## Answer to Math Puzzles in Fermat's Room

12/19/2013

(Read the original questions in Math
Puzzles in the Film *Fermat's Room*)

**Answer:**

1. The numbers are in** alphabetical
order**. (See the first one or two letters in their spelling: Eight, Five, Four,
Nine, One, Seven, Six, Three, Two, Zero)

2. **First you turn both hourglasses
over to let the sands drop**. At this point, the time can be counted (judging by
the amount of sands left in the upper and lower parts of the glass) in two hourglasses
are respectively and ;

**After 4 minutes** passes, the time (amount of sands) left in
two hour glasses are: and . At this point, **turn only the 4-minute hourglass over** to
continue the time counting;

**After another 3 minutes**, the time (amount of sands) left in
two hour glasses are: and . This time, **turn only the 7-minute hourglass over** to
continue;

**After 1 more minute**, the time (amount of sands) left in two
hour glasses are: and . At this point, **again turn only the 7-minute
hourglass over** to count for the last one minute.

When there is no sand left in the upper part of the 7-minute hourglass ( and ), you can tell exactly nine minutes has passed. (4 + 3 + 1 + 1 = 9)

3. The answer is that **you only need to take
one candy out of the box labeled "Mixture"** to know everything. FIG. 1
is a table which lists every possible situation for each box.

(FIG. 1)

There is one premise given in the puzzle that is very important: NONE of the labels on the boxes are correct. So you can be sure that the box labeled "Mixture" does not contain a mixture of both candies—this box contains either purely chocolate candies or purely mint candies. Suppose the candy you take from this box is chocolate, then this box should be labeled "Chocolate". Since the box of Chocolate is verified, you know that the box labeled "Mint" contains neither mint candies nor chocolate candies, so this box can only be a box of Mixture. And therefore, the only other box left is sure to be a box of Mint.

The logic is the same if the candy you take from the box wrongly labeled
"Mixture" is a mint candy (see FIG. 2): Given that this box cannot be
Mixture and you take a mint candy out of it, it is sure to be a box of Mint.
Once the box of Mint is verified, you will not get any mint candy from the box
wrongly labeled "Chocolate", so this box can be nothing but Mixture.
Then you will know the only other box left is a box of Chocolate.

(FIG. 2)

4. **You only need to enter the room
once** to decide which switch controls the lamp bulb. This puzzle is actually not
all about math, but involves common sense. The crucial factor here is not the
light of the bulb you can see, but the temperature you can feel from the bulb. The
quickest way to solve the puzzle is to turn the first switch on and waif for a
while. Then turn it off and turn on the second switch, and enter the room immediately.
If you find the bulb is on (there's light giving out from it), you can be sure
that it is controlled by the second switch; if the bulb (light) is out, you can
touch it to feel if it is hot. If the bulb is out but feels hot, then it must
be the first switch that controls the bulb; if the bulb is out and feels cold
as a common glass product should be, it can be certain that none of the first
or the second switches work, thus it must be controlled by the third switch.

5. Suppose the mother's age is x and the son's age is y, you can make a binary linear equation:

x – y = 21

x + 6 = 5 x (y + 6)

Solving the equation, you get y = -3/4.

-3/4 x 12 = -9, that is, **the son is -9 months old**, which
means he's just formed in the mother's womb—the son has not been born yet. So at
this time the father is together with the mother.

6. It is true that you are still not able to solve a linear equation with three variables by knowing only the product and the sum of them (actually you even don't know the exact number of the sum). In this case, you have to list all the possible combination of the three ages, the product of which is 36:

1 x 1 x 36 = 36, 1 + 1 + 36 = 38

1 x 2 x 18 = 36, 1 + 2 + 18 = 21

1 x 4 x 9 = 36, 1 + 4 + 9 = 14

1 x 6 x 6 = 36, 1 + 6 + 6 = 13

2 x 2 x 9 = 36, 2 + 2 + 9 = 13

2 x 3 x 6 = 36, 2 + 3 + 6 = 11

3 x 3 x 4 = 36, 3 + 3 + 4 = 10

Observe the sums of these combinations. There are two combinations that have the same sum 13. If the sum is a number other than 13, then A can directly get the correct age combination from the list, and he would not have asked B for more information. So you can infer here that A's room number is exactly 13, so he could still not get to the answer when B told him the sum of the three ages. Knowing that A's room number is 13, you can be sure that the three children's age combination is either (1, 6, 6) or (2, 2, 9).

B's last hint seems to have nothing to do with his
children's ages, but actually it gives a very important premise in this puzzle:
"My oldest child plays the piano" suggests that B has only one oldest
child. In other words, there is no such case that two of his children are not
only at the same age but are both older than the youngest one as well. So the
three children's ages will not be (1, 6, 6). The only one combination left is (**2,
2, 9**). This is the correct answer.

7. No matter which guard you are
going to ask, the only one question you should ask is: "**If I ask the other
guard which of the doors leads to freedom, which door do you think he will be pointing
to?**" If the guard you ask is the honest one, he will point to the door
that is blocked because he knows the other guard will lie to you; if the guard
you ask is the liar, he will also point to the blocked door due to the same
reason. So you just need to pick the opposite door, which exactly will lead you
to freedom.

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