设 $M=d_4 d_3 d_2 d_1$, 这里 $d_4\geqslant d_3\geqslant d_2\geqslant d_1$, 令 $N=d_1 d_2 d_3 d_4$. 假设 $M > N$.
计算 $T=M-N$. 并将 $T$ 的各个数字仍按照从大到小排列. 重新得到 $M_2=d_4^{(2)} d_3^{(2)} d_2^{(2)} d_1^{(2)}$, 仍简记为 $M_2=d_4 d_3 d_2 d_1$.
试证: 经过有限步后得到 $T=6174$.
值得注意的是对于 $T=6174$, 将其数字进行排序得 $M=7641$, 于是 $N=1467$, 相减得 $T=7641-1467=6174$.
Note:
6174+4716=10890,
1089+9801=10890.
1
Let $M=dcba$, $d\geqslant c\geqslant b\geqslant a$; $N=abcd$. Suppose that $N < M$. We calculate $T=M-N$.
It is easy to show that T looks like $uzwv$, where $u+v=10$ and $z+w=8$.
\[
\begin{aligned}
&5+5=6+4=7+3=8+2=9+1=10,\\
&4+4=5+3=6+2=7+1=8+0=8.\\
\end{aligned}
\]
Hence, we only need to check the 25 cases:
9810, 9711, 9621, 9531, 9441;
8810, 8721, 8622, 8532, 8442;
8730, 7731, 7632, 7533, 7443;
8640, 7641, 6642, 6543, 6444;
8550, 7551, 6552, 5553, 5544.
All of above numbers' figures have been sorted.