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 The Nine Digits Page 8with some Ten Digits (pandigital) exceptions | |||
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When I use the term ninedigital in these articles I always refer to a strictly zeroless pandigital (digits from 1 to 9 each appearing just once).
Topic 8.3 [ September 11, 2025 ]
The sum of all nine- and pandigital numbers.
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The manual approach by just counting them up is just unmanageable. There are \(9!\) terms which is \(362880\). There is however a feasible approach by counting first the units, then the tens, the hundreds, the thousands and so on. All nine digits \(1, 2, 3, 4, 5, 6, 7, 8\) and \(9\) occur several times as last digit (units). A possible order of the series There are \(8! = 8 * 7 * \ldots * 2 * 1 = 40320\) possibilities of those series of numbers. In a like manner one can reason that there are \(40320\) numbers ending with digit \(2\), We can find out that all ninedigits are discussed that way since So, the sum of all ninedigits with units \(1\) is easy to calculate now The second-to-last digit in a number indicates the tens digit. In the same way as with the units, The total sum of all nine-digit numbers considered is thus The following Pari/GP code confirms our result
The output is also \(201599999798400\) |
This manual approach is analogue to the ninedigital case. The manual approach by just counting them up is just unmanageable. There are \(9*9!\) terms which is \(3265920\). All ten digits \(0, 1, 2, 3, 4, 5, 6, 7, 8 and 9\) occur several times as last digit (units). A possible order of the series There are \(9! = 9 * 8 * 7 * \ldots * 2 * 1 = 362880\) possibilities of those series of numbers. In a like manner one can reason that there are \(362880\) numbers ending with digit \(1\), We can find out that all pandigitals are discussed that way since So, the sum of all pandigitals with units 0 is easy to calculate now The second-to-last digit in a number indicates the tens digit. In the same way as with the units, The total sum of all nine-digit numbers considered is thus But this is not the end of the story as pandigitals are not supposed to start with a leading zero. All we have to do is calculate the difference
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What is the sum of all ninedigitals ?Sources :
Formula para calcular la suma de las N combinaciones posibles con M digitos consecutivos
Reken mee : de som van alle negencijferige getallen waarin elk van de cijfers van 1 t.e.m. 9 slechts één keer voorkomt (Jaargang 28 pag 22-24)
OEIS A050278 - Pandigital numbers: numbers containing the digits 0-9. Version 1: each digit appears exactly once.
OEIS A065583 - Sum of numbers which in base n have (n-1) distinct nonzero digits.
OEIS A071268 - Sum of all digit permutations of the concatenation of first n numbers.
OEIS A213880 - a(n) = sum of n-digit numbers with distinct nonzero digits.
Topic 8.2 [ August 24, 2025 ]
Nine- and pandigital's equal to the sum of two, three and four positive and/or negative fifthpowers.
| Case of ninedigitals | Case of pandigitals | ||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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One beautiful solution with only TWO fifthpowers ! \(\Large{\bbox[lightyellow,5px, border:1px green solid]{{\color{blue}{312496875}}={\color{purple}{50^5-5^5}}}}\) Ninedigital solutions with THREE fifthpowers. Here is a complete list, one column with positive terms and another with
Ninedigital solutions with FOUR fifthpowers.
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One beautiful solution with only TWO fifthpowers ! \(\Large{\bbox[mintcream,5px, border:1px green solid]{{\color{blue}{2109574368}}={\color{purple}{78^5-60^5}}}}\) Pandigital solutions with THREE fifthpowers. Here is a complete list, one column with positive terms and another with
Pandigital solutions with FOUR fifthpowers.
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Topic 8.1 [ August 24, 2025 ]
Nine- and pandigital's equal to the sum of five positive and/or negative fifthpowers. Building a longer and longer list of solutions.
Regarding the ninedigitals we know from Topic 4.5 that there is only one solution with five positive terms
so that when all these five terms are concatenated we get another nice ninedigital !
