by weight can be determined using no more than 1000000 operations and that this is the only coin
whose position by weight can be determined using this machine. (Baltic Way, 1999) [hide=”Solution”]
Divide the 1999 coins into a set of 3 coins and a set of 1996 coins. We use the machine on the set of
3 coins to determine the middle coin.
We replace the middle coin with a coin from the set of 1996
coins, and repeat. After 1997 weighings, we can identify the heaviest and lightest coin among all 1999
coins (although we don’t know which is which). Discard these two extreme coins. We are left with
1997 coins. We repeat the procedure above with these 1997 coins, using 1995 weighings, to identify
the secondheaviest and secondlightest coin. We then discard these two coins, and repeat, until we
are left with the 1000th coin. Thus, we can identify the 1000th coin with 1997 + 1995 +
· · ·
+ 1 = 999
2
weighings.
We cannot, however, identify any other coin.
To prove this, label the coins
a
1
,
a
2
,
. . .
,
a
1999
. For each weighing, we weigh some three coins
a
i
t
,
a
j
t
, and
a
k
t
, and determine the middle coin
a
m
t
. We can list the results of these weighings in a table.
Coins
Middle Coin
a
i
1
, a
j
1
, a
k
1
a
m
1
a
i
2
, a
j
2
, a
k
2
a
m
2
.
.
.
.
.
.
a
i
n
, a
j
n
, a
k
n
a
m
n
Suppose that we decide what the
k
th coin is based on these results.
If we reverse the order of the
coins (from lightestheaviest to heaviestlightest), then the results above are consistent with this new
ordering. Thus, we cannot distinguish between the
k
th coin and (2000

k
)th coin. [/hide] [b]Olympiad
Combinatorics, Problem 5[/b] On an infinite chessboard, a game is played as follows.
At the start,
n
2
pieces are arranged on the chessboard in an
n
×
n
block of adjoining squares, one piece in each
square.
A move in the game is a jump in a horizontal or vertical direction over an adjacent occu
pied square to an unoccupied square immediately beyond.
The piece which has been jumped over
is removed.
Find those values of
n
for which the game can end with only one piece remaining
on the board.
(IMO, 1993) [hide=”Solution”] We color the squares with three colors (say colors
1, 2, and 3), with the following periodic pattern.
[asy] unitsize(1 cm); fill((0,0)–(1,0)–(1,1)–(0,1)–
cycle,white); fill(shift(1,0)*((0,0)–(1,0)–(1,1)–(0,1)–cycle),gray(0.7)); fill(shift(2,0)*((0,0)–(1,0)–(1,1)–
(0,1)–cycle),black); fill(shift(0,1)*((0,0)–(1,0)–(1,1)–(0,1)–cycle),gray(0.7)); fill(shift(1,1)*((0,0)–(1,0)–
(1,1)–(0,1)–cycle),black); fill(shift(2,1)*((0,0)–(1,0)–(1,1)–(0,1)–cycle),white); fill(shift(0,2)*((0,0)–(1,0)–
(1,1)–(0,1)–cycle),black); fill(shift(1,2)*((0,0)–(1,0)–(1,1)–(0,1)–cycle),white); fill(shift(2,2)*((0,0)–(1,0)–
(1,1)–(0,1)–cycle),gray(0.7)); for(int i = 0; i ¡= 3; ++i) draw((i,0)–(i,3)); draw((0,i)–(3,i)); [/asy] Let
n
i
be the number of pieces on squares with color
i
.
For each move, each
n
i
changes by 1.
It fol
lows that the parity of each
n
i
changes.
If there is one piece remaining at the end, then one
n
i
is
odd and the other two are even.
However, if
n
is divisible by 3, then all the
n
i
are equal at the
start, which means that any time, all the
n
i
must have the same parity. Therefore, when
n
is divis
ible by 3, we cannot leave exactly one piece.
We claim that
n
is not divisible 3, then we can leave
6