| | fully recovered ...
feeling great~ actually ... sleep jor 12 hrs from 8pm to 8am ... then skipped all lesson (sick ma~) ... then play raise of nations for 21 hrs  now totally recovered ... hahaha~~
the best medicine is always to have fun~
anyway ... remember the math problem posted 2 days ago?
-- Is there exist a number with all digits 1 that is divisible by 499? --
here's the answer --->
Let 1111.... = summation of 10^k = 499t + r ... where ( 0 < r < 499),
then there is 498 choices of r from 1 to 498 but we have infinitely many choices of k.
Therefore, (by the pigeon-hole principle) there exist a number u and v such that submation of 10^u and submation of 10^v both have the same remainder r upon divided by 499,
so submation of 10^u - submation of 10^v is divisible by 499 (suppose u>v)
It remains to show that the difference of the sum is in form of 11111111...........
Consider 1111 - 11= 1100, then by dividing it by 100 we can get 11.
Similarly, summation of 10^u - summation of 10^v must be in form of (1111....1)10^p for some integer p.
Which now we have proved that : (111.....)10^p = 499t
Note that 10^k can only be prime factorized to 2^p * 5^p,
but both 2 and 5 is not divisible by 499 so 499 and 10^k is relatively prime,
so (111.....) must be divisible by 499 by some choice of p.
Hence there exist (111...111) such that it is divisible by 499
How many of u guys get the answer right ar?? (or by another method ...)
anyway ... Ultimate Logging System Professional will now be closed temperately for maintenance ... it should be restored after midnight tonight (US time) ... sorry for the inconvenient ...
please support the new system ~~
<--- Ultimate Logging System Professional --->
donation (USA) can go through me (by cash or check) or PayPal ... please try it out for free (1 week) ar~~ |
| | Posted 4/7/2005 9:53 AM - 1 view - 3 comments
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