Solve this IQ quiz
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Well, this quiz was given my friend, and he havent tell me the ans to it yet.
So i thought of sharing it to the ppl here as well
Here goes:
You have twelve coins. You know that one is fake. The only thing that distinguishes the fake from the real ones is that its weight is imperceptibly different. You have a perfectly balanced scale that only tells which side weighs more than the other side.
What is the smallest number of times you must use the scale in order to always find the fake coin?
Use only the twelve coins themselves and no others, no other weights, no cutting coins, no pencil marks on the scale. etc.
These are modern coins, so the fake coin is not necessarily lighter.
Presume the worst case scenario, and don't hope that you will pick the right coin on the first attempt. -
3 times
first time - 6 on each side..
second time - 3 on each side
3rd time, just choose any 2 to weigh. If they are the same, the other coin will be the lighter 1.. -
divide n conquer
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BRAVO!!!Originally posted by BufPuf:3 times
first time - 6 on each side..
second time - 3 on each side
3rd time, just choose any 2 to weigh. If they are the same, the other coin will be the lighter 1..
This is mine...
If you knew the fake coin was lighter, then the solution would have an easy explanation. But you do not. So....
Number the coins 1 through 12.
1. Weigh coins 1,2,3,4 against coins 5,6,7,8.
1.1. If they balance, then weigh coins 9 and 10 against coins 11 and 8 (we know from the first weighing that 8 is a good coin).
1.1.1. If the second weighing also balances, we know coin 12 (the only one not yet weighed) is the counterfeit. The third weighing indicates whether it is heavy or light.
1.1.2. If (at the second weighing) coins 11 and 8 are heavier than coins 9 and 10, either 11 is heavy or 9 is light or 10 is light. Weigh 9 against 10. If they balance, 11 is heavy. If they don't balance, you know that either 9 or 10 is light, so the top coin is the fake.
1.1.3 If (at the second weighing) coins 11 and 8 are lighter than coins 9 and 10, either 11 is light or 9 is heavy or 10 is heavy. Weigh 9 against 10. If they balance, 11 is light. If they don't balance, you know that either 9 or 10 is heavy, so the bottom coin is the fake.
1.2. Now if (at first weighing) the side with coins 5,6,7,8 are heavier than the side with coins 1,2,3,4. This means that either 1,2,3,4 is light or 5,6,7,8 is heavy. Weigh 1,2, and 5 against 3,6, and 9.
1.2.1. If (when we weigh 1,2, and 5 against 3,6 and 9) they balance, it means that either 7 or 8 is heavy or 4 is light. By weighing 7 and 8 we obtain the answer, because if they balance, then 4 has to be light. If 7 and 8 do not balance, then the heavier coin is the counterfeit.
1.2.2. If (when we weigh 1,2, and 5 against 3,6 and 9) the right side is heavier, then either 6 is heavy or 1 is light or 2 is light. By weighing 1 against 2 the solution is obtained.
1.2.3. If (when we weigh 1,2, and 5 against 3, 6 and 9) the right side is lighter, then either 3 is light or 5 is heavy. By weighing 3 against a good coin the solution is easily arrived at.
1.3 If (at the first weighing) coins 1,2,3,4 are heavier than coins 5,6,7,8 then repeat the previous steps 1.2 through 1.2.3 but switch the numbers of coins 1,2,3,4 with 5,6,7,8.
LOL LOL!!!
PAISEH HOR!! -
wa lao ehOriginally posted by mhcampboy:BRAVO!!!
This is mine...
If you knew the fake coin was lighter, then the solution would have an easy explanation. But you do not. So....
Number the coins 1 through 12.
1. Weigh coins 1,2,3,4 against coins 5,6,7,8.
1.1. If they balance, then weigh coins 9 and 10 against coins 11 and 8 (we know from the first weighing that 8 is a good coin).
1.1.1. If the second weighing also balances, we know coin 12 (the only one not yet weighed) is the counterfeit. The third weighing indicates whether it is heavy or light.
