Determine as many positive integers, N, not ending in 0, as you can, such that N^5 contains two copies of N.

{9, 99, 999, 9999, ... } appears to be the solution set.

For example, 999^5 = 995009990004999
-------------------------
for i in (range(1,100000000)):
    if i%10 == 0:
        continue
    s = str(i)
    c = str(i**5)
    if c.count(str(i)) >=2:
        print(i,i**5)

Posted by Larry on 2023-09-19 12:06:33