1. The sum of the page numbers on opposite sides of a sheet of paper must be odd because the pages are numbered consecutively making one of them odd and the other even and thus their sum odd. If we add up 25 such odd numbers, we'll always get an odd number, but 2010 is even.
2. First notice that the sum of the first ten numbers is 55. That is, 1+2+3+4+5+6+7+8+9+10=55. Consider now what happens if we change +3 to -3. The sum is decreased by 6. If we change +8 to -8, the sum is decreased by 16.
In fact, whenever we change a sign from + to -, we decrease the sum by an even number. Starting at the odd number of 55 and subtracting even numbers from it will always lead to an odd number, but 0 is even.
3. Since this is a checkerboard, alternate squares are colored red and black, and so the two missing diagonally opposite squares are both the same color, say red. A normal checkerboard has 32 red and 32 black squares, but this one has only 30 red and 32 black squares. This makes covering it with 31 dominoes impossible since each domino will cover one red and one black square.
4. In order for the series to end in either six or seven games, it is necessary for it go more than five games. Thus after five games one of the teams, say it's A, is ahead three games to two. (If one of the teams were ahead four games to one, the series would be over after five games.)
So it's three games to two in favor of A. If A wins, the series lasts exactly six games, and if B wins, the series goes into the seventh game. Since each team is equally likely to win, it's equally likely that the series will go six or seven games.
5. The best strategy is for the last person in line (prisoner number 12) to answer "red" if he sees an even number of red hats in front of him and blue if he sees an odd number. Unfortunately for him, he'll be right only 50% of the time, but his answer enables all of the other prisoners to survive.
If the last prisoner (12) answers "red," the prisoner in front of him (number 11) knows that 12 saw an even number of red hats. If 11 also sees an even number of red hats in front of him, he knows that his hat is blue and answers accordingly. If 11 sees an odd number of red hats in front of him, he knows his hat is blue and answers accordingly. And so on for prisoners number 10, 9, 8, and so on.
6. Since Obama wishes to move the court in a more liberal direction, the only way this might come about in the short term is to pick not necessarily the most eloquent or learned judge, but rather that judge who would be most likely to persuade Justice Kennedy, the swing justice among the five more conservative justices. This would switch the court's vote in certain cases from 5-4 to 4-5.
Simplistic and nowhere near as clear-cut as the straight math problems, this analysis is at least tenuously related to the issue of parity. Incidentally from what I've read about her, Judge Diane Wood would be the most likely replacement for the recently retired Justice Stevens to bring this change to the court.
John Allen Paulos, a professor of mathematics at Temple University in Philadelphia, is the author of the best-sellers "Innumeracy" and "A Mathematician Reads the Newspaper," as well as, most recently, "Irreligion: A Mathematician Explains Why the Arguments for God Just Don't Add Up." He's on Twitter and his "Who's Counting?" column on ABCNews.com appears the first weekend of every month.