另类的"好数"数论问题

pizza49

n!进制下的m位数若满足其平方的后m位数刚好是其本身的话，则称为“好数”。这样的m位好数有多少个，并求出这些m位数之和？（n!是n的阶乘，例如6进制的两位好数有$13_{(6)}$,因为$13_{(6)}*13_{(6)}=213_{(6)}$）
注：任何位数最前的数字不能为0，比如说不能出现0111这样的数。

northwolves

1. Listmillion[k_]:=Select[Range@1000000,Take[IntegerDigits[#^2,k],-Length@IntegerDigits[#,k]]==IntegerDigits[#,k]&];
   
2. For[n = 2, n < 11, n++, Print[{n, Listmillion[n!]}]]

复制代码

{2,{1}}
{3,{1,3,4,9,28,81,136,1216,6561,16768,29889,76545,203392,636417}}
{4,{1,9,16,64,513,13312,151552,180225}}
{5,{1,16,25,40,81,96,105,576,6400,6976,7425,8001,13825,165376,512001,677376}}
{6,{1,81,145,225,496,576,640,6400,180225,186625,331776,338176,512001}}
{7,{1,225,721,945,1296,2016,2080,2241,2800,2961,3025,3745,4096,4320,4816,512001}}
{8,{1,4096,5761,8065,9856,12160,13825,17920,22401,26496,28161,30465,32256,34560,36225}}
{9,{1,58240,72576,76545,130816,134785,149121,155520,207361,213760,228096,232065,286336,290305,304641}}
{10,{1,512001,518400,856576}}