1.1.2. If (at the second weighing) coins 11 and 8 are heavier than coins 9 and 10, either 11 is heavy or 9 is light or 10 is light. Weigh 9 against 10. If they balance, 11 is heavy. If they don't balance, you know that either 9 or 10 is light, so the top coin is the fake.
1.1.3 If (at the second weighing) coins 11 and 8 are lighter than coins 9 and 10, either 11 is light or 9 is heavy or 10 is heavy. Weigh 9 against 10. If they balance, 11 is light. If they don't balance, you know that either 9 or 10 is heavy, so the bottom coin is the fake.
1.2. Now if (at first weighing) the side with coins 5,6,7,8 are heavier than the side with coins 1,2,3,4. This means that either 1,2,3,4 is light or 5,6,7,8 is heavy. Weigh 1,2, and 5 against 3,6, and 9.
1.2.1. If (when we weigh 1,2, and 5 against 3,6 and 9) they balance, it means that either 7 or 8 is heavy or 4 is light. By weighing 7 and 8 we obtain the answer, because if they balance, then 4 has to be light. If 7 and 8 do not balance, then the heavier coin is the counterfeit.
1.2.2. If (when we weigh 1,2, and 5 against 3,6 and 9) the right side is heavier, then either 6 is heavy or 1 is light or 2 is light. By weighing 1 against 2 the solution is obtained.
1.2.3. If (when we weigh 1,2, and 5 against 3, 6 and 9) the right side is lighter, then either 3 is light or 5 is heavy. By weighing 3 against a good coin the solution is easily arrived at.
1.3 If (at the first weighing) coins 1,2,3,4 are heavier than coins 5,6,7,8 then repeat the previous steps 1.2 through 1.2.3 but switch the numbers of coins 1,2,3,4 with 5,6,7,8.
LOL LOL!!!
PAISEH HOR!!
yrs damn complicated leh
mine answer is the best liao i think..hahahahaha -
thats wat i thought.. ur never smart
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its nt fun!
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ask the money changers which 1 is fake
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Cheh! got reference one. I tot you so smart lolOriginally posted by mhcampboy:BRAVO!!!
This is mine...
If you knew the fake coin was lighter, then the solution would have an easy explanation. But you do not. So....
Number the coins 1 through 12.
1. Weigh coins 1,2,3,4 against coins 5,6,7,8.
1.1. If they balance, then weigh coins 9 and 10 against coins 11 and 8 (we know from the first weighing that 8 is a good coin).
1.1.1. If the second weighing also balances, we know coin 12 (the only one not yet weighed) is the counterfeit. The third weighing indicates whether it is heavy or light.
1.1.2. If (at the second weighing) coins 11 and 8 are heavier than coins 9 and 10, either 11 is heavy or 9 is light or 10 is light. Weigh 9 against 10. If they balance, 11 is heavy. If they don't balance, you know that either 9 or 10 is light, so the top coin is the fake.
1.1.3 If (at the second weighing) coins 11 and 8 are lighter than coins 9 and 10, either 11 is light or 9 is heavy or 10 is heavy. Weigh 9 against 10. If they balance, 11 is light. If they don't balance, you know that either 9 or 10 is heavy, so the bottom coin is the fake.
1.2. Now if (at first weighing) the side with coins 5,6,7,8 are heavier than the side with coins 1,2,3,4. This means that either 1,2,3,4 is light or 5,6,7,8 is heavy. Weigh 1,2, and 5 against 3,6, and 9.
1.2.1. If (when we weigh 1,2, and 5 against 3,6 and 9) they balance, it means that either 7 or 8 is heavy or 4 is light. By weighing 7 and 8 we obtain the answer, because if they balance, then 4 has to be light. If 7 and 8 do not balance, then the heavier coin is the counterfeit.
1.2.2. If (when we weigh 1,2, and 5 against 3,6 and 9) the right side is heavier, then either 6 is heavy or 1 is light or 2 is light. By weighing 1 against 2 the solution is obtained.
