MovieChat Forums > La habitación de Fermat (2007) Discussion > What was your favorite puzzle?

What was your favorite puzzle?

posted 15 years ago by locoowl
23 replies | jump to latest

I was most intrigued by the "enigmas" presented during this file. Some of them I had heard or variations of them - the shepard, the wolf, the hen, the cabbage; the guards from the Land of the True and the Land of the False - but some I had not heard of.

The one that fascinated me the most was the puzzle about the ages of the three children. I actually have figured it out (just because we were given the answer does not mean we understand how the answer is derived).

The one about what the father was doing was rather coarse, but hey, this is cinematic realism, right?

Which ones were your favorites?

[–] skymyr 15 years ago

Could you please explain to me the puzzle about the ages of the children?

Thanks!

[–] locoowl 15 years ago

Could you please explain to me the puzzle about the ages of the children?

The teacher has 3 children. One of his students asks their ages. The teacher's replay:

If you multiply their ages together, the result is 36.
If you add their ages together you get your [the student's] house number.

The student replies that he needs more information before he can determine the answer. The teacher then replies:

Oh yes! The oldest one plays the piano.

*******************************************************
The key to this is how many combinations of 3 numbers which when multiplied together will total 36?

36 = 6 X 3 X 2 ; Sum = 11
36 = 4 X 3 X 3 ; Sum = 10
36 = 6 X 6 X 1 ; Sum = 13
36 = 9 X 2 X 2 ; Sum = 13
36 = 9 X 4 X 1 ; Sum = 14
36 = 12 X 3 X 1 ; Sum = 16
36 = 18 X 2 X 1 ; Sum = 21

There are 2 combinations which have the same sum. Since the student claims he needs more information, then one of those two must be the answer. When the teacher says that "the oldest child plays the piano," we now have our answer. The only combination where there is one oldest child is the one with 9, 2, and 2. Oh and the student's house number is 13.

Hope that helps!

[–] houndtang75 15 years ago

That doesn't quite follow - you could still say there was an 'oldest child' in most of those combinations.

[–] grambax 15 years ago

But both student and teacher know the student's house number, so it can only be one of the two where the sum is 13. And only one of those has an oldest child. In the other, the older children are six year old twins. (OK, technically one of the twins would probably be a few minutes older, but its a logic problem, not reality, so we can assume that that isn't an option, as the answer must be calculable).

I though it was a really good film BTW. I thouroughly enjoyed it. Shame to see it playing to tiny audiences in the UK. It deserves far better.

[–] locoowl 15 years ago

That doesn't quite follow - you could still say there was an 'oldest child' in most of those combinations

Sorry! I should have been a bit clearer. The reason the student has to have more information is that there are two combinations of factoring 36 with three numbers which have the same sum: 13.

One of those combinations is 6X6X1 and the other is 9X2X2. Only one of these has an "oldest" child, as the other combination has the elder children being the same age. Sure there is an oldest child in all the other combinations but we are not concerned with them.

Hope that helps!

[–] richie_patrese 15 years ago

Thanks for this - I'd been wondering what the significance of the piano was...of course it's a red herring but now I understand why.

[–] fraser65 15 years ago

I really liked that puzzle - of course it could have been possible to have an elder child even if he had two kids who were 6, say if one was 6 years and 1 day old, and the other was 6 years and 364 days old.

Another possible combination of ages that factor up to 36 is 1x1x36 (of course, that would still not be a valid answer to the riddle).

Yeah, I'm anal, I know.

I'd heard most of those riddles before, I love riddles involving lateral thinking rather than just mathematics. Awesome film!

[–] locoowl 15 years ago

of course it could have been possible to have an elder child even if he had two kids who were 6, say if one was 6 years and 1 day old, and the other was 6 years and 364 days old.

Yes, we can come up with all sorts of exceptions and extra baggage, but that defeats the simplicity of the mathematical/logic puzzle. Things are sort of idealized in that world.

Cheers!

