數學 : 求n的最小值

Let n=10000a+1998=1999b where a, b are positive integers.

10000a=1999b-1998
=(2000b-b)-(2000-2)
=2000(b-1)-(b-2) ----------- (1)
=> (b-2) is divisible by 2000.

Let b=2000c+2 where c is positive integer.
Putting b into (1):
10000a=2000(2000c+1)-(2000c)
=2000(1999c)+2000
=> 5a=1999c+1
=> 1999c+1 is divisible by 5
=> c ends with digit 1 or 6
=> c = 1, 6, 11, 16, ...
smallest value of c = 1
=> smallest value of a= (1999+1)/5=400
=>smallest value of n=10000(400)+1998=4001998

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n=_____1998=1999x?

1999x______2=____________8

So:

1999x2002=_____1998

Ans:4001998

正整數n的末四位是1998，並且為1999的倍數，求n的最小值
Sol
土法煉鋼
10000a+1998=1999b
9995a+5a+1998=1999b
5a+1998=1999(b－5a)
5a+1998=1999c
5a=1999c－1998
5a=2000c－2000－c+2
－c+2=5d
c=2－5d
5a=1999(2－5d)－1998=－9995d+2000
a=400－1999d
a>=0
400－1999d>=0
400>1999d
0.2002>d
d的最大值=0
a的最小值=400
n=400*10000+1998=4001998

n ≡ 1998 (mod 10000) ≡ 0 (mod 1999)。

1999 與 10000 互質，

今有 1 = (400)(10000) - (2001)(1999)

顯然 n 所有整數解為：

n ≡ -(1998)(2001)(1999)

≡ -(4000000 - 1)(1999 -1)

≡ 4001998 (mod 19990000)。

最小滿足條件之正整數為 n = 4001998。

2012-05-12 17:09:48 補充：
The above is a systematicand STANDARD way that can be applied to every case in which the Chinese Remainder Theorem is valid. It can also be generalized to solve any finite number of modulo equations (if the Chinese Remainder Theorem is valid).

n=1999x≡1998 (mod 10000)…(1)
10000x≡0 (mod 10000)…(2)
(1)×5=>9995x≡9990 (mod 10000)…(3)
(2)-(3)=>5x≡-9990 ( mod 10000)≡10(mod 10000)…(4)
(4)×399=>1995x≡3990 (mod 10000)…(5)
(1)-(5)=>4x≡-1992 (mod 10000)…(6)
(4)-(6)=>x≡2002 (mod 10000)
Minimum n=1999×2002=4001998

n = xxxxxxx1998
= abc------zwy x 1999

9 x y = ?8 => y = 2

n - 1999 x 2 = xxxxxx1998 - 3998
= xxxxxx0000 - 2000
= abc------zw0 x 1999

所以 xxxxxx000 - 200
=(xxxxxx00 - 20) x 10
= abc----zw x 1999

9 x w = ?0 => w = 0

(xxxxxx00 - 20) x 10 = abc----z0 x 1999
xxxxxx00 - 20 = abc----z x 1999
(xxxxxx0 - 2) x 10 = abc----z x 1999

9 x z = ?0 => z = 0
(xxxxxx0 - 2) x 10 = abc----0 x 1999
xxxxxx0 - 2 = abc--mnl x 1999
gggggg8 = abc--mnl x 1999

9 x l = ?8 => l = 2

gggggg8 - 3998 = (abc--mn2 - 2) x 1999
gggggg8 - 3998 = (abc--mn2 - 2) x 1999

上面都是找出 abc------mnlzwy 的necessary condition，
(也是 sufficient condition,自己check吧,但應該不用check也知)
y , w , z , l 都是 「fixed」。
求最小的n 就是求最小的 abc------ mnlzwy
所以如果我set abc--mn 全0，gggggg8 =3998
那麼 0 = (2-2) x 1998 = 0。
沒有違反等式。

所以 n = 2002 x 1998 = (2000 + 2) x (2000 - 2)
= 4000000 - 4
= 3999996
我用了最笨的方法吧...

2012-05-12 12:16:28 補充：
=.= 錯左

2012-05-12 12:18:58 補充：
所以 n = 2002 x 1999 = (2000 + 2) x (2000 - 1)
= 4000000 + 2000 - 2
= 4001998
寫錯左 x 1998 =.=

1999 x (????????2) = ??????????1998
首先第一組數要係 2 字尾, 第二組數要係 1998 字尾
那麼第一組數的最後 4 個數字必為 2002
所以 n = 1999 x 2002, or 1999x12002, 1999x 22002 ......
n 的最小值 = 4001998