推 oranger:終於搞懂了!! 謝謝解答:) 09/01 00:51
※ 引述《oranger (從新出發)》之銘言：
: 167.
: If x, y, and z are integers and xy + z is an odd integer, is x an even
: integer?
: (1) xy + xz is an even integer.
: (2) y + xz is an odd integer.
: Ans:A
: 176.
: If x and y are positive integers, is the product xy even?
: (1) 5x - 4y is even.
: (2) 6x + 7y is even.
: Ans: D
: 請大家幫忙解答了 謝謝!!
: 我想請問大家在遇到像這種算奇數偶數的題目時，
: 該怎麼解題
: 我每次只要碰到這種題目都會死的很難看 /_\ 好苦惱
167
由題目可知有兩種情況:(i) xy = odd and z = even (ii) xy = even and z = odd
(1) 考慮(i)，可得xz = odd，但z = even，故不可能得出xz = odd，矛盾
考慮(ii)，可得xz = even，又z = odd，則x必定為even ..........sufficient
(2) 考慮(i)，可得 x = odd，代入(2)選項符合
考慮(ii)，x = odd or even代入(2)選項也符合，故無法確定......insufficient
176
(1) 4y = even，and we know 5x-4y = even，then 5x must be even
since 5 is odd，so we get x is even
therefore，xy is even..............sufficient
(2) 6x = even，and we know 6x+7y = even，then 7y must be even
since 7 is odd，so we get y is even
therefore，xy is even..............sufficient
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