Who is the most logical and intellectual on TSR???

Scroll to see replies

1) Put 6 on one side of the scale, and the other 6 on the other side. The 6 coins on the lower side of the scale has the fake, lighter coin. Discard the other 6, and keep these 6.

2) Now you're left with 6 coins. Put 3 on one side of the scale, and the other 3 on the other side. The side of the scale which is lower has the fake coin. Discard the other 3 coins.

3) Now you're left with 3 coins. Take any two coins, place one on each side of the scale. If the scales are equal, then the remaining coin is the fake one. If they're unequal, the lighter one is the fake one.

Simples.

Edit: Whats with the neg rep? Can a guy not get something wrong? Jeez, such robots.. =/

(edited 13 years ago)

Reply 101

alharrison4

Original post by HistoryRepeating

Divide the coins in half, 6 coins in pile A and 6 in pile B

1) Split Pile A in half and weigh 3 on each side of the scale. If they weigh the same they are all gold and the fake is in B, if they weight differently the fake is in A. Lets say the fake is in B, so A is all gold (note: this doesnt make a difference to the proof, if the fake is in A just swap A and B in all the below reasoning, I've just done it for ease)

2) Take the 6 B (with the fake in them), split it in half. weigh one half against any 3 of the A, gold coins. If they weigh the same they are all gold and can be added to A and the fake is in the other half, if they weigh differently one of them is fake AND you know whether the fake is heavier or lighter than gold.

3) Take the half (3 coins) with the fake in and pick any two of them to weight against each other. If they weigh the same the third (unweighed one) is the fake, if they weigh differently, the one that is heavier or lighter (as found out in step 2 above) is the fake.

Coming up with this solution took me 4 minutes with a pen and paper from first principles, I've never done a problem like this before and I didnt cheat.

Brilliant! I'm impressed.

EDIT: Even if it is 'flawed' :smile:

Reply 102

Blueflare

I really shouldn't have read this thread at 5 in the morning.

Reply 103

HistoryRepeating

Original post by F1Addict

Fake could be heavier, ergo fail.

Reply 104

HistoryRepeating

1) Weigh 1,2,3,4 vs 5,6,7,8

If = then you have 2 more weighs to find which of 9,10,11,12 it is, which is easy (weigh 9,10 vs 11,Gold, if = then its 12, if < or > weigh 9vs10 (equal = 11, same as before = 9, different = 10))
If < goto step 2,
If > goto step 2 but in reverse.

2) You now know that EITHER one of 1,2,3,4 is fake and heavy or one of 5,6,7,8 is fake and light. Weigh 1,5,6 against 2,7,8

If =, fake is 3 or 4 and is heavy, weigh against each other.
If < then either 1 is fake and heavy or one of 7,8 is fake and light, goto 3
If > then either 2 is fake and heavy or one of 5,6 is fake and light, goto 3 in reverse and with 2,5,6 instead.

3. Weigh 1 and 7 against two gold ones discarded earlier.

If =, 8 is fake and light
If <, 1 is fake and heavy
If >, 7 is fake and light.

This is I think the simplest (and therefore most elegant!) solution, requiring 4v4, 3v3 and 2v2 rather than 3 lots of 4v4. Took about 20 more minutes, I think I could do it with pen and paper in an interview but a bit too slow to be impressive.

(edited 13 years ago)

Reply 105

kristinaalovesu

Wow, seeing all the numbers here gave me a cluster headache.

Reply 106

fat_hampster

Original post by HistoryRepeating

Pretty similar to my solution, though step 2 is slightly different.

I'd disagree that this is the most elegant solution though, I think rupertthebear/munn's solution is more elegant. Their solution is easier to explain, and also easier to adapt to similar problems. It also separates the weighing and the analysis. Our solutions require analysis between the attainment of weighings.

Reply 107

username133326

I think anyone who wastes enough time on this to come to the correct conclusion is still a loser...

Reply 108

harr

Original post by munn

Most (if not every) time I've read it before though it has required you to find the weight, but i think that's because it's impossible to find the fake without finding the weight at the same time

1 2 3 4 v 5 6 7 8
1 2 5 6 v 3 9 10 11
1 3 5 10 v 2 4 8 9

Always finds the fake (the possible results are lll,llr,lrl,l-r,rll,rl-,r--,r-r,-rr,-rl,-r-,---) but if the fake is the 12th coin then it doesn't tell you if the weight is heavier or lighter. Admittedly I've just constructed this example to be awkward; I expect most obvious solutions will tell you whether the fake is heavier or lighter.

Reply 109

Markh1000

Original post by F1Addict

Fail.

Certainly not you.

Original post by harr

i stand corrected
I suppose it's possible to determine the fake from 13 (possibly 14 but i'm not sure and i reckon it'll take much longer) coins using my method then

Reply 112

Converse

Original post by Aurora.

Probably should have seen that coming :p:

Yeah, you should have. :smile:

Reply 113

F1Addict

Original post by HistoryRepeating

Fake could be heavier, ergo fail.

Original post by Markh1000

Fail.

Yeah i know, I should've read the problem properly, instead of just skimming it. :P
Life goes on.. :biggrin:

Reply 114

vahik92

if anyone wants my answer,the way i solved it, here it is:

1. weigh 1,2,3,4 and 5,6,7,8

if 1234 = 5678, go to step 2, otherwise go to step 3

2. in here, we only have 4 coins left - 9,10,11,12 - which is easy, weigh any 2 of these e.g. 9 with 10, if they are balanced, than we have 11, 12 left with one more weighing so weighing one of these will give us the answer. similarly if 9 and 10 arent balanced, weigh any of 9 or 10 and you get which is the fake.

3. this is a bit hard. from step one we weighted 1,2,3,4 and 5,6,7,8. p.s. note, there is no difference whether 1,2,3,4 is heavier or lighter, i will write the case where 1,2,3,4 is heavier but if it is lighter, change the places of the two groups and you still will get the answer.

so, 1,2,3,4 > 5,6,7,8.

now, we set aside two coins from first group and one coin from the second group. lets say, 1,4,8. because the coin number are not equal, we put any of the real coins (9,10,11,12) instead of 1 so we get - 9,2,3 and 5,6,7. now, we swap 3 and 7 with their places. we have 9,2,7 and 5,6,3. weighting this gives:

3a. 9,2,7 > 5,6,3
because in this case we set aside 1,4,8 and the coins are unbalanced, we can exclude the latter three . we can also exclude 3 and 7, because even having swapped their places, the sign hasn't changed - the first group is still heavier. therefore we are only left with 2,5 and 6. this is what we do.we take off 2. move 5 to the left side. put one real coin on each side. so we get, 9, 5 and 10,6. we weigh this. if they are equal, fake is 2. if the left side is still heavier, then the fake is 6 because otherwise if it was 5, moving 5 to the left would make left side heavier. and if the sign changes - left side gets lighter, then it is 6.

3b. 9,2,7 < 5,6,3
this is easy. we can exclude 1,4,8. now, we knew that left side was originally heavier. but after swapping the places of 3 & 7, the sign has changed which means the fake is one of these two. easy - you weigh any of these two - e.g. 3 with 10. if they're balanced, fake is 10, otherwise its 3.

3c. 9,2,7 = 5,6,3
in here we deduce that the fake is one from 1,4,8. we have 3 coins, and one possible weighting. we know how to find the fake out of 3 coins by weighing 1 time only from step 3a. solved.