Date: Tue, 04 May 1999 12:56:25 -0500 Sender: Emily Subject: puzzled My name is Emily, I am in the 6th grade and my teacher gave me some extra probelms to see if I could solve them. She can't find the KEY and I can't figure out the probelm so she suggested I e-mail you guys to see if you could help us. Here's the Problem: The numbers 1,2,3...,1999 are written on the blackboard in the classroom. Evertime the teacher enters the room he chooses two numbers on the blackboard, say a & b, with a> or equal to b, then he erases them and writes the difference a-b somewhere on the blackboard. After this procedure is carried on 1998 times, there will be only one number left on the blackboard. Prove the last remaining number must be even. I hope you can help us! Hi Emily This is a fun problem. Did you try it with a smaller list of integers? If you start with only 1,2,3 on the board then you end with an even integer. If you start with 1,2,3,4 then you also end with an even integer, but if you start with 1,2,3,4,5 then you end with an odd integer. The key is whether the number of odd integers in the original list is even or odd.    In your list 1,2,3...,1999 there are 999 even integers and 1000 odd integers. Now notice that even minus even is even even minus odd is odd odd minus even is odd odd minus odd is even Thus when the teacher chooses the first two integers to subtract: If both the integers chosen are even then the number of odd integers on the board remains unchanged. If an even and an odd integer are chosen then the number of odd integers on the board remains unchanged. If both the integers chosen are odd then the number of odd integers on the board is reduced by two. In any case after the first step the number of odd integers on the board is still even.    The same is true at every stage, that is the number of odd integers on the board is always even. Hence when you have only one integer remaining on the board, this integer must be even. Cheers Chris and Harley Go to Math Central To return to the previous page use your browser's back button.