1.2.3. If (when we weigh 1,2, and 5 against 3, 6 and 9) the right side is lighter, then either 3 is light or 5 is heavy. By weighing 3 against a good coin the solution is easily arrived at.
1.3 If (at the first weighing) coins 1,2,3,4 are heavier than coins 5,6,7,8 then repeat the previous steps 1.2 through 1.2.3 but switch the numbers of coins 1,2,3,4 with 5,6,7,8.
LOL LOL!!!
PAISEH HOR!! -
Originally posted by anakinfoo:Cheh! got reference one. I tot you so smart lol
I am smart. By searching for the solution... -
Yea bravo, u are smart no doubt in actually finding a solution through other websOriginally posted by mhcampboy:
I am smart. By searching for the solution... -
Bufpuf's answer only works if u know the fake coin's weight comparative to real coin's
u mentioned, 1st step, weigh 6 on each side
then weigh 3 on each side, but lemme ask u which side of the 6 are u gonna take in the 1st place? the lighter or the heavier 1
Already there's loophole from the 2nd step onwards le..so it's not the right answer -
more detailed explanation here:-Originally posted by BufPuf:3 times
first time - 6 on each side..
second time - 3 on each side
3rd time, just choose any 2 to weigh. If they are the same, the other coin will be the lighter 1..
split 12 coins into 2 batches of 6 each.
- place 1st batch of 6 coins with 3 on each side...on the scale one plate may be heavier than the other. since we know the one coin may be heavier or lighter than the other 11 on weight per unit. (label 3 coins 'A' and the other 3 'B'). if they balance take down the weight of 3 genuine coins average to find the weight of one genuine coin.
- place the 2nd batch of 6 coins with 3 coins each onto the 2 plates on the balance scale. if the scale balances these then this batch of 6 coins are generic and no fake one among them. (label 3 coins as 'C' and the other 3 as 'D'). if they balance take down the weight of 3 genuine coins average to find the weight of one genuine coin.
Note: which scenario occurs for 1st time and 2nd time in precedence is not important.
analyse the 4 groups of 3 coins each with a greater or smaller reading in weight. that group shall be where the fake coin is. then just pick any good coin and weigh it againist the group of 3 coins suspected to hide a fake coin.
it will just take 3 times of weighing on the balance scale even if you know tthe weight of the geniune coin and is lucky to hit the jackpot on the very first try and no more than 5 times if you score snakes eye.
recording of the weight of the coins is permissible since the thread starter only highlighted the times of weighing and never mention anything about recording weights. -
Again, it's not workable, coz u're not allowed to use anything else besides the 12 coin and the scale, meaning labelling coin is not allowed whether u use pencil marker or wadever. He may not mention anything about recording but he did say use only the 12 coins and the scale and no othersOriginally posted by pipipopo:more detailed explanation here:-
split 12 coins into 2 batches of 6 each.
- place 1st batch of 6 coins with 3 on each side...on the scale one plate may be heavier than the other. since we know the one coin may be heavier or lighter than the other 11 on weight per unit. (label 3 coins 'A' and the other 3 'B'). if they balance take down the weight of 3 genuine coins average to find the weight of one genuine coin.
- place the 2nd batch of 6 coins with 3 coins each onto the 2 plates on the balance scale. if the scale balances these then this batch of 6 coins are generic and no fake one among them. (label 3 coins as 'C' and the other 3 as 'D'). if they balance take down the weight of 3 genuine coins average to find the weight of one genuine coin.
Note: which scenario occurs for 1st time and 2nd time in precedence is not important.
analyse the 4 groups of 3 coins each with a greater or smaller reading in weight. that group shall be where the fake coin is. then just pick any good coin and weigh it againist the group of 3 coins suspected to hide a fake coin.
it will just take 3 times of weighing on the balance scale even if you know tthe weight of the geniune coin and is lucky to hit the jackpot on the very first try and no more than 5 times if you score snakes eye.
recording of the weight of the coins is permissible since the thread starter only highlighted the times of weighing and never mention anything about recording weights.
