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 Average Of Digits of powers (Posted on 2020-08-20)
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.)
 re(3): Misunderstanding of Part 3 | Comment 6 of 7 |
(In reply to re(2): Misunderstanding of Part 3 by Larry)

From approximately where I left off, only the integral ones are listed:

` 709200      3.0     502964640000    356702522688000000      252973429090329600000000  715050      3.0     511296502500    365602564112625000      261424113468732506250000  736200      3.0     541990440000    399013361928000000      293753637051393600000000  750060      3.0     562590003600    421976258100216000      316507512150648012960000  750510      3.0     563265260100    422736210357651000      317267753235520652010000  800640      3.0     641024409600    513229783302144000      410912293703028572160000  806400      3.0     650280960000    524386566144000000      422865326938521600000000  811800      3.0     659019240000    534991819032000000      434306358690177600000000  853200      3.0     727950240000    621087144768000000      529911551916057600000000  901350      3.0     812431822500    732285423210375000      660045466210671506250000  917010      3.0     840907340100    771120439945101000      707125154634057068010000  926100      3.0     857661210000    794280046581000000      735582751138664100000000  996633      6.0     993277336689    989932971896368137      986599867579993065482721 `

and so L=996633 with an AOD of 6.

 Posted by Charlie on 2020-08-23 18:40:22

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