# What's the answer to these logic puzzles:

All are from the movie “Fermat’s Room”:

A candy merchant receives 3 opaque boxes. One box contains mint candies, another contains anise candies, and the last box contains a mixture of mint and anise. The boxes are labeled Mint, Anise, and Mixed. All of the boxes are labeled incorrectly. What is the minimum number of candies the merchant will have to sample to correctly label each box?

The fifth enigma is stated as follows: “A student asks his teacher, ‘How old are your three daughters?’ The teacher replies, ‘If you multiply their ages you get 36. If you add their ages you get my house number.’ ‘I am missing a detail,’ protests the student. ‘Oh yes,’ says the teacher, ‘the older one plays piano.’ How old are the 3 daughters?”

The seventh enigma is an elementary algebra problem with a twist. A mother is 21 years older than her son. In 6 years, the son will be one-fifth his mother’s age. The enigma asks what the father is currently doing.

Some guesses:

1. Don’t know how you can answer the first one without knowing the % of mint and anise in the mixed box

2. 9, 2, 2

3. jerking off to internet porn?

Assuming the merchant knows that all the labels are incorrect, one.

The minimum is three - he takes two from one box and gets one mint, one anise. Now he knows that one is mixed, and so he needs to take just one from one of the other boxes.

Of course, that’s the minimum - he might not get so lucky.

This assumes the mixture is 50/50

The first one seems the easiest to me (I completely missed the clue about twins in #2, gj there):

ug…I saw a flaw in my reasoning, I’ll try again.

The first one is an old chestnut I’ve seen many a time.

Only one draw is necessary – one draw from the box labelled “mixed” will tell you what solid flavor is in there (ie, if a mint one is drawn, that clearly has to be the “all mint” box).

Once that one is properly labelled, the other two are forced (to continue my example, the one labelled ‘anise’ can’t be anise, and can’t be mint, so that one is in reality the “mixed” one).

The other two I’m not sure about – why couldn’t the ages by 3,3, and 4, or 2,2,9?

(on edit: gikes! beaten by Superhal!)

because 3 and 4 year olds don’t play the piano?

3 boxes with incorrect “labels”:
#1. “Mixed” (Must be anise or mint)
#2. “Mint” (Must be anise or mixed)
#3. “Anise” (Must be mint or mixed.)

Obviously, we must sample the “mixed” box, as this will tell us exactly what is actually in it, as it can’t be mixed. Let’s say anise.

Now we have:
#1. Anise.
#2. “Mint” (must be Mint or Mixed.)
#3. “Anise” (Must be Mint or Mixed.)

Box #2 must be Mixed, as the label must be incorrect. That leaves Mint as the “anise” box.

First one:

The minimum is one. Taste a candy from the box labeled “Mixed.” It will be either mint or anise. You know that this box is labeled incorrectly, so put either the “Mint” or “Anise” label on that box. Whichever one of “Mint” and “Anise” is left is the “Mixed” box. The last box gets the final label, whichever flavor the “Mixed” box was not.

Second one:

Yes, 9, 2, and 2 is correct. But another possible answer for the factors of is 6, 6, and 1, which is why the student needed to know that there is only one “older” daughter.

Third one:

I did the algebra, and the son is currently aged minus 9 months. What the father is doing right now is left as an exercise to the reader.

It doesn’t matter - even if the mixture is 99/1, three samples is still the minimum required to make the determination.

>>Yes, 9, 2, and 2 is correct. But another possible answer for the factors of is 6, 6, and 1, which is why the student needed to know that there is only one “older” daughter.

But, how is 4, 3, 3 eliminated?

nm i re read what you said.

Hi.

As someone upthread suggested, couldn’t you have “3, 3, and 4”?

This still leaves toy with just one “older” daughter, and while it’s unlikely that a four year old can play the piano, it’s not outside the realm of possibility.

It may be known in the movie that the house number is 13.

Well 4 year olds don’t play the piano really either. Instead of 9, 2, and 2 I see no reason why it couldn’t be 9, 4, and 1. Or 6, 3, and 2.

This depends on whether he’s the biological father.

Or 36, 1, 1, for that matter.

Here’s all the puzzles:
Each of the enigmas relayed through the PDA have a one-minute deadline and a simple solution—that is, they do not rely upon any advanced mathematical knowledge. After the candy puzzle, the second enigma is a series of 169 ones and zeros which, when placed in a 13-by-13 square and interpreted as pixels, form an image of a skull.

The third enigma describes a sealed room containing one light bulb. Outside of the room there are three switches, only one of which operates the bulb. The puzzle solver begins outside the room, able to operate the switches in any way he sees fit, but when the door is opened for the first time, he must determine which switch operates the light. The enigma is to determine how this is to be done.

The fourth enigma is to describe how one can use a four-minute hourglass and a seven-minute hourglass to measure a period of nine minutes.

The fifth enigma is stated as follows: “A student asks his teacher, ‘How old are your three daughters?’ The teacher replies, ‘If you multiply their ages you get 36. If you add their ages you get my house number.’ ‘I am missing a detail,’ protests the student. ‘Oh yes,’ says the teacher, ‘the older one plays piano.’ How old are the 3 daughters?”

The sixth enigma is a Knights and Knaves puzzle. Two doors are guarded by two men, one of whom always lies and one of whom always tells the truth; however, the puzzle solver does not know which man is which. One of the doors leads to freedom and one to captivity. The enigma is to determine a single question that, if asked of one of the guards, would reveal the door to freedom with certainty.

The seventh enigma is an elementary algebra problem with a twist. A mother is 21 years older than her son. In 6 years, the son will be one-fifth his mother’s age. The enigma asks what the father is currently doing.