**Problem 6**

Let _{} represent an arbitrary
arrangement of the numbers _{}. Prove that if n is
odd then the product _{} is even.

[Problem submitted by Vin Lee, LACC Associate Professor of Mathematics. Source:
the 1906 Eotvos Competition, a high school
mathematics competition held in

**Solution A for Problem 6:**

Since n is
odd there is a positive integer k such that n = 2k + 1.

Note that _{}, _{}, _{},…, _{}.

So, of the numbers _{}, exactly _{} are odd numbers; and
of the numbers _{}, exactly _{} are odd. So, of the numbers _{} exactly _{} are odd.

Since the
product _{} has exactly n factors
and contains _{} odd numbers, at least
one of these factors is the difference of two odd numbers; that is, at least
one factor is even. Therefore, the
product is even.

**Solution B for Problem 6:**

Let us fill
each blank in following with a number from the set {1, 2,
3, 4,….n}.

(-1) __(-2)__ (-3) __(-4)__ (-5) __(-6)__…..
(-n)

In order to
make the final product odd, each odd number must be filled in one of the blanks
underlined above. When n =2k+1, there are only k underlined-blanks but k+1 odd
numbers. Therefore an odd final product is impossible.