\(\Large{\bbox[lightyellow,5px,border:1px green solid]{{\color{blueviolet}{816725493}}={\color{green}{9}}^5+{\color{green}{17}}^5+{\color{green}{26}}^5+{\color{green}{43}}^5+{\color{green}{58}}^5~~\to~~{\color{green}{917264358}}~~\text{ is also ninedigital ! }}}\)
| Case of ninedigitals | Case of pandigitals | |
|---|---|---|
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Firstly how many ninedigital solutions are there in general with five only positive terms ? \({\color{maroon}{123486957}}=1^5+4^5+19^5+22^5+41^5\) \({\color{maroon}{123495687}}=7^5+22^5+24^5+24^5+40^5\) \({\color{maroon}{123764895}}=12^5+28^5+28^5+31^5+36^5\) \({\color{maroon}{123765489}}=15^5+21^5+27^5+27^5+39^5\) \({\color{maroon}{123789564}}=2^5+5^5+6^5+32^5+39^5\) \({\color{maroon}{123798564}}=1^5+5^5+7^5+32^5+39^5\) \({\color{maroon}{123864597}}=9^5+19^5+19^5+31^5+39^5\) \({\color{maroon}{124358697}}=4^5+4^5+14^5+24^5+41^5\) \({\color{maroon}{124598376}}=9^5+18^5+31^5+32^5+36^5\) \({\color{maroon}{124689357}}=2^5+19^5+27^5+31^5+38^5\) ⋮ Some ninedigitals can be expressed in two ways, some of which are multigrades. The multigrades here happen when using exponent \(1\) & \(5\) and they sum up to equal values. Only these three solutions exist. \(\bbox[mintcream,3px,border:1px solid]{{\color{maroon}{617492358}}\mathbf{\color{blue}{\;=\;}}16^5+25^5+29^5+46^5+52^5\mathbf{\color{blue}{\;=\;}}21^5+21^5+27^5+49^5+50^5}\) \(16+25+29+46+52\mathbf{\color{blue}{\;=\;}}168\mathbf{\color{blue}{\;=\;}}21+21+27+49+50\) \(\bbox[mintcream,3px,border:1px solid]{{\color{maroon}{764251389}}\mathbf{\color{blue}{\;=\;}}16^5+32^5+44^5+45^5+52^5\mathbf{\color{blue}{\;=\;}}23^5+31^5+33^5+51^5+51^5}\) \(16+32+44+45+52\mathbf{\color{blue}{\;=\;}}189\mathbf{\color{blue}{\;=\;}}23+31+33+51+51\) \(\bbox[mintcream,3px,border:1px solid]{{\color{maroon}{895762143}}\mathbf{\color{blue}{\;=\;}}2^5+38^5+41^5+49^5+53^5\mathbf{\color{blue}{\;=\;}}13^5+29^5+37^5+48^5+56^5}\) \(2+38+41+49+53\mathbf{\color{blue}{\;=\;}}183\mathbf{\color{blue}{\;=\;}}13+29+37+48+56\) ⋮ \({\color{maroon}{985327164}}=13^5+17^5+37^5+50^5+57^5\) \({\color{maroon}{986275413}}=4^5+30^5+47^5+47^5+55^5\) \({\color{maroon}{986354172}}=15^5+22^5+22^5+42^5+61^5\) \({\color{maroon}{986527413}}=4^5+4^5+16^5+37^5+62^5\) \({\color{maroon}{986542173}}=1^5+7^5+16^5+37^5+62^5\) \({\color{maroon}{986713542}}=5^5+38^5+49^5+50^5+50^5\) \({\color{maroon}{987362451}}=20^5+35^5+41^5+50^5+55^5\) \({\color{maroon}{987562134}}=11^5+17^5+39^5+51^5+56^5\) \({\color{maroon}{987615432}}=3^5+12^5+18^5+37^5+62^5\) \({\color{maroon}{987623415}}=4^5+24^5+28^5+54^5+55^5\) Secondly I would like you to start a list of solutions with ninedigitals equal to five mixed signs