Also, it's a balance scale, not a weighing machine, u won't know the value of the weights, so how u gonna even find average -
Originally posted by biangz:
then we must read the thread carefully if the scale is not able to produce a measuring indicator how can our eyes have enough magnification power to focus and tell a difference between 2 weights?
Again, it's not workable, coz u're not allowed to use anything else besides the 12 coin and the scale, meaning labelling coin is not allowed whether u use pencil marker or wadever. He may not mention anything about recording but he did say use only the 12 coins and the scale and [b]no others
Also, it's a balance scale, not a weighing machine, u won't know the value of the weights, so how u gonna even find average[/b]
okay assuming we can tell and have bionic eyes..then here's the probability:-
the minimum number of times to pick out the fake coin is still 3 times assuming you got bulls eye for all the try outs.
1st time: weigh 6 coins found 3 versus 3 balance
or
1st time: weigh 6 coins found 3 heavier and 3 lighter (depends on how you address each group of 3)
either way by now we are able to identify the 6 coins that hide 1 fake coin with the above '1st time' scenario alternate.
2nd time: use 3 coins from the group that you know to contain all genuine coins to weigh against any group (3 coins) of the other 6 coins. assuming you hid bulls eye, the other 3 coins now unbalance the scale and you know which group (of 3s) the fake coin is hiding in.
3rd time: pick one coin from any of the genuine group of coins and pick any coin from the group of 3 coins known to hold one fake coin. assuming you are so lucky again and hit another bulls eye, you got your fake coin now and assuming youhave bionic eye you can tell whether the fake coin is heavier or lighter than a genuine coin from the balance. also since no markers are allowed, from 2nd time onwards we can always keep the genuine coin sampling on the left/ right scale balance so we know where to compare the fake coin withh the genuine coin and tell a different in weight from our 'bionic eye' by the drop/ raise in height. -
i still see 4 weighings no matter wat i do for the worst case scenario, meaning each time u balance the coins, one side will always b heavier than the other
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what biangz how many times must explain to those without common sense and never put heart into any situation or bother to sit down and use brain to think.
12 coins...1 of the 12 is fake and can be identified by been greater or smaller in weight.
split 12 coins into 2 groups of 6 each...from here on never mix the 2 groups together.
take first group and split them into 2 groups of 3. weigh.
1st time:
if:-
the balance evens out, that means the fake coin is in the other group of 6 coins.
if:-
the balance shows a difference, the fake coin is in this group. thus take note of the location of the difference. (e.g the heavier 3 coins are on the left or right scale plate)
2nd time:
either way you can get to this step even if your 1st time weighing is either result of that mentioned above since you have found which group the fake coin is hiding in.
if:-
now assuming your 1st time the balance evens out, just take any group of 3 coins from the suspect 6 coins and weigh it against any 3 coins of the 6 coins that are genuine and if the balance unevens and you got the 3 coins suspected to contain one fake. even if it balances you will still know where the fake coin is since its definitely with the other group.
if:-
now assuming your 1st time shows a difference, just remove the heavier 3 coins and take note of the position of the lighter 3 coins. you know the remaining 6 coins are genuine and just pick any group of 3 coins from them and place it on the balance. if it balance you know the fake coins are heavier and is the group of 3 coins you just take up from the scale. if it still unbalnce it means the 3 coins remaining on the scale hides a fake cooin and the fake coin is lighter than a genuine one.
3rd time:
either way from step 2 now it should not affect your way to step 3 and you should know which group of 3 coins hide a fake one and if the fake coin is indeed heavier or lighter.
now assumine you are a lucky star and hit bulls eye, you now take any one genuine coin and weigh it against any one of the suspected from the group of 3 fake coins and find the weight unbalance.
now assumiming you are down on luck and hit snakes eye you will only get an unbalanced scale on the 3rd coin of the group of 3 coins suspected to hide a fake one then you will take the most 5 times of weighing in total to find the fake coin.
so please use common sense and think it will take 3x or 5x just to find 1 fake coin in a group of 12 coins by displacing the balance 3 - 5 times. -
Haiyo, this question is even blogged by one of the Mathematicians.
