- UID
- 557225
- 在线时间
- 小时
- 注册时间
- 2010-8-18
- 最后登录
- 1970-1-1
- 主题
- 帖子
- 性别
- 保密
|
the best way to approach things like this is to break down the compound statements into statements about the parity of the individual variables.
to do that, you'll almost certainly have to split the statements into cases.
"xy + z is odd"
two numbers can add to give an odd sum only if they have opposite parity. hence:
case 1: xy is odd, z is even
there's only one way this can happen:
x = odd, y = odd, z = even. (1)
case 2: xy is even, z is odd
there are 3 ways in which this can happen:
x = even, y = even, z = odd (2a)
x = odd, y = even, z = odd (2b)
x = even, y = odd, z = odd (2c)
this is a bit awkward, but, once you've divided the question prompt up into cases, all you have to do is look at your results, check the cases, and you'll have an answer.
--
statement (1)
the easiest way to handle expressions like this is to factor out common terms. you can handle the statement without doing so, but it's more work that way.
pull out x:
x(y + z) is even.
this means that at least one of x and (y + z) is even.
* if x is even, regardless of the parity of (y + z), then the answer to the prompt question is "yes" and we're done.
* the other possibility would be x = odd and (y + z) = even. this is impossible, though, as it doesn't satisfy any of the cases above.
therefore, the answer must be "yes".
sufficient.
--
statement (2)
this means that y and xz have opposite parity.
* y = even, xz = odd --> this means x = odd, y = even, z = odd. that's case (2b), which gives "no" to the question.
at this point you're done, because STATEMENTS CAN'T CONTRADICT EACH OTHER, se you know that "yes" MUST be a possibility with this statement (as statement #1 gives exclusively "yes" answers).
if you use this statement first, you'll have to keep going through the cases.
insufficient.
ans = a |
|
京公网安备11010202008513号 京ICP证101109号 京ICP备12012021号
发表于 2011-8-29 19:45:33
收藏
收藏
