Problem A1
Let f : R2 → R be a function such that f(x, y)+f(y, z)+f(z, x) = 0 for all real
numbers x , y, and
z. Prove that there exists a function g : R → R such that f(x, y) = g(x)-g(y)
for all real numbers
x and y.
Solution : The function g(x) = f(x, 0) will do. Taking x = y = z in the
equation, we find that
3f(x, x) = 0 for all x. Then, taking z = y, we find that f(x, y) + f(y, x) = 0
for all x and y, that
is, f(y, x) = -f(x, y) for all x and y. Finally, taking z = 0, we see that
f(x, y) + f(y, 0) + f(0, x) = 0 ,
that is,
f(x, y) = f(x, 0) - f(y, 0)
for all x and y.
Remark: The most general function g(x) that will work is f(x, 0)+c, where
c is a constant. This
function clearly works, and if g(x)-g(y) = h(x)-h(y) for all x and y, then
g(x)-h(x) = g(y)-h(y)
for all x and y. Since the left-hand side of this equation is independent of y
and the right-hand side
is independent of x, both sides are independent of both variables , and hence
equal to (the same)
constant.
Problem A2
Alan and Barbara play a game in which they take turns
filling entries of an initially empty 2008×
2008 array. Alan plays first. At each turn, a player chooses a real number and
places it in a vacant
entry. The game ends when all the entries are filled. Alan wins if the
de terminant of the resulting
matrix is nonzero, Barbara wins it if is zero . Which player has a winning
strategy?
Solution: Barbara has a winning strategy. If Alan puts the number c into
position (i, j) on his
first turn, then Barbara puts that same number c into position (k, j) on her
first turn, where k ≠ i,
but k is otherwise arbitrary. At this point, rows i and k are distinguished from
the other 2006 rows
of the matrix. Thereafter, if Alan puts a number d into position (i, l) on his
turn, Barbara follows
by putting that same number d into position (k, l) and conversely. If Alan puts
an entry into any
row other than i or k, then Barbara puts any number whatever into any vacant
position whatever
not in either row i or row k. This will always be possible, since if column l is
vacant in one of rows
i and k when Alan's turn arrives, it is vacant in both, and if there is one
vacant location outside
those rows when Alan's turn arrives, there are two vacant locations outside
them.
In this way, rows i and k are always identical after Barbara's turn, and in
particular at the
end of play, so that the determinant is zero.
Problem A3
Start with a finite sequence 
of positive
integers. If possible, choose two indices j < k
such that 
does not divide
, and replace
and by
gcd 
and lcm
respectively.
Prove that if this process is repeated, it must eventually stop and the final
sequence does not depend
on the choices made. (Note: gcd means greatest common divisor and lcm means
least common
multiple.)
Solution: Let the least common multiple of 
,
where
are prime numbers and 
are positive integers.
Then we can write 
, where
are non negative integers . We set up a
one-to-one correspondence between sequences
and m × n matrices, as follows:
Thus, is the product
of the elements of the jth column of the corresponding matrix. When the
pair , j < k, is
replaced by 
), the effect on the
matrix is that the
entries in columns j and k of row i are either left alone (if
) or interchanged (if
).
The net effect, then, is to reduce the number of inversions in the sequence
by at least 1
in the latter case, and to leave the number of inversions unchanged in the
former. Since there can
be at most n(n - 1) inversions in any row, this procedure must terminate after
at most n(n - 1)
applications. The end result is a matrix with the same entries in each row as
the original matrix,
but now arranged in ascending order . Obviously, that matrix is determined
uniquely by the initial
matrix.
Problem A4
Define f : R → R by
converge?
Solution: It does not converge. To see this, we define a sequence of
points by 
,
so that 
and an → ∞ as n → ∞. We observe that
the function ln x maps
the interval 
in a one-to-one continuous manner onto
Notice that
f(x) approaches 
as x approaches
from
either side. Hence f(x) is continuous
and obviously increasing for all positive values of x . It fol lows that
converges if and only if the integral
converges, which is to say, the sum
converges.
However, using the substitution u = ln t, 
,
we find that for n ≥ 1,
Thus all the terms in this series are equal to the first
term, which is
The series therefore diverges.
