Riddles Options

[vikram]

Vikram, just a quick idea:
Assuming that coin is heavier,
1) put six coins each in the two sides of a weighing machine. One goes down. Discard the six on the lighter pan.
2) Of these six, put three each on the two pans. One will go down. Discard the lighter 3.
3) Put just TWO - one each on the two pans. If one of them goes down, that is the heavy one. If both balance, then the one that's left outside is the heavy one.

Wat if coin is lighter ??

[Mandrake]

Ditto procedure. Only in that case, discard the heavier group

[vikram]

Ditto procedure. Only in that case, discard the heavier group

not correct but great reply

[Mandrake]

Not correct?

But doesn't it give the right coin?

[amicable]

Can't we weigh it 4 times?

I got solution to one condition, that too I am not sure if it's correct:

Here's my guess:

1. Take 8 coins. Weigh it keeping 4 on each side.

2. If they are similar in weight then the counterfeit coin is one of the 4 coins we have not considered

i. So, take 2 from that, keep one on each side. If they are similar then, the coin we want is from the other two we didnt consider.
a. Replace a coin from the weighing machine with one coin kept aside. If they are same, then the last coin left is counterfeit coin.
b. Else, the coin we just replaced is the counterfeit coin.

ii. If in 2nd weighing there is weight difference then its one of those two coins. Replace one with one of the two coins kept aside.
a If the weight is still not the same, then the coin we didnt replace is the counterfeit coin
b. If the weight is the same, then the coin we replaced is the counterfeit coin.

3. Dont know what to do if condition 2 fails. I can identify if I get to weigh it 4 times

Is this partial answer correct??

[vikram]

your partial answer is partially correct ...but you have to examin diff scenarios....good try

[vivekpm]

Can I mark the coins? Or label it after weighing. I mean does your puzzle allow ways to distinguish coins which are heavy or light or equal in first weighing?

Cheers,

[vikram]

Can I mark the coins? Or label it after weighing. I mean does your puzzle allow ways to distinguish coins which are heavy or light or equal in first weighing?

Cheers,

Yes surly you can do this

[vivekpm]

Can I mark the coins? Or label it after weighing. I mean does your puzzle allow ways to distinguish coins which are heavy or light or equal in first weighing?

Cheers,

Yes surly you can do this

Then here is the solution:

Make 3 sets of 4 coins each (Set1, Set2 and Set3)
Weigh Set1 and Set2.

1. If Set1=Set2 :
We know that Set 3 is defective.
Divide Set3 into two sets of two coins each (Set3a and Set3b)
Take Set3a and weight it against two coins from either Set1 or Set2.

a.If they are equal:
Set3b is defective.
Take one coin from Set3b and weigh it against one from from either Set1 or Set2 or Set 3a.

i.If they are equal:
Coin left out in Set3b is defective

ii.If not equal:
Last coin used from Set3b is defective

b. If they are not equal:
Set 3a is defective.
Take one coin from Set3a and weigh it against one from from either Set1 or Set2 or Set 3b.

i.If they are equal:
Coin left out in Set3a is defective

ii.If not equal:
Last coin used from Set3a is defective.

2. If Set1 NOT EQUAL TO Set2:
Set 3 is fine. Label all coins from Set 3 as "ND" (Not Defective)
Label all coins from the pan which is heavier (containing either Set1 or Set2) as "H"
Label all coins from the pan which is lighter as "L"
Take 2H and 2L coins (lets say Set4) and weigh them against 3 "ND" coins and 1L coin (Say Set5).

a.If Set4 is lighter,
One of the 2L coins used in Set4 is defective and is the lighter coin.
Weigh one of these against any ND coin and we will get the defective one.

b.If Set4 is heavier,
One of the 2H is defective and is the heavy coin.
Take one of the 2H coins used above and weigh it against one of the ND coin and you will get the defective coin.

c.If Set4 is neither lighter or heavier (that is equal to 3 ND coins and 1L coins),
Then one of the left out 2H or 1L coins are defective.
Weigh 2H coins against each other. The one which goes down is defective.
If they are equal, the lighter coin left out is defective.

Cheers,

[vikram]

Can I mark the coins? Or label it after weighing. I mean does your puzzle allow ways to distinguish coins which are heavy or light or equal in first weighing?

Cheers,

Yes surly you can do this

Then here is the solution:

Make 3 sets of 4 coins each (Set1, Set2 and Set3)
Weigh Set1 and Set2.

1. If Set1=Set2 :
We know that Set 3 is defective.
Divide Set3 into two sets of two coins each (Set3a and Set3b)
Take Set3a and weight it against two coins from either Set1 or Set2.

a.If they are equal:
Set3b is defective.
Take one coin from Set3b and weigh it against one from from either Set1 or Set2 or Set 3a.

i.If they are equal:
Coin left out in Set3b is defective

ii.If not equal:
Last coin used from Set3b is defective

b. If they are not equal:
Set 3a is defective.
Take one coin from Set3a and weigh it against one from from either Set1 or Set2 or Set 3b.

i.If they are equal:
Coin left out in Set3a is defective

ii.If not equal:
Last coin used from Set3a is defective.

2. If Set1 NOT EQUAL TO Set2:
Set 3 is fine. Label all coins from Set 3 as "ND" (Not Defective)
Label all coins from the pan which is heavier (containing either Set1 or Set2) as "H"
Label all coins from the pan which is lighter as "L"
Take 2H and 2L coins (lets say Set4) and weigh them against 3 "ND" coins and 1L coin (Say Set5).

a.If Set4 is lighter,
One of the 2L coins used in Set4 is defective and is the lighter coin.
Weigh one of these against any ND coin and we will get the defective one.

b.If Set4 is heavier,
One of the 2H is defective and is the heavy coin.
Take one of the 2H coins used above and weigh it against one of the ND coin and you will get the defective coin.

c.If Set4 is neither lighter or heavier (that is equal to 3 ND coins and 1L coins),
Then one of the left out 2H or 1L coins are defective.
Weigh 2H coins against each other. The one which goes down is defective.
If they are equal, the lighter coin left out is defective.

Cheers,

HEY !!! You did it ....GRAET !!!

[Mandrake]

Vikram I am still not clear why my solution is not correct. After all, it does give the right answer

[vivekpm]

Vikram I am still not clear why my solution is not correct. After all, it does give the right answer

I guess it is because you should know before hand whether the defective coin is lighter or heavier. What if you don't know that.

Cheers,

[vikram]

No Mandrak !! your answer fails on the firststep...
both half are not going to be equal...so which half will you select...lighter one or the heavier one??....

Did you got it now !!

[Mandrake]

Theek hai - concede defeat

(What to do? Examiner is saying that he only doesn't know whether defective coin is heavy or light Then how did he know one coin was defective? )

[vikram]

OK here is another one !!

If you use a certain formula on 13, you end up with 7.

Under the same formula, 2352 becomes 16, 246 becomes 14, 700 turns into 16, and 1030 becomes 14.

What would 9304 become?