作者yhubi (yhubill)
看板GMAT
標題[問題] PP計量問題
時間Tue May 6 23:28:30 2008
12. 781-!-item-!-187;#058&000341
Sets A, B, and C have some elements in common.
If 16 elements are in both A and B, 17 elements are in both A and C,
and 18 elements are in both B and C,
how many elements do all three of the sets A, B, and C have in common?
(1) Of the 16 elements that are in both A and B, 9 elements are also in C.
(2) A has 25 elements, B has 30 elements, and C has 35 elements.
這題(1)我了解 9 即是答案 對吧
那第(2)為什麼不行呢 我是這樣算的
20+30+35-16-17-18-2(ABC)=25+30+35
ABC算出來不是整數 所以(2)不行 是這樣所以不充分嗎
列式解題方向是這樣嗎?
197. 19016-!-item-!-187;#058&012397
The integers m and p are such that 2 < m < p and m is not a factor of p.
If r is the remainder when p is divided by m, is r > 1 ?
(1) The greatest common factor of m and p is 2.
(2) The least common multiple of m and p is 30.
第一個我知道 r一定大於1
但第二個r應該都等於1呀 所以應該是充分??
6=5*1+1 6 5最小公倍數30
10=3*3+1 10 3 最小功倍數30 還有其他嗎? 負數不在其中吧??
先感謝回答的各位
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◆ From: 218.168.68.72
1F：→ yhubi:197題有 15=6*2+3 不充分 解答完成 05/06 23:44
※ 編輯: yhubi 來自: 218.168.68.72 (05/06 23:48)
2F：推 reznor:15=6*2+3 => m=2, p=15 不是已經不符合2 < m < p的條件嗎? 05/07 00:42
3F：推 fly12404:197好怪... 05/07 01:07
4F：推 dddaaa:最小公倍數 =30 不代表m*p=30 05/07 08:10
5F：→ dddaaa:所以{m,p}可以為{6,15}{6,30}{5,6}{10,15}{6,10} 05/07 08:12
6F：→ dddaaa:以上組合餘數都不一樣 05/07 08:14
7F：→ dddaaa:這樣應該有回答到吧˙˙˙ 05/07 08:15
8F：→ yhubi:感謝 之前腦袋卡住想不到 05/07 11:04