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标题: ask a math JJ of last week [打印本页]

作者: 1036 时间: 2003-5-13 09:24 标题: ask a math JJ of last week

Can any da niu teach me how to solve the following JJ ?
from Robert math JJ summary 505-511-2003.

thanks
14. n is positive integer, sequence n 里的第n个数是n digits.并且每个digit’s number is 1. 求前40项和的百位数是几？
[确认]：38
[思路]：the sequence is like this: 1, 11, 111, 11111,…….

作者: 1036 时间: 2003-5-13 09:47

sorry, I guess I still don't understand the question above.

I guessed this question is asking: S= 1+11+111+1111+... 1...1(40 times)

then this Q is : ask S 's bai wei shu ? For example, 2130 's bai wei shu is 1.
I guess I am wrong?

Ben ben

作者: Echo 时间: 2003-5-29 13:25

不是吧，百位只有0～9的说
应该是：因为满10就进位了，所以前40个和的百位和38加上十位和进上来的4（39加个位和40进上来的4＝43，4进到百位）应该是2（38加4＝42，4进到千位去了）
所以最后应该是2

作者: isuffering 时间: 2003-6-2 20:32

Discussion:
My solution is as follows:

S
=1+11+111+...+11..11(40 digits)
=1+
  1+10+
  1+10+10^2+
  +........ +
  1+10+10^2+....+10^39 =
=1*40+10*39+(10^2)*38+(10^3)*x
=40+390+3800+x*1000
=4230+x*1000
=230+4000+X*1000
=0+10*3+(10^2)*2+(10^3)*Y+...

So the hundred number is 2

Got it?

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