terms. \({\color{blue}{123456789}}=v^5+w^5+x^5+y^5+z^5\) \({\color{maroon}{123458976}}=(-84)^5+(-228)^5+276^5+314^5+(-332)^5\) \({\color{maroon}{123459687}}=11^5+87^5+(-102)^5+(-187)^5+188^5\) \({\color{maroon}{123459768}}=9^5+(-52)^5+(-164)^5+(-192)^5+207^5\) \({\color{maroon}{123459876}}=(-214)^5+(-448)^5+532^5+553^5+(-597)^5\) \({\color{maroon}{123468597}}=59^5+(-115)^5+(-451)^5+(-616)^5+640^5\) \({\color{maroon}{123469587}}=824^5+1091^5+(-1114)^5+1138^5+(-1162)^5\) \({\color{maroon}{123497856}}=(-121)^5+(-493)^5+503^5+670^5+(-673)^5\) \({\color{maroon}{123568974}}=152^5+(-445)^5+(-461)^5+(-497)^5+585^5\) \({\color{maroon}{123589467}}=24^5+30^5+124^5+276^5+(-277)^5\) \({\color{maroon}{123648975}}=(-19)^5+30^5+67^5+94^5+(-97)^5\) ⋮ \({\color{maroon}{987621534}}=(-82)^5+(-176)^5+193^5+370^5+(-371)^5\) \({\color{maroon}{987623451}}=(-320)^5+375^5+(-384)^5+(-445)^5+465^5\) \({\color{maroon}{987624153}}=98^5+729^5+1308^5+1440^5+(-1592)^5\) \({\color{maroon}{987625431}}=181^5+522^5+639^5+800^5+(-861)^5\) \({\color{maroon}{987635142}}=80^5+(-89)^5+(-141)^5+(-171)^5+183^5\) \({\color{maroon}{987641235}}=22^5+109^5+(-177)^5+(-422)^5+423^5\) \({\color{maroon}{987651324}}=(-106)^5+164^5+241^5+251^5+(-286)^5\) \({\color{maroon}{987652413}}=(-130)^5+(-814)^5+(-1004)^5+(-1427)^5+1488^5\) \({\color{maroon}{987653142}}=143^5+295^5+(-352)^5+(-436)^5+452^5\) \({\color{maroon}{987654231}}=(-228)^5+(-266)^5+311^5+420^5+(-426)^5\) \({\color{blue}{987654321}}=v^5+w^5+x^5+y^5+z^5\) |
Firstly how many pandigital solutions are there in general with five only positive terms ? \({\color{maroon}{1023459867}}=4^5+13^5+20^5+53^5+57^5\) \({\color{maroon}{1023469587}}=13^5+28^5+34^5+41^5+61^5\) \({\color{maroon}{1023469587}}=15^5+18^5+21^5+30^5+63^5\) \({\color{maroon}{1023479568}}=14^5+16^5+29^5+51^5+58^5\) \({\color{maroon}{1023685497}}=9^5+9^5+28^5+39^5+62^5\) \({\color{maroon}{1023846957}}=2^5+15^5+20^5+53^5+57^5\) \({\color{maroon}{1023897456}}=2^5+3^5+45^5+52^5+54^5\) \({\color{maroon}{1024536789}}=34^5+37^5+45^5+51^5+52^5\) \({\color{maroon}{1024539867}}=30^5+38^5+39^5+41^5+59^5\) \({\color{maroon}{1024576389}}=16^5+24^5+45^5+47^5+57^5\) ⋮ Some pandigitals can be expressed in two ways, some of which are multigrades. The multigrades here happen when using exponent \(1\) & \(5\) and they sum up to equal values. Three examples from the collection. \(\bbox[gold,3px,border:1px solid]{{\color{maroon}{1243879506}}\mathbf{\color{blue}{\;=\;}}5^5+24^5+47^5+55^5+55^5\mathbf{\color{blue}{\;=\;}}9^5+27^5+40^5+49^5+61^5}\) \(5+24+47+55+55\mathbf{\color{blue}{\;=\;}}186\mathbf{\color{blue}{\;=\;}}9+27+40+49+61\) \(\bbox[gold,3px,border:1px solid]{{\color{maroon}{5062178349}}\mathbf{\color{blue}{\;=\;}}2^5+12^5+63^5+63^5+79^5\mathbf{\color{blue}{\;=\;}}20^5+24^5+30^5+61^5+84^5}\) \(2+12+63+63+79\mathbf{\color{blue}{\;=\;}}219\mathbf{\color{blue}{\;=\;}}20+24+30+61+84\) \(\bbox[gold,3px,border:1px solid]{{\color{maroon}{8932546107}}\mathbf{\color{blue}{\;=\;}}4^5+43^5+61^5+67^5+92^5\mathbf{\color{blue}{\;=\;}}12^5+21^5+75^5+75^5+84^5}\) \(4+43+61+67+92\mathbf{\color{blue}{\;=\;}}267\mathbf{\color{blue}{\;=\;}}12+21+75+75+84\) Note