Can You Solve 'The Hardest Logic Puzzle In The World'?

[MassiveJim]

I quite like the extra detail on last week's one. Hadn't thought of that as further possibilities.

For this one:

Spoiler

We know xyz = 36 and x+y+z=n (where n is known to both logicians)

We get possible options:

1*1*36 = 36 / 1+1+36 = 38
1*2*18 = 36 / 1+2+18 = 21
1*3*12 = 36 / 1+3+12 = 16
1*4*9 = 36 / 1+4+9 = 14
1*6*6 = 36 / 1+6+6 = 13
2*2*9 = 36 / 2+2+9 = 13
2*3*6 = 36 / 2+3+6 = 11
3*3*4 = 36 / 3+3+4 = 10

This means n is either 10,11,13,13,14,16,21,38.

If we assume that both logicians knew their apartment number (and so knew n), we can assume n must be 13 (as this is the only number that has more than one solution). For any other value of n, the logician would would have been able to solve the ages of the children with just those first two pieces of information.

If we now take:

1*6*6 = 36 / 1+6+6 = 13
2*2*9 = 36 / 2+2+9 = 13

The third piece of information only tells us that there is an 'oldest' child. This must mean that the children are 2,2 and 9.

oh ok then

 

[Zbornak]

I quite like the extra detail on last week's one. Hadn't thought of that as further possibilities.

For this one:

Spoiler

We know xyz = 36 and x+y+z=n (where n is known to both logicians)

We get possible options:

1*1*36 = 36 / 1+1+36 = 38
1*2*18 = 36 / 1+2+18 = 21
1*3*12 = 36 / 1+3+12 = 16
1*4*9 = 36 / 1+4+9 = 14
1*6*6 = 36 / 1+6+6 = 13
2*2*9 = 36 / 2+2+9 = 13
2*3*6 = 36 / 2+3+6 = 11
3*3*4 = 36 / 3+3+4 = 10

This means n is either 10,11,13,13,14,16,21,38.

If we assume that both logicians knew their apartment number (and so knew n), we can assume n must be 13 (as this is the only number that has more than one solution). For any other value of n, the logician would would have been able to solve the ages of the children with just those first two pieces of information.

If we now take:

1*6*6 = 36 / 1+6+6 = 13
2*2*9 = 36 / 2+2+9 = 13

The third piece of information only tells us that there is an 'oldest' child. This must mean that the children are 2,2 and 9.

Very clever! I like that one - would never have got it though.

 

[wmb]

I quite like the extra detail on last week's one. Hadn't thought of that as further possibilities.

For this one:

Spoiler

We know xyz = 36 and x+y+z=n (where n is known to both logicians)

We get possible options:

1*1*36 = 36 / 1+1+36 = 38
1*2*18 = 36 / 1+2+18 = 21
1*3*12 = 36 / 1+3+12 = 16
1*4*9 = 36 / 1+4+9 = 14
1*6*6 = 36 / 1+6+6 = 13
2*2*9 = 36 / 2+2+9 = 13
2*3*6 = 36 / 2+3+6 = 11
3*3*4 = 36 / 3+3+4 = 10

This means n is either 10,11,13,13,14,16,21,38.

If we assume that both logicians knew their apartment number (and so knew n), we can assume n must be 13 (as this is the only number that has more than one solution). For any other value of n, the logician would would have been able to solve the ages of the children with just those first two pieces of information.

If we now take:

1*6*6 = 36 / 1+6+6 = 13
2*2*9 = 36 / 2+2+9 = 13

The third piece of information only tells us that there is an 'oldest' child. This must mean that the children are 2,2 and 9.

Exactly what I'd scribbled out on a pad.

 

[Bill Door]

Although I am sure that is the correct answer, the 3rd piece of information seems insufficient to deduce the ages with absolute certainty.

Spoiler

Although unlikely, it would be possible to have two children aged 6 but with one being 9-11 months older than the other. In this situation it would still be normal to refer to one as being the eldest.

 

[rubberduck]

Ok here goes:

God problem

Spoiler

Question 1 To God A - If I asked you if God B is random, would you say ja?

Random: If God A is random, then you will randomly get Ja/Da


True: If God A is true, then by asking him a question in the format 'If I ask you X, would your answer be Ja', then we get:

Assuming ja=yes and da=no.
If he responds with ja(yes), this means the truthful answer to X is ja(yes)
If he responds with da(no), this means the truthful answer to X is da(no)

Assume ja=no and da=yes.
If he responds with ja(no), this means the truthful answer to X is da(yes)
If he responds with da(yes), this means the truthful answer to X is ja(no)

False: If God A is false, then by asking him a question in the format 'If I ask you X, would your answer be Ja', then we get:

Assuming ja=yes and da=no.
If he responds with ja(yes), it means that his answer to X is da(no). But because he is lying, this means the truthful answer to X is ja(yes)
If he responds with da(no), it means that his answer to X is ja(yes). But because he is lying, this means the truthful answer to X is da(no)

Assume ja=no and da=yes.
If he responds with ja(no), it means that his answer to X is ja(no). But because he is lying, this means the truthful answer to X is da(yes)
If he responds with da(yes), it means that his answer to X is da(yes). But because he is lying, this means the truthful answer to X is ja(no)


As you can see, in both of the above cases the God A will say 'Ja' when God B is the random god.

So if God A say 'Ja', you know that God B is the random god, or you are talking to the random god already. Either way you know that God C cannot be random.

Similar applies to if he answers 'Da'. This tells you that you're either already talking to the random God, or that God B is not random. Either way, you can deduce that B is not random.


Question 2 To the non-random God identified above (ie True or False god)

If I asked you if you're true, would you say ja?

Similar to above logic, both gods will answer Ja=Yes and Da=No.

Question 3 To the same god above

If I asked you if God A is random, would you say ja?

Again, Ja=Yes, Da=No.

You now know the identity of all gods.

I think you googled the answer.