This leaves B being lighter than the rest of the weights.
Scenario 4) Uses the same process of elimination as scenario 3 to prove that B is heavier.
Scenario 5) Weigh A, B, C & D against E, F, G & H If A, B, C & D are lighter than E, F, G & H this means either one of the left weights is lighter or one of the right weights is heavier. By elimination this also means I, J, K & L are all even. Remove C, D & H, move B to the right side and F & G to the left side and add one of the remaining weights ( I ), which we know is normal from step one, to the right side. If A, F & G are equal in weight to E, B & I then we can eliminate A, F, G, E & B. This leaves the weights that we removed. Either C or D is lighter or H is heavier. Remove A, F, G & I and place C & H on the left side leaving two weights on the right. As we know all the others are equal you could use any two of the other weights on the right side. If C & H are lighter than E & B then we eliminate H as it has changed sides without affecting the balance. We can also eliminate D as it was left off and the scales are out of balance. This leaves C being lighter than the rest of the weights.
Scenario 6) Uses the same process of elimination to prove that C is heavier.
Scenario 7) Weigh A, B, C & D against E, F, G & H If A, B, C & D are lighter than E, F, G & H this means either one of the left weights is lighter or one of the right weights is heavier. By elimination this also means I, J, K & L are all even. Remove C, D & H, move B to the right side and F & G to the left side and add one of the remaining weights ( I ), which we know is normal from step one, to the right side. If A, F & G are equal in weight to E, B & I then we can eliminate A, F, G, E & B. This leaves the weights that we removed. Either C or D is lighter or H is heavier. Remove A, F, G & I and place C & H on the left side leaving two weights on the right. As we know all the others are equal you could use any two of the other weights on the right side. If C & H are equal to E & B we can eliminate C & H leaving D as being the lighter weight.
Scenario 8) Uses the same process of elimination as scenario 7 to prove that D is heavier.
Scenario 9) Weigh A, B, C & D against E, F, G & H If A, B, C & D are heavier than E, F, G & H this means either one of the left weights is heavier or one of the right weights is lighter. By elimination this also means I, J, K & L are all even. Remove C, D & H, move B to the right side and F & G to the left side and add one of the remaining weights ( I ), which we know is normal from step one, to the right side. If A, F & G are heavier than E, B & I then we can eliminate F, G & B as they have changed sides without affecting the balance. We can also eliminate C, D & H as they were removed without affecting the balance. This just leaves A & E. Remove F & G and move E to the left side leaving two weights on the right. As we know all the others are equal you could use any two of the other weights on the right side. If A & E are lighter than B & I then we eliminate A as it has stayed on the same side but the balance has reversed. This leaves E being lighter than the rest of the weights.
Scenario 10) Uses the same process of elimination as scenario 9 to prove that E is heavier.
Scenario 11) Weigh A, B, C & D against E, F, G & H If A, B, C & D are heavier than E, F, G & H this means either one of the left weights is heavier or one of the right weights is lighter. By elimination this also means I, J, K & L are all even. Remove C, D & H, move B to the right side and F & G to the left side and add one of the remaining weights ( I ), which we know is normal from step one, to the right side. If A, F & G are lighter than E, B & I then we can eliminate A & E as they are still on the same sides but the balance has reversed. We can also eliminate C, D & H as they were removed and it is still out of balance. This just leaves F, G & B. 11.1)Ok in my original solution I removed A & G so I'll go with that for now. Remove A & G and move B to the left side leaving two weights on the right. As we know all the others are equal you could use any two of the other weights on the right side. If F & B are lighter than E & I then we eliminate G as it was removed and the scales are still out of balance. We also eliminate B as it has changed sides but the balance hasn't changed. This leaves F being lighter than the rest of the weights. 11.2) Just for consistancy's sake let's remove F as this was my next step in previous solutions. ABCD > EFGH AFG < EBI GB = EI Remove A & F and move B to the left side leaving two weights on the right. As we know all the others are equal you could use any two of the other weights on the right side. If G & B are equal to E & I then we eliminate G & B as they are equal in weight to E & I. This leaves F which we know from steps 1 & 2 must be a lighter weight.
Scenario 12) Uses the same process of elimination as scenario 11 to prove that F is heavier.
Scenario 13) Uses the same process of elimination as 11.2 to prove that G is lighter.
Scenario 14) Uses the same process of elimination as 11.2 to prove that G is heavier.
Scenario 15) Weigh A, B, C & D against E, F, G & H If A, B, C & D are heavier than E, F, G & H this means either one of the left weights is heavier or one of the right weights is lighter. By elimination this also means I, J, K & L are all even. Remove C, D & H, move B to the right side and F & G to the left side and add one of the remaining weights ( I ), which we know is normal from step one, to the right side. If A, F & G are equal in weight to E, B & I then we can eliminate A, F, G, E & B. This leaves the weights that we removed. Either C or D is heavier or H is lighter. Remove A, F, G & I and place C & H on the left side leaving two weights on the right. As we know all the others are equal you could use any two of the other weights on the right side. If C & H are lighter than E & B we can eliminate C as it is on the same side as in step 1 but the balance has reversed. We can also eliminate D as it has been removed and the scales are still out of balance. This leaves H as being the lighter weight.
Scenario 16) Uses the same process of elimination as scenario 15 to prove that H is heavier. Scenario 1-16 also prove that I'm some kind of obsessive freak  .
