大家来看看这个数字魔术的原理！

古月圣

http://funnies.com/mathtrick.html

cristiano~7

如果一開始選擇的數是大數
假設選擇 w, x, y, z 代入 1000 100 10 1 的數位
1000(w - ?) + 100(x - ?) + 10(y - ?) + 1(z - ?) > 0
把 w, x, y, z 其中至少一對調換就是形成一個新的4位數
比如說一開始選擇 4321 之後調換成 1234
1000(w - z) + 100(x - y) + 10(y - x) + 1(z - w) > 0
999w - 999z + 90x - 90y 是 9的倍數，可以被9整除
可以知道1000, 100, 10, 1 之間不管如何調換，相減后它的和都是可以被9整除
1000-100=900,1000-10=990,1000-1=999,100-10=90,100-1=99,10-1=9

可以被9整除的數的性質：每個數位的數字加起來也是9的倍數
例如 5589 = 5+5+8+9 = 27 = 9 * 3
所以只要我們知道其他每個數位的數字是什麼，用一個略大的9的倍數來減
就可以找到被圈起來的數字是什麼

如果我們一開始選擇8765,之後調換成6578
8765 - 6578 = 2187

2187,當我們圈2,1+8+7=16, 18 - 16 = 2
2187,當我們圈1,2+8+7=17, 18 - 17 = 1
2187,當我們圈8,2+1+7=10, 18 - 10 = 8
2187,當我們圈7,2+1+8=11, 18 - 11 = 7

katami

古月圣 于 24-3-2005 09:39 PM  说 :
http://funnies.com/mathtrick.html

[size=-1]Let's see what the magician does to ur mind.
Say u have 3 digit number (e.g. 921) as the original number.
There r 3! = 6 sets of numbers if the digits are not overlapping
(e.g.
          921 <- orig number <- (i)
    The remainig 6 - 1 = 5 numbers are
          912 <- swap the last 2 <- (ii)a
          291 <- swap the first 2 <- (ii)b
          129 <- swap the first and last <- (ii)c
          219 <- replace first 2 with last 2 <- (iii)a
          192 <- replace  last 2 with first 2 <- (iii)b

(i) - (ii)a,b,c always gives u a zero in a digit of ur results
(e.g. 0xy, xy0, x0y, where x,y is [0-9]). U give x or y to the magician.

[size=-1]Let's investigate the effect of swapping 2 digits and subtracting them
(e.g.           
                  a  b              9 2          ...(I) : orig
                  b  a            - 2 9          ...(II): swap
                ------------------------------------------
           (a-1)-b 1{b}-a   (9-1)-2 1{2}-9     ...(III): result
      
   When u add the 2 digits of result in (III), it is always = 9
   (a-1)-b + 1{b}-a = 1{b} -b + 1 = 10 - 1 (as b is [0-9]) = 9

[size=-1]
Let's also investigate the effect of replacing first/last 2 with last/first 2 and subtracting them
(e.g.        
                  a b c             9 2 1      ... (IV) : orig
                  c a b             1 9 2      ... (V)  : replace last 2 with first 2
                  b c a             2 1 9      ... (VI) : replace first 2 with last 2
                 -------------------------------------

if a > c, b > a, c < b
      
  (IV)-(V)  =>  a-c [b-1]-a 1[c]-b ...(VII)
  (IV)-(VI) =>  a-b [b-1]-c 1[c]-a ...(VIII)

if a > c, b < a, c < b

  (IV)-(V)  =>  [a-1]-c 1[b-1]-a 1[c]-b ...(IX)
  (IV)-(VI) =>  [a-1]-b 1[b-1]-c 1[c]-a ...(X)

     When u add the 3 digits of result in (VII)(VIII), it is either 9 or 18
     (VII)   a-c   +  [b-1]-a + 1[c]-b = 1[c]-c-1 = 10 - 1 = 9
     (VIII)  a-b   +  [b-1]-c + 1[c]-a = 1[c]-c-1 = 10 - 1 = 9 (since c,b is [0-9])
     (IX)  [a-1]-c + 1[b-1]-a + 1[c]-b = 1[b-1]-b + 1[c] -c = 18
     (X)   [a-1]-b + 1[b-1]-c + 1[c]-a = 1[b-1]-b + 1[c] -c = 18 (since c,b is [0-9])

Tha magician does not need to know the original numbers and the re-arragned numbers.
By giving the digits of ur subtracted results, he will do the following.

                          The magician reads ur mind
If u give 1 digit    -->  (9 - ur digit)
If u give 2/3 digits -->  adds the 2/3 digits (= x)
                             if x < 9, then ( 9 - x)
                             if 18 < x < 9, then (18 - x)
                             if 27 < x < 18,t hen (27 -x)

The number 9, 18, 27 are called magic numbers of this rule as it characterize the rule.

[ Last edited by katami on 27-3-2005 at 01:23 PM ]