自然数等于它的各位数同次幂之和的解

A Way for Finding out a Natural Number Which Equals the Sum of Each of the Digits to the Same Power

本文首先证明,对于任一个大于0的整数m,自然数要等于它的各位m次幂之和,自然数的位数p与幂次m必须满足不等式: 1+[mlg2]≤p≤1+[mlg9+1g(m-1)] 才有可能,然后,在这基础上,再求出m=1,2,……,11的解,最后附FORTRAN77源程序一份,供上机计算时参考.

this paper first proves the following inequality: Where p is the number of digits of a natural number; m is the power of each digit. Then, solutions to this problem are provided for cases where m=1, 2, 3, ……, 11 Finally, a source program in FORTRAN—77 for finding out these natural numbers is given.