[–] Eddie C. 13 years ago

Twins may be born the same day, but not at the same time. It's possible that the father considers the twin born first to be the eldest---but most would consider them to be exactly the same age.

[–] dayvit78 7 years ago

I missed the logic part of that and assumed the reason the 9 year old one was correct was because 6 year olds can't play the piano. But maybe they can, what do I know? :)

[–] YarkoTheWarbler 15 years ago

Mine was:

You are outside a room, beside a closed door. There are 3 switches on the wall.

You are told "There is a single common light bulb on the wall behind these switches, it is now off. One of these switches controls that light. You have 5 minutes to flick these switches as much as you want. Then we open the door and you tell me which switch controls the light on the room"

Excellent for thinking outside the box.

[–] alucardmatenrow 15 years ago

I loved the last one, but i can not remember it exactly, does someone has?

[–] zwixxx 15 years ago

Switches ABC

(1) Switch A on, wait 4min59s
(2) Turn A off, Turn B on, Enter Room
-> if light on then B
-> if light off but bulb hot then A
-> if light off then C

QED :)

Similar to the 3 switches & 3 bulbs puzzle

Switches ABC, Bulbs XYZ

(1) Turn A on, wait 4min59s
(2) Turn A off, Turn B on, Enter Room
-> whichever bulb is on is controlled by B
-> whichever bulb is off but hot is controlled by A
-> and whichever bulb is off and cold is controlled by C.

[–] rockstar-40 9 years ago

Hope it wasn't a fluorescent bulb though...

[–] pjhllnd 14 years ago

The only one I didn't get was the age of the 3 children, I wish I put more time into that now.

Someone asked what the final riddle was, it went someting like this:

A mother is 21 years older than her son. In 6 years, the son will be one-fifth his mother's age.

The enigma asked what the father is currently doing.

[–] badnomad 14 years ago

SIMPLE algebraic equation:

A mother is 21 years older than her son...
M=21+S

In 6 years, the son will be one-fifth his mother's age....
S+6=(M+6)/5

Substitute M for 21+S...
S+6=((21+S)+6)/5 (add 21+6)...
S+6=(27+S)/5 (multiply both sides of the equation by 5)...
5S+30=27+S (subtract 27 from both sides)...
5S+3=S (subtract 5S from both sides)....
3=-4S (divide both sides by -4)....
-3/4 = S

S is currently -3/4 years or 9 month BEFORE being born. So the dad was copulating!

[–] JohanPar 14 years ago

My favourite puzzle in the movie was the one with brilliant mathematical minds not being able to solve puzzles manageable for high school students. I had heard these before and found it a bit ridiculous.

[–] house1985 14 years ago

At first, they weren't giving too much attention to the puzzles, but when the walls started moving and they realized they were going to die unless they solved the riddles, they were in panic and scared... try to think clearly in such a state. Even the brightest of the minds wouldn't be have to think straight.

Plus, don't forget the time pressure.

[–] mariusica 14 years ago

The best one was the one with the candies. Everybody at work was intrigued by it.
"A merchant receives 3 boxes of candies, one with chocolate, one with mint and one mixed. Each box is labeled wrong. How many candies does he have to take out of the boxes in order to label all correctly ?". Loved it!

[–] orland1995 9 years ago

I just watched the film and loved it but did not actually understand the solution to this puzzle. How can 1 candy be enough to label all boxes correctly?

[–] a-mar 9 years ago

By relying on the fact that every label must be wrong and then using process of elimination.

Your boxes are:
A 1
B 2
C 3

And I tell you that those labels are wrong. So you take one piece of candy 1 from A, and find that it tastes like C.

Now where does candy 2 go? C is taken, and it can't be B because that label was wrong. Therefore 2 must be A.

That leaves only one box left. Candy 3 must go in B.

[–] orland1995 9 years ago

ahh, thanks! I did not understand that every label is wrong at the start.

[–] dayvit78 7 years ago

yea, in these type of math problems, it's actually very important to pay attention to every word... it becomes a reading comprehension problem more than a math problem. Otherwise, you'll assume too much or too little and won't be able to solve it.