I lost the link liao, after I reformat my computer. It's a Science blog. -
is it? so easy thing someone even created a blog for it.....throw face away manOriginally posted by ndmmxiaomayi:Haiyo, this question is even blogged by one of the Mathematicians.
I lost the link liao, after I reformat my computer. It's a Science blog. -
Norh, the link is there, click on the word "paiseh hor" is the answer and it's 3 times lolOriginally posted by mhcampboy:BRAVO!!!
This is mine...
If you knew the fake coin was lighter, then the solution would have an easy explanation. But you do not. So....
Number the coins 1 through 12.
1. Weigh coins 1,2,3,4 against coins 5,6,7,8.
1.1. If they balance, then weigh coins 9 and 10 against coins 11 and 8 (we know from the first weighing that 8 is a good coin).
1.1.1. If the second weighing also balances, we know coin 12 (the only one not yet weighed) is the counterfeit. The third weighing indicates whether it is heavy or light.
1.1.2. If (at the second weighing) coins 11 and 8 are heavier than coins 9 and 10, either 11 is heavy or 9 is light or 10 is light. Weigh 9 against 10. If they balance, 11 is heavy. If they don't balance, you know that either 9 or 10 is light, so the top coin is the fake.
1.1.3 If (at the second weighing) coins 11 and 8 are lighter than coins 9 and 10, either 11 is light or 9 is heavy or 10 is heavy. Weigh 9 against 10. If they balance, 11 is light. If they don't balance, you know that either 9 or 10 is heavy, so the bottom coin is the fake.
1.2. Now if (at first weighing) the side with coins 5,6,7,8 are heavier than the side with coins 1,2,3,4. This means that either 1,2,3,4 is light or 5,6,7,8 is heavy. Weigh 1,2, and 5 against 3,6, and 9.
1.2.1. If (when we weigh 1,2, and 5 against 3,6 and 9) they balance, it means that either 7 or 8 is heavy or 4 is light. By weighing 7 and 8 we obtain the answer, because if they balance, then 4 has to be light. If 7 and 8 do not balance, then the heavier coin is the counterfeit.
1.2.2. If (when we weigh 1,2, and 5 against 3,6 and 9) the right side is heavier, then either 6 is heavy or 1 is light or 2 is light. By weighing 1 against 2 the solution is obtained.
1.2.3. If (when we weigh 1,2, and 5 against 3, 6 and 9) the right side is lighter, then either 3 is light or 5 is heavy. By weighing 3 against a good coin the solution is easily arrived at.
1.3 If (at the first weighing) coins 1,2,3,4 are heavier than coins 5,6,7,8 then repeat the previous steps 1.2 through 1.2.3 but switch the numbers of coins 1,2,3,4 with 5,6,7,8.
LOL LOL!!!
PAISEH HOR!!
btw, it's classified under very difficult segment sia, down there got alot more puzzle, some are really hard, some doable got time then go grind brain abit lor
quite fun de -
y so ma fan?? i juz go mama shop spend all e coins lor...
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also can lah.Originally posted by tt3:y so ma fan?? i juz go mama shop spend all e coins lor... -
1. Pick any 4 coins and put 2 on each side of the balance
2. Repeat Step 1 until the weight balancing device is no longer in balance.
3. one of the 4 coins that were placed on the balance last is the fake coin.
4. take one of these 4 coins and weigh it against one of the remainder coins. -
Yes. Too bad I lost the link, if not, can show you guys.Originally posted by pipipopo:is it? so easy thing someone even created a blog for it.....throw face away man -
Why don't just weigh any ramdom 2 coins and if they balance, it's confirmed both are real.
Then, just remove 1 of the 2 coins and then one by one take each remaining coin and taking turns, place it on the scale. When the scale eventually imbalances, confirm that the coin that was added to the scale is the fake one.