<p>I've been working solving the Mathematics sections on the McGraw-Hill 2013 SAT study guide. On page 52, number 11, I faced a confusing math problem. It goes as following:
How many different positive three-digit integers begin with an odd digit and end with an even digit?
A-125
B-180
C-200
D-225
E-250
What I did is count 2,4,6,8 as the even last digit integers and thus every positive first digit number (1,3,5,7,9) has 40 options. For example, 102,104,106,108 (that's four) and 112,114,116,118 (another four) and etcetera! Then I multiplied 40 by the first odd possible first digit integers (5). In the end I got 200. I check the answer and it's wrong. The correct answer is 250. What McGraw Hill did is count zero as an even integer, and thus it would be 50 times 5. This is really stupid and annoying. The blue book does not count zero as an even number, yet this book counts zero as an even integer. Can someone please clarify this: Is zero an even integer? I really don't know whether to count it as negative or not. I might face questions like this on the SAT and I can't afford losing valuable credit on an absurd problem as such.</p>
<p>0 is definitely even. Remember an integer n is even iff n = 2k for some integer k, and 0 = 2*0. Also 0 is nonnegative.</p>
<p>Also, just to be clear on this: the blue book definitely counts zero as even. Is it possible that you were flip-flopping “even” and “positive” in your head? Because, as MITer94 said, zero is non-negative. But it is also non-positive! It’s just zero, all by itself, neither negative nor positive and that makes some people think that it is also neither odd nor even. Even strong students make that mistake. I know of a 2390 scorer who owes the missing points to that little error.</p>
<p>Yeah, I was just thinking, if the blue book screwed up on the parity of zero, then they really screwed up.</p>