that when the equations of the multigrades are not equal then the difference of the sums of the base terms are multiples of \(30\) ! Here is an example with \(3*30=90\) with pandigital \(5986421307\) and \(14^1+21^1+22^1+63^1+87^1=207\) minus \(40^1+51^1+54^1+75^1+77^1=297\) which gives a difference of \(90\). Can someone explain this observation mathematically ? ⋮ \({\color{maroon}{9875034126}}=1^5+51^5+54^5+81^5+89^5\) \({\color{maroon}{9875064123}}=14^5+22^5+74^5+75^5+88^5\) \({\color{maroon}{9875103426}}=1^5+1^5+14^5+76^5+94^5\) \({\color{maroon}{9875134026}}=26^5+43^5+72^5+79^5+86^5\) \({\color{maroon}{9875423601}}=30^5+38^5+52^5+81^5+90^5\) \({\color{maroon}{9875642310}}=6^5+39^5+63^5+68^5+94^5\) \({\color{maroon}{9876153024}}=4^5+14^5+16^5+76^5+94^5\) \({\color{maroon}{9876204315}}=17^5+22^5+46^5+77^5+93^5\) \({\color{maroon}{9876341025}}=30^5+48^5+54^5+56^5+97^5\) \({\color{maroon}{9876531024}}=7^5+9^5+18^5+76^5+94^5\) Secondly I would like you to start a list of solutions with pandigitals equal to five mixed signs terms. \({\color{blue}{1023456789}}=v^5+w^5+x^5+y^5+z^5\) \({\color{maroon}{1023459768}}=(-51)^5+75^5+203^5+206^5+(-235)^5\) \({\color{maroon}{1023476598}}=(-186)^5+(-401)^5+554^5+976^5+(-985)^5\) \({\color{maroon}{1023479568}}=59^5+80^5+(-190)^5+(-267)^5+276^5\) \({\color{maroon}{1023486957}}=(-181)^5+191^5+241^5+327^5+(-341)^5\) \({\color{maroon}{1023495876}}=(-620)^5+1056^5+(-1215)^5+(-1268)^5+1363^5\) \({\color{maroon}{1023497658}}=(-174)^5+(-1043)^5+1165^5+1188^5+(-1268)^5\) \({\color{maroon}{1023568974}}=13^5+785^5+1130^5+1744^5+(-1788)^5\) \({\color{maroon}{1023645798}}=(-1126)^5+1260^5+(-1273)^5+(-1332)^5+1439^5\) \({\color{maroon}{1023694587}}=(-46)^5+(-210)^5+267^5+334^5+(-348)^5\) \({\color{maroon}{1023758649}}=(-2)^5+(-56)^5+(-283)^5+(-357)^5+377^5\) ⋮ \({\color{maroon}{4768523109}}=6061^5+10665^5+10920^5+11786^5+(-13953)^5\) ⋮ \({\color{maroon}{9876203415}}=430^5+(-713)^5+850^5+1456^5+(-1468)^5\) \({\color{maroon}{9876204315}}=(-69)^5+(-373)^5+549^5+1436^5+(-1438)^5\) \({\color{maroon}{9876450123}}=22^5+382^5+411^5+1186^5+(-1188)^5\) \({\color{maroon}{9876451230}}=525^5+(-566)^5+(-588)^5+(-710)^5+769^5\) \({\color{maroon}{9876502314}}=340^5+492^5+(-573)^5+(-617)^5+652^5\) \({\color{maroon}{9876520413}}=122^5+(-146)^5+652^5+781^5+(-836)^5\) \({\color{maroon}{9876524031}}=372^5+(-1431)^5+1501^5+1784^5+(-1815)^5\) \({\color{maroon}{9876530124}}=(-185)^5+(-267)^5+450^5+544^5+(-578)^5\) \({\color{maroon}{9876540321}}=(-479)^5+(-664)^5+(-1202)^5+(-1534)^5+1620^5\) \({\color{maroon}{9876541320}}=-19^5+72^5+(-768)^5+(-1323)^5+1340^5\) \({\color{maroon}{9876542301}}=(-180)^5+(-219)^5+324^5+381^5+(-405)^5\) \({\color{blue}{9876543210}}=v^5+w^5+x^5+y^5+z^5\) |
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Patrick De Geest - Belgium
- Short Bio - Some PicturesE-mail address : pdg@worldofnumbers.com