2
$\begingroup$

I was reading the following problem in a book: $2^{29}$ has exactly $9$ digits, all of these digits are distinct, what is the missing digit?

I tried to solve it, and I did it correctly.


Suddenly, the following two questions popped into my mind:

  • Find the number of pairs of natural numbers $m$ and $n$ $(n \ne 1)$ such that $m^n$ has exactly $9$ digits, all of these digits are distinct.

  • Find the least $9$-digit number that can be expressed as $m^n$ where $m$ and $n$ are natural numbers and $n \ne 1$.


These questions may be solved using a software, but I am not sure if we can use a purely mathematical way.

Do not provide a solution. I am just asking for useful hints/formulae/techniques, then I will try to solve them.


Your help would be appreciated. THANKS!

$\endgroup$
3
  • $\begingroup$ How did you solve the case $2^{29}$ ? $\endgroup$ – TheSilverDoe Nov 18 '20 at 17:38
  • 2
    $\begingroup$ Digits are essentially random, there is no known connection between "all digits are distinct" and other properties of a number. So I don't see another option than to check all eligible powers one by one. I wouldn't want to do that manually, but a computer wouldn't find it hard, I guess. $\endgroup$ – user436658 Nov 18 '20 at 18:52
  • $\begingroup$ I agree with @ProfessorVector - there's really no reason to expect any sort of general connection when digits are so closely tied to their base. Unless you're maybe looking at a base (for the exponent) with some sort of "special relationship" with $10$ (and it's not even clear what such a "special relationship" would be), there's really unlikely to be any general connection of the type you're looking for. Sorry. $\endgroup$ – Lieutenant Zipp Nov 18 '20 at 19:28
0
$\begingroup$

My computer says: $$10124^{2}=102495376$$ $$10128^{2}=102576384$$ $$10136^{2}=102738496$$ $$10214^{2}=104325796$$ $$10278^{2}=105637284$$ $$11826^{2}=139854276$$ $$12363^{2}=152843769$$ $$12543^{2}=157326849$$ $$12582^{2}=158306724$$ $$12586^{2}=158407396$$ $$13147^{2}=172843609$$ $$13268^{2}=176039824$$ $$13278^{2}=176305284$$ $$13343^{2}=178035649$$ $$13434^{2}=180472356$$ $$13545^{2}=183467025$$ $$13698^{2}=187635204$$ $$14098^{2}=198753604$$ $$14442^{2}=208571364$$ $$14676^{2}=215384976$$ $$14743^{2}=217356049$$ $$14766^{2}=218034756$$ $$15353^{2}=235714609$$ $$15681^{2}=245893761$$ $$15963^{2}=254817369$$ $$16549^{2}=273869401$$ $$16854^{2}=284057316$$ $$17252^{2}=297631504$$ $$17529^{2}=307265841$$ $$17778^{2}=316057284$$ $$17816^{2}=317409856$$ $$18072^{2}=326597184$$ $$19023^{2}=361874529$$ $$19377^{2}=375468129$$ $$19569^{2}=382945761$$ $$19629^{2}=385297641$$ $$20089^{2}=403567921$$ $$20316^{2}=412739856$$ $$20513^{2}=420783169$$ $$20754^{2}=430728516$$ $$21397^{2}=457831609$$ $$21439^{2}=459630721$$ $$21744^{2}=472801536$$ $$21801^{2}=475283601$$ $$21877^{2}=478603129$$ $$21901^{2}=479653801$$ $$22175^{2}=491730625$$ $$22456^{2}=504271936$$ $$22887^{2}=523814769$$ $$23019^{2}=529874361$$ $$23113^{2}=534210769$$ $$23178^{2}=537219684$$ $$23439^{2}=549386721$$ $$23682^{2}=560837124$$ $$23728^{2}=563017984$$ $$23889^{2}=570684321$$ $$24009^{2}=576432081$$ $$24237^{2}=587432169$$ $$24276^{2}=589324176$$ $$24441^{2}=597362481$$ $$24807^{2}=615387249$$ $$25059^{2}=627953481$$ $$25279^{2}=639027841$$ $$25572^{2}=653927184$$ $$25941^{2}=672935481$$ $$26152^{2}=683927104$$ $$26351^{2}=694375201$$ $$26409^{2}=697435281$$ $$26733^{2}=714653289$$ $$27105^{2}=734681025$$ $$27129^{2}=735982641$$ $$27209^{2}=740329681$$ $$27273^{2}=743816529$$ $$27984^{2}=783104256$$ $$28171^{2}=793605241$$ $$28256^{2}=798401536$$ $$28346^{2}=803495716$$ $$28582^{2}=816930724$$ $$28731^{2}=825470361$$ $$29034^{2}=842973156$$ $$29106^{2}=847159236$$ $$29208^{2}=853107264$$ $$30384^{2}=923187456$$ $$2^{29}=536870912$$ So there are quite a few squares and just one higher power with that property.

$\endgroup$
1
  • $\begingroup$ I was looking for a purely mathematical way, without using a software, but it seems that there is no way. I just thought that these two questions are interested. Thank you very much for providing this. Your answer may be helpful to verify my solution (If I could solve). (+1) $\endgroup$ – Hussain-Alqatari Nov 20 '20 at 8:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.