SOLUTION TO PROBLEM R-6

Note that if a single difference a_i-b_i is even, the product $(a_1-b_1)(a_2-b_2)\cdots (a_{n}-b_{n})$ will be even. In other words, for this product to be odd every difference must be odd.

For a_i-b_i to be odd the pair $\{a_i, b_i\}$ must include one even and one odd integer. If every pair includes one even and one odd integer, then the number of evens in $\{a_1, a_2, a_3, \ldots , a_n\}$ must equal the number of odds in $\{b_1, b_2, b_3, \ldots , b_n\}$. Similarly the number of odds in $\{a_1, a_2, a_3, \ldots , a_n\}$ must equal the number of evens in $\{b_1, b_2, b_3, \ldots , b_n\}$. But $\{a_1, a_2, a_3, \ldots , a_n\}=\{b_1, b_2, b_3, \ldots , b_n\}$. So the number of evens equals the number of odds.

So for the product to be odd, we must have an even number of elements! In other words, if n is odd (like n=7) then the product will be even.

If n is even, the product can be odd. Suppose that our set includes as many evens as odds. For simplicity we focus on the case where the integers are either 1 of 2. If the first ordering is $1, 1, \ldots , 1, 2, 2, \ldots ,2$ while second is $2, 2, \ldots ,2, 1, 1, \ldots , 1$, the product will be

$$(-1\cdot-1\cdot-1\cdots-1)\cdot(1\cdot1\cdot1\cdots1)=\pm1$$

In either case, this is odd.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Feb 22 10:49:11 CST 2002