davidbin

请看 new update above

davidbin

你可能指的是以下方法
1+C1/10+C2/9+C3/8+C4/7+C5/6 = 1+10+36+56+35+6 = 144

davidbin

I discovered the pattern below
P0  P1  P2  P3  P4  P5  P6  P7  P8  P9  P10  P11   ....  PN
1    2    3    5    8   13  21  34  55  89  144  233
PN = PN-1 + PN-2 (N>=2)
if 11 digits, the answer will be 233

仙都峰居士

davidbin 发表于 18-4-2012 17:08
I discovered the pattern below
P0  P1  P2  P3  P4  P5  P6  P7  P8  P9  P10  P11   ....  PN
1    2  ...

1. 
   
2. I discovered the pattern below
   
3. P0  P1  P2  P3  P4  P5  P6  P7  P8  P9  P10  P11   ....  PN
   
4. 1    2    3    5    8   13  21  34  55  89  144  233
   
5. PN = PN-1 + PN-2 (N>=2)
   
6. if 11 digits, the answer will be 233
   
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8. 
   
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10. 点评
    
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15. 静雯  You are so great!  发表于 前天 18:10
    
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你们都很厉害！希望你们能为教育版多作贡献。建议班主们考虑为此贴加精，以示鼓励。

静雯

Problem 14
5 girls, A, B, C, D and E divided a certain number of marbles amongst themselves in the following way: A took 1 marble and 1/5 of the remaining marbles, then B took 1 marble and 1/5 of the remaining marbles. C, D and E did the same. At least how many marbles were there in the beginning?

davidbin

抛砖引玉
3121

(((((x-1)4/5-1)4/5-1)4/5-1)4/5-1)4/5 must be an integer
上式可推出
X=(3125N+2101)/256 ;N is an integer
3125N + 2101 = (256X12 + 53 ) N + 256X8 + 53 = 256X12N + 256X8 + 53(N+1) 能被256整除
N = 255
X = 3121

滑板王子

静雯 发表于 3-2-2012 09:42
谢谢各位版亲的热情支持与辛勤劳动。

这个方法最好了！明了

静雯

One SMO Junior problem:

The diagram below shows that ABCD is a parallelogram and both AE and BE are straight lines. Let F and G be the intersections of BE with CD and AC respectively. Given that BG=EF, find the quotient DE/AE.

davidbin

DE/AE = EF/EB = a/(2a+b)
DH/FG= DC/FC
a/b= DC/FC
(a-b)/b = (DC-FC)/FC = DF/FC = EF/BF = a/(a+b)
a2-ab-b2 = 0
(a- 1.618b)(a+0.618b) = 0
a= 1.618b
DE/AE = a/(2a+b) = 1.618b/4.236b = 0.382

davidbin

davidbin 发表于 30-5-2012 19:41
DE/AE = EF/EB = a/(2a+b)
DH/FG= DC/FC
a/b= DC/FC

没当过老师， 因家中有小学生，最近刚把小学知识复习了一下。
不知SMO Junior  针对的是那个年级的学生。看来要复习中学知识了