第一题：
不学法语的100-69=31
不学德语的100-79=21
不学日语的100-89=11
不学英语的100-99=1
至少不学一门语言的人最大可能是31+21+11+1=64
所以答案是100-64=36
哈哈，很久没有动脑筋了(自己的孩子还没有学到这么深),不知现在还有有没有小学生的水平.:L
试一下第一题,答错不要笑咱啊!
Let’s take only english and Japanese first. 99 students take english, 89 students take Japanese. so 1 student does not take English and 11 don’t take japanese. the 1 student that does not take English has a possibility of taking only Japanese or does not take both jap and eng. So if this student does not take both, the 1 can be deducted from the 89 student who take Japanese. Until now, at least 88 students take both English and Japanese.
Using the same logic, we add in German. Now 88 students take both English and Japanese. 12 students may take only 1 subject or take none at all. 79 students take German. the remaining 21 do not. So we deduct 21 and 12 from the total number of 100 students, leaving a balance of 67 students, who take all 3 subjects.
Now we add in French. 69 take and 31 do not take. Add 31 to 21 and 12 = 64. This is the number of students that study 3 languages or less. 100-64=36%.
At least 36% of students take all 4 subjects. Answer x= 36%
再来试一下第二题
The answer is 16.
given 29 to start with, if Peter deducts 16, Jane is only left with 13. She can either deduct 4 or 9, and then Peter would be able to deduct the other, leaving a balance of 0. Peter will win at the third try.
这是几年级的奥数题啊？
2 3 5 7 11 13 17 19 23 29 31 37 41 || 43 47 53 59 61 67 71 73 79 83 89 97 101
41X43= 1763 2007
故分割线画在43前，左右1——1配对，(2X101),….(41X43)共有13组
反向解法相当漂亮，正向也不难
A=69 B=79,C=89,D=99
least A & B = 69+79 -100 = 48
least C & D = 89 + 99 -100 = 88
least A&B&C&D = 48 + 88 -100 = 36
Valid formula for 正/反向 is same
A+B+C+D – 300
老同志过了下数学瘾，该下课去了