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)