中科新闻网公司产品目录模板:神秘的"6174"－陷阱数
来源：百度文库 编辑：中科新闻网 时间：2024/05/08 03:50:27
再把4176按上面方法重复一遍：7641-1467=6174
再作，奇怪的事就出现了，7641-1467＝6174；又回到了6174。
这个是偶然的吗？不，再随便举一个四位数，按上面的方法连续去做，最终那个6174还是幽灵般出现。对于任何一个数字不完全相同的四位数，最多运算7步，必然落入陷阱中
这个陷阱数已由印度数学家给出证明。
谁能提供一下证明的过程。
注意，是任意四位不完全相同的数！！！
What is the Kaprekar Number?
The number 6174 is called the Kaprekar number. The Indian mathematician D.R.Kaprekar made the following discovery in 1949.
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(1) Take a four-digit number with different digits (acbd with .a<b<c<d)..
(2) Form the largest and the smallest number from these four digits (dcba and abcd)..
(3) Find the difference of these digits. Maybe this is 6174 (dcba - abcd = 6174?).
If it is not, form the largest and the smallest number from the difference and subtract these numbers again. You may have to repeat this procedure.
The end result is always 6174, but there are no more than 7 steps.
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1st example: Take the number 1746.
1st step: 7641 - 1467 = 6174
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2nd example: Take the number 5644.
1st step: 6544 - 4456 = 2088
2nd step: 8820 - 0288 = 8532
3rd step: 8532 - 2358 = 6174
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3rd example: Take the number 7652.
1st step: 7652 - 2567 = 5085
2nd step: 8550 - 0558 = 7992
3rd step: 9972 - 2799 = 7173
4th step: 7731 - 1377 = 6354
5th step: 6543 - 3456 = 3087
6th step: 8730 - 0378 = 8352
7th step: 8532 - 2358 = 6174
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The problem is solved. (Spektrum der Wissenschaft, Erstausgabe 1978)
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Manick Srinivasan and Ramkumar Ramamoorthy sent me more computer results: If you allow numbers with noughts at the beginning like 0342 or 0045, then the Kaprekar number fits for any four digit number except for 1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999.
They counted the numbers with the same steps:
There totally are 8991=9000-9 four digit numbers.
最笨的办法，编个程序，将所有的四位数（最高位不是0）计算一遍，就可以证明了，而且这样的证明是严密的。从某个角度理解：有限量的数学问题，都可以计算机证明。
对呀~~我也发现了，不知道是为什么，先占个位，等高手来回答吧，我顺便学习一下。
任意四位数？不是吧？我取1111，2222该怎么算呢？好象都是0！
最高位是0就不可以了.笨