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This is about the sum of digits of the $n$-th powers of a number which is equal to the number itself. A very trivial example is the number $1$: $$1^0=1,\quad1^3=1 \quad 1^{40}=1.$$ there are "many"such numbers,and one of that numbers is 666,which is its 47th and 51st powers.My question:is 666 the largest repunit number with that property? and is there only finitely many such numbers? (before this i,ve conjectured that 1 and 666 are the only such number,which is wrong.)

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    $\begingroup$ $18^3 = 5832, 18^6 = 34012224, 3 \cdot 6 = 18$. $\endgroup$ – Arthur Jan 5 '15 at 20:32
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    $\begingroup$ $9^2=81$; am I missing something? $\endgroup$ – egreg Jan 5 '15 at 20:32
  • $\begingroup$ Having accepted Mark Bennet's counterexample for the original question, you should probably post the latest edit, asking if $666$ is the largest number with the digit-sum property, as a new question. If you do, however, please make some effort to answer the question yourself, and indicate the range of numbers and powers you've examined. (The sequence in Xoff's answer looks pretty dense among the numbers up to $200$. Offhand I wouldn't expect it to taper off by the time it gets to the Beast.) $\endgroup$ – Barry Cipra Jan 5 '15 at 21:23
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    $\begingroup$ I have rolled this post back. It is highly inappropriate to ask a question, get it answered, and then (almost) completely change the question. This makes it very confusing for other users and the effort of the person who answered go to waste. If you have a related but different question, please ask a new question and link to this one. $\endgroup$ – RghtHndSd Jan 6 '15 at 20:43
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    $\begingroup$ I don't find anything curious about the last fact in your post. If you do some weird, unthought of operation to the two numbers appearing in the exponents, you get a number that, when you divide by 11, and add that to the number multiplied by 10, you get the first number? You're getting a bit numerological here. $\endgroup$ – djechlin Jan 6 '15 at 20:45
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There are many such numbers such that the sum of digits of a power equals that number !

$7^{4}$, $8^{3}$, $9^{2}$, $17^{3}$, $18^{3}$, $18^{6}$, $18^{7}$, $20^{13}$, $22^{4}$, $25^{4}$, $26^{3}$, $27^{3}$, $27^{7}$, $28^{4}$, $28^{5}$, $31^{7}$, $34^{7}$, $35^{5}$, $36^{4}$, $36^{5}$, $40^{13}$, $43^{7}$, $45^{6}$, $46^{5}$, $46^{8}$, $53^{7}$, $54^{6}$, $54^{8}$, $54^{9}$, $58^{7}$, $63^{8}$, $64^{6}$, $68^{7}$, $71^{9}$, $80^{17}$, $80^{19}$, $81^{9}$, $82^{10}$, $85^{10}$, $86^{13}$, $90^{19}$, $90^{20}$, $90^{21}$, $90^{22}$, $90^{28}$, $91^{14}$, $94^{10}$, $97^{10}$, $98^{11}$, $103^{13}$, $104^{13}$, $106^{10}$, $106^{13}$, $107^{11}$, $107^{13}$, $107^{15}$, $108^{11}$, $108^{12}$, $117^{10}$, $118^{14}$, $126^{13}$, $127^{14}$, $133^{16}$, $134^{13}$, $134^{15}$, $135^{13}$, $135^{14}$, $136^{15}$, $140^{25}$, $142^{16}$, $143^{17}$, $146^{13}$, $152^{15}$, $154^{14}$, $154^{15}$, $155^{19}$, $157^{19}$, $160^{28}$, $163^{16}$, $169^{16}$, $169^{22}$, $170^{31}$, $170^{33}$, $171^{17}$, $171^{19}$, $172^{15}$, $172^{18}$, $173^{19}$, $181^{16}$, $181^{18}$, $181^{19}$, $181^{20}$, $187^{16}$, $189^{19}$, $193^{22}$, $199^{15}$, $199^{21}$

Some of them like $54$ are even such that $54^6$ and $54^9$ have the property with $6\times 9=54$

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    $\begingroup$ Searching OEIS on $7,8,9,17,18,20,22$ leads to oeis.org/A124359 $\endgroup$ – Barry Cipra Jan 5 '15 at 20:34
  • $\begingroup$ How I missed 7^4 and 9^2 as potentially the most obvious examples I will never know $\endgroup$ – Mark Bennet Jan 5 '15 at 20:48
  • $\begingroup$ How can i miss those many such numbers,DUH.. $\endgroup$ – Gary B Jan 5 '15 at 21:35
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Note that $8^3=512$ for a counterexample

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  • $\begingroup$ Why the downvote? This answers the question. $\endgroup$ – Mark Bennet Jan 5 '15 at 20:22

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