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Average Of Digits of powers (Posted on 2020-08-20) Difficulty: 3 of 5
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)

No Solution Yet Submitted by Larry    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Hints/Tips re(2): Misunderstanding of Part 3 | Comment 5 of 7 |
(In reply to re: Misunderstanding of Part 3 by Steven Lord)

There is a 6-digit integer, L, such that the average of it's digits is an integer that is not '3'.
The same is true for L^2, L^3, and L^4
(all 4 have the same average digit)

  Posted by Larry on 2020-08-23 15:58:35
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