(1) 
Turn on switch 1 for a while and turn it off. Turn on
switch 2. Go to the other room. The bulb that is off but feels warm is controlled
by switch 1. The bulb that is on is controlled by switch 2. The remaining bulb
is controlled by switch 3.
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(2) 
They both contain the same amount of foreign substance.
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(3) 
Burn rope 1 at both ends, at the same time burn rope 2
at one end. As soon as rope 1 burns out (30 minutes), burn rope 2 at the other
end. When rope 2 burns out, the time is 45 minutes.
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(4) 
i) Ball numbered as 1~9.
weigh 123 and 456, if balanced
then weigh 7 and 8, if balanced
then 9 is bad.
else weigh 7 and 9, if balanced
then 8 is bad.
else 7 is bad.
else (suppose 123 is heavier), weigh 14 and 25, if balanced
then weigh 3 and 7, if balanced
then 6 is bad.
else 3 is bad.
else (suppose 14 is heavier), weigh 1 and 7, if balanced
then 5 is bad.
else 1 is bad.
ii) Ball numbered as A, B, C, D, ...
weigh ABCD and EFGH, if balanced
then weigh IJK and ABC, if balanced
then L is bad
else (suppose IJK is heavier), weigh I and J, if balanced
then K is bad.
else the heavier one is bad.
else (suppose ABCD is heavier), weigh ABE and CDF, if balanced
then weigh A and G, if balanced
then H is bad.
else G is bad.
else
if ABE is heavier, we know that the bad one is A, B or F;
if ABE is lighter, we know that the bad one is E, C or D.
suppose ABE is heavier, weigh AF and IJ, if balanced
then B is bad.
else if AF is heavier,
then A is bad.
else F is bad.
Following the instruction above, one can also tell whether the bad one is
heavier or lighter without any extra weighing.
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(5) 
Fasten one end of the rope on the tree on the bank.
Carry the other end and go along the bank of the lake. As it is a lake,
eventually, you will come back to the tree and at that time the rope is already
circled around the tree on the island.
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(6) 
Tom will always win if Jerry goes first. Every time Jerry goes,
Tom shall take an appropriate step that leaves PAIRS (either 1 or 2) of piles with the
same number of chess pieces in each pile of a pair (say 3344, 0044, 0011, ...). Finally,
the 00XX will become 0000.
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(7) 
Orientation of the word "coin" will be the same as before
(not upsidedown).
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(8) 
For rooms 2, 3, 5, and 6, A can win B. For rooms 1 and 4, A can
never win. (You figure out how.)
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(9) 

(10) 
For squareshaped cake, we need N+M cuts. For round cake, N+M1
cuts. If we have N, M, K, then the answers are N+M+K and N+M+K2 respectively. (You figure
out how.)
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(11) 
The tangent of lowerleft angle of the red triangle is 3/8 while
that of the green triangle is 2/5. They are not equal, which means the "diagonal" of the
big triangle is not a straight line. It bends down a bit in the first picture, and bends
up a bit in the second. The difference of these two tiny bends makes you twist your brain
until now.
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(12) 
I found 12 naturally recognizable faces with their noses circled in red.
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(13) 
I can only find 4 with their heads circled in red.
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I can find 5 for this one. (Thanks to Ana from Portugal for the original pciture.)



(14) 
