→ yyc2008 : 可以再寫得淺顯一點嗎? 10/29 23:55
推 k32314282 : 想問怎麼知道A,B要這樣取 10/30 09:31
→ playerOrz : 受教了 10/30 16:52
推 cheesesteak : 感謝解答 10/30 20:56
推 k32314282 : 誰可以解答一下嗎？@@ 10/31 09:34
Inspired by the fact 1/7=.142857xxxx
And 142+857=999. We have the following solution:
Let n=10^2011
Then the origin product equals to (10n-1)^2/27=nA+ B,
where A=(100n-28)/27 and B=(8n+1)/27, both are integers.(B<n)
Note that A+B=4n-1=39999999..9
We see that
the sum of digits of nA+B
= the sum of digits of A+ the sum of digits of B
= the sum of digits of (A+B)
= 3+9*2011
= 18102.
※ 引述《cheesesteak (牛排‧起司)》之銘言：
: 1. 1/x+1/(y+z)=1/(3y)+1/(z+x)=1/(4z)+1/(x+y)=1/5
: 3x+2y+z=? Ans:33
: 2. 1111...1 * 3333...3 = N
: (2012個1) (2012個3)
: 求N所有位數和=? Ans:18102
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 118.161.26.49
※ 文章網址: http://www.ptt.cc/bbs/Math/M.1414594779.A.E93.html