Let AOD(i) be the sum of digits of i divided by the number of digits of i, or the Average Of Digits.

N is the smallest integer > 1 such that the AOD(N^k) is the same for k in {1,2,3,4}

M has the same requirements as N, except AOD(M) must be an integer.

L has the same requirements as M, except that L is the smallest such integer with AOD(L) equal to some integer other than AOD(M).

Find:

1. N, AOD(N)

2. M, AOD(M)

3. L, AOD(L)

i did it w/ a brute force program. (I have looked up to 55104 for L)

Here is the output:

lord@rabbit-3 p5 % aod1

           

   N^k, k=1,2,3,4    Ave Digits

-----------------------------------

               00801 3.00     3

          0000641601 3.00     6

     000000513922401 3.00     9

00000000411651843201 3.00    12

               07092 4.50     4

          0050296464 4.50     8

     000356702522688 4.50    12

00002529734290903296 4.50    16

               07362 4.50     4

          0054199044 4.50     8

     000399013361928 4.50    12

00002937536370513936 4.50    16

               08010 2.25     4

          0064160100 2.25     8

     000513922401000 2.25    12

00004116518432010000 2.25    16

               08064 4.50     4

          0065028096 4.50     8

     000524386566144 4.50    12

00004228653269385216 4.50    16

               08118 4.50     4

          0065901924 4.50     8

     000534991819032 4.50    12

00004343063586901776 4.50    16

               08532 4.50     4

          0072795024 4.50     8

     000621087144768 4.50    12

00005299115519160576 4.50    16

               09261 4.50     4

          0085766121 4.50     8

     000794280046581 4.50    12

00007355827511386641 4.50    16

 Last N = 55104

Edited on August 20, 2020, 11:37 pm

Posted by Steven Lord on 2020-08-20 21:06:28