# Reply To: Puzzles

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TRICK (sort of) PUZZLES – ANSWERS

In the interests of full disclosure, here is a list of answers for the many trick puzzles (and the many not-trick puzzles) I have posted in this thread. Question numbers marked with a * were based on, or taken from, the book ‘Mathematical Magic Show’ (see post #92159).

(Note that crossword grids are designed to be viewed with a monospaced font.)

1*) Zero. If three letters are correct, so is the fourth.

2*) Any three points in space are coplanar. To prove uniqueness, we note that the points are not collinear (otherwise the tetrahedron would be a line!), which suffices.

3*) Many, many solutions exist! Here is a list of the solutions which people have posted:

a. (@Yello, #91785; @f6exb_the_frenchy, #91916) We can measure the temperature difference between the top and bottom of the building, and then use an approximation of 1°C per 100m (or 1°C per 500ft) to estimate the height.

b. (@f6exb_the_frenchy, #91896) Use the thermometer as “la croix du bûcheron”.

c. (@bridges, #91928) Use some form of trigonometry using both the thermometer and the height of the building. (Many solutions exist along these lines.)

d. (@bridges, #91928) Tie the thermometer to the end of a long string, as a weight. Lower this string from the top of the building and find the length of string needed to get the thermometer to touch the ground.

Some extra solutions which I know of are below:

e. Similar to (d) but when the thermometer reaches the bottom, time how long it takes for the string to vibrate back and forth. This time is only dependent on the length of the string.

f. (#91801) Drop the thermometer from the top of the building and time how long it takes to fall.

g. Find the owner of the building and ask them to give you the height of the building, in exchange for the thermometer.

h. Exchange the thermometer for a really, really long tape measure at a hardware store. Then use the tape measure to measure the building’s height directly. (This is risky!)

4) There are no other examples, since one of the numbers will be a multiple of 3 at least 9 (which is not prime).

5) It is actually possible without crossing the North Pole! But not for the reasons you might expect… There will exist some line of latitude L near the North Pole with the following property: say that x is the proportion of L traversed by travelling 1km East from a point on L. Then if you travel 1km North, to a line M, the proportion of M you will traverse travelling 1km East is now x plus one full revolution. Any point on L now works as a starting point.

(Essentially travelling 1km East by the time you get to M cancels out the 1km West travelled along L but also results in one extra full revolution of M. This means that the lines of longitude you start and end on are identical.)

6) Eat half of each pill.

7) Three days. Each hen lays a single egg in the time period.

8) There aren’t any stairs! (This was a stupid question…)

9*) She dealt the rest of the pack backwards, starting from D.

10*) In the most stupid crossword ever:

CODE

CODE

CODE

CODE

11*) Zero.

12) Yes, it is possible. Pento needs to start 1km north of any line of latitude that is (1/n)km long, where n is a positive integer.

13) The magician looked at the two numbers at the ends of the chain, which give the missing domino when combined. The reason why dominoes needed to be removed beforehand is because if the volunteer removed a double themselves, both sides of the chain would be the same and all the magician could deduce from the ends of the chain is that a double was removed (but they cannot deduce which one – try it yourself!).

14) No. n and n² have the same parity (oddness/evenness) but left and right pages have different parities.

15) In this case, x=1, which means that x-1 = 2x-2 = 0. So all fractions in the equation shown have a 0 on the denominator, which is illegal.

16) We will consider the general case where ABCD is a parallelogram (but you can reduce this to the rectangle case). Note that ABCD has rotational symmetry of order 2 about its centre O. We claim that O is the required point of concurrency. It suffices to show that A,O,C are collinear, and similarly for the other lines.

Rotate ABCD about O by 180°. Then A and C swap places, so line AC must stay constant (and only line AC). But O also stays constant, so must lie on AC. The other lines have similar stories.

17) Both clues are repeated! The grid is:

POSTS

O_I_I

SIEVE

T_V_V

SIEVE

18) All clues are anagrams (precisely, cyclic permutations) of each other:

TEA

EAT

ATE

19) There are only two distinct words, ANNA and NAAN:

NAAN

ANNA

ANNA

NAAN

20) Yes, it is now possible. The most efficient strategy (I think) is for you to respond with 65. (91 should also work.) It is now possible for you to engineer your moves such that one of the following games occurs:

13 → 65 → 01 → 86 → 43

13 → 65 → 01 → 86 → 02 → 94 → 47

13 → 65 → 05 → 95 → 01 → 86 → 43

13 → 65 → 05 → 95 → 01 → 86 → 02 → 94 → 47

13 → 65 → 05 → 95 → 19 → 57 → 01 → 86 → 43

13 → 65 → 05 → 95 → 19 → 57 → 01 → 86 → 02 → 94 → 47

13 → 65 → 05 → 95 → 19 → 57 → 03 → 87 → 01 → 86 → 43

13 → 65 → 05 → 95 → 19 → 57 → 03 → 87 → 01 → 86 → 02 → 94 → 47

13 → 65 → 05 → 95 → 19 → 57 → 03 → 87 → 29 → 58 → 01 → 86 → 43

13 → 65 → 05 → 95 → 19 → 57 → 03 → 87 → 29 → 58 → 01 → 86 → 02 → 94 → 47

13 → 65 → 05 → 95 → 19 → 57 → 03 → 87 → 29 → 58 → 02 → 62 → 01 → 94 → 47

13 → 65 → 05 → 95 → 19 → 57 → 03 → 87 → 29 → 58 → 02 → 62 → 31 → 93 → 01 → 94 → 47

You win in all cases.

We now show that the first player should always win. Essentially they just need to start with a big enough prime number (any prime above 50 will do): I will choose 89. The second player must respond 1, and then the game could be forced to be one of these:

89 → 1 → 86 → 43

89 → 1 → 86 → 2 → 94 → 47

The first player wins in both cases.