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 Hungary since 1894.]

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.