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)
N = 801, AOD(N) = 3.
(followed by 7092, 7362; AODs=4.5, 8010, AOD=2.25)
M = 801, AOD(M) = 3
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