數學問題:整除數學

Let (2003n + 2002) / (20n + 2) = m is an positive integer , then
n(20m - 2003) + 2m - 2002 = 0
n = (2002 - 2m) / (20m - 2003)
10n + 1
= [(20020 - 20m) + 20m - 2003] / (20m - 2003)
= 18017 / (20m - 2003)
= 43 x 419 / (20m - 2003) is an positive integer.
∴ 20m - 2003 = 43 (rejected) or 419 (rejected)
or 18017 ⇒ m = 1001
Hence n = (2002 - 2*1001) / (20*1001 - 2003) = 0 only.

2013-01-05 11:38:39 補充：
Note : 2003n + 2002 ≠ 0 so m is an positive integer.

2013-01-05 13:21:01 補充：
Missing 20m - 2003 = 1 (rejected)

the proof is concise (y)
but i think '0' is not a positive integer

All positive integer n....
that mean
n>0
20n+2>0
2003n+2002>0
[(20n+2)/(2003n+2002)]>=1....
20n+2>=2003n+2002
-2000>=1983n
n<=-2000/1983
n<0
矛盾....冇解
......................................................................................................................
不知是否打多
如果可以接受n是負數
不難得知
當=-1
分母是-1
所以
[(20n+2)/(2003n + 2002)]=-18/-1=18

OR BY上面[慢野但有唯一性]
如果20n+2<0
2003n+2002<0
就有20n+2<=2003n+2002
n>=-2000/1983
n>=-1
n>=-1,n<0 [BY上面]
所以只可以n=-1

喜歡的可以證埋
[(20n+2)/(2003n+2002)]>0
不盡述
希望你真係打錯題目.....
有個簡單D,打埋啦都係=.=
20n+2>=0
n>=-1/10>0
2003n+2002>0

20n+2<=0
n<0<=-1
2003n+2002<0

2013-01-05 11:55:25 補充：
.... 係咁解架咩....T_T
又衰英文......T_T
我唔會講X口架.....
我死了...
SOR丫..自由....

2013-01-05 14:15:51 補充：
同001基本一樣....
(2003n+2002)/(20n+2)
n must=2k
[2003(2k)+2002]/[20(2k)+1]
=(2003k+1001)/(20k+1)
=20[(2003k+1001)/(20k+1)]
=[40060k+20020]/(20k+1)
=18017/(20k+1)