__________________
Inspire you of think the elephant dint Inspire you of think the elephant dint |
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06-02-2008, 08:19 AM
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#490 | |||||||||||||||||||||||
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All Star
Join Date: Feb 2005
Posts: 726
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The bit in bold (my emphasis) is the important implication that was missing from your original answer. However, I didn't mention this in my initial response because I never made that logical jump (my bad). You can have the extra mark back, and your gold star.
__________________
Question everything -- especially that which you already believe to be true. |
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06-02-2008, 10:40 AM
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#491 |
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League Commissioner
Join Date: Mar 2003
Location: Bay Area, CA
Posts: 11,352
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06-03-2008, 11:16 AM
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#492 |
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MVP
Join Date: Nov 2006
Location: Manchester
Posts: 1,906
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^^ haha classic
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27" imac 3.4ghz i7 16gb Ram 1TB HD GTX680 Mx imac core duo 1.83ghz: OS X 10.6.8, 2gb ram, 1Tb internal |
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06-03-2008, 01:47 PM
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#493 |
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MVP
Join Date: Apr 2007
Location: Sherwood, Arkansas, USA
Posts: 1,320
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Here's an easy one for all you Mac whizzes. For years now when I reconcile my monthly bank statement, it is always off a few cents. I use calculator.app to do the addition/subtraction. If I put in all the right numbers, why is my balance always off?
One clue. |
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06-03-2008, 01:50 PM
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#494 |
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MVP
Join Date: Jun 2007
Location: Skellefteċ, Sweden
Posts: 1,173
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Number of valid digits in it's math?
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/Bengt-Arne Fjellner IT-Administrator Luleċ university, Sweden. Some say: "You learn as long as you live". My way: "You live as long as you learn". |
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06-03-2008, 03:48 PM
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#495 |
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MVP
Join Date: May 2004
Location: Melbourne, AUS
Posts: 1,576
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Because you need to reinstall Calculator?
Or is this some weird result of switching from RPN mode?
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Inspire you of think the elephant dint Inspire you of think the elephant dint Last edited by fazstp; 06-03-2008 at 03:58 PM. |
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06-03-2008, 05:59 PM
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#496 | |||||||||||||||||||||||
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All Star
Join Date: Feb 2005
Posts: 726
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The display values are rounded to the nearest integer? (i.e. it isn't showing values after the decimal points). 2454510.50 (displays as 2454511) + 44499.50 (displays as 44500) = 2499010.00 (displays as 2499010)
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Question everything -- especially that which you already believe to be true. |
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06-03-2008, 06:41 PM
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#497 |
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MVP
Join Date: Apr 2007
Location: Sherwood, Arkansas, USA
Posts: 1,320
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If you open calculator and then select view > precision it gives you the options of 0 to 16. The default, I think, is 6... anyway that's where mine was set. I always assumed this was the number of decimal places it would display or something. Apparently not.... at a setting of 6, it will not add a 7 digit and a 5 digit number correctly.... see the tape view previously posted. Change the setting to 12 or something and it adds correctly.
I pay most of my bill using Bill Pay and since I was sitting at the computer anyway, I used calculator.app to record the new balance after each check. That little quirk in calculator.app has cost me many, many hours over the years trying to find errors to the point I just quit looking if it wasn't more than a dollar off. Try it out. I never use decimals in the calculator when doing $s and cents.... just an extra step since I can usually remember the last two digits are cents. |
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06-03-2008, 07:33 PM
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#498 |
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MVP
Join Date: May 2004
Location: Melbourne, AUS
Posts: 1,576
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That's freaking weird. I thought precision only applied to what came after the decimal point? Also 1,000,000 + 5 = 1,000,000. Shouldn't it round up to 1,000,010?
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Inspire you of think the elephant dint Inspire you of think the elephant dint |
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06-03-2008, 08:24 PM
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#499 | |||||||||||||||||||
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MVP
Join Date: Apr 2007
Location: Sherwood, Arkansas, USA
Posts: 1,320
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Silly me, I just thought the darn thing was a simple calculator.... nothing more, nothing less... and used it for just that for years. When I picked up the incorrect total I started poking around to see what else was there... and found the scientific calculator complete with memory keys, a programmer calculator, conversions, etc.. Total surprise after all these years on a Mac.  Guess I'm a little slow.
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06-04-2008, 05:32 AM
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#500 | |||||||||||||||||||||||
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All Star
Join Date: Feb 2005
Posts: 726
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If you consider mantissa and exponent representation (0.m x 10^e), then the pieces begin to fall into place. If the precision denotes the maximum number of digits that m can have, then: 2,454,511 = 0.245451 x 10^7 (= 2,454,510) 44,500 = 0.445 x 10^5 (= 44,500) (= 0.004450 x 10^7) So the result is 0.249901 x 10^7 (= 2,499,010). 1,000,000 = 0.1 x 10^7 (= 1,000,000) 5 = 0.5 x 10^1 (= 5) (= 0.000000 x 10^7 as the 5 which would be in the 7th place is chopped off, i.e. no rounding is performed). Therefore, the result is 0.1 x 10^7 which is 1,000,000 as returned by Calculator.app. This could be completely wrong, but I'm convinced!
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Question everything -- especially that which you already believe to be true. |
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Linear Mode
