Answer
:
The 8 cheeses can be removed in 33
moves, 10 cheeses in 49
moves, and 21
cheeses in 321 moves.
I will give my general method of solution in the
cases of 3, 4, and 5 stools.
Write out the following table to any
required
length:—
Stools. 
Number of
Cheeses. 
3 
1 
2 
3 
4 
5 
6 
7 
Natural
Numbers. 
4 
1 
3 
6 
10 
15 
21 
28 
Triangular
Numbers. 
5 
1 
4 
10 
20 
35 
56 
84 
Triangular Pyramids. 

Number of Moves. 
3 
1 
3 
7 
15 
31 
63 
127 
4 
1 
5 
17 
49 
129 
321 
769 
5 
1 
7 
31 
111 
351 
1023 
2815 
The first row contains the natural
numbers.
The second row is
found by
adding the natural numbers together from the beginning.
The numbers in
the third row are obtained by adding together the numbers in the second
row from the beginning.
The fourth row contains the successive powers
of
2, less 1.
The next series is found by doubling in turn each number of
that series and adding the number that stands above the place where you
write the result.
The last row is obtained in the same way.
This table
will at once give solutions for any number of cheeses with three
stools,
for triangular
numbers with four stools, and for pyramidal numbers with
five stools. In these cases there is always only one method of
solution—that is, of piling the cheeses.
In the case of three stools, the
first and fourth rows tell us
that 4
cheeses may be removed in 15 moves, 5 in 31, 7 in 127.
The second and
fifth rows show that, with four stools, 10 may be removed in 49, and 21
in 321 moves.
Also, with five stools, we find from the third and sixth
rows that 20 cheeses require 111 moves, and 35 cheeses 351
moves.
But
we
also learn from the table the necessary method of piling.
Thus, with
four
stools and 10 cheeses, the previous column shows that we must make
piles
of 6 and 3, which will take 17 and 7 moves respectively—that
is, we
first pile the six smallest cheeses in 17 moves on one stool; then we
pile the next 3 cheeses on another stool in 7 moves; then remove the
largest cheese in 1 move; then replace the 3 in 7 moves; and finally
replace the 6 in 17: making in all the necessary 49 moves.
Similarly we
are told that with five stools 35 cheeses must form piles of 20, 10,
and
4, which will respectively take 111, 49, and 15 moves.
If the number of cheeses in the case
of four stools is not
triangular,
and in the case of five stools pyramidal, then there will be more than
one way of making the piles, and subsidiary tables will be
required.
This
is the case with the Reve's 8 cheeses.
But I will leave the reader to
work out for himself the extension of the problem.
