Problems & Puzzles: Puzzles

Puzzle 940. Digits on p^k Carlos Rivera poses the following puzzle: The digital decimal expression of a powered prime (p^k) cannot exhibit the following behavior: “All the 10 digits 0-9 to be present in the same quantity”. At the much in p^k  can be present only 9 from these 10 digits in the same quantity. Here are some examples: 1) 30059^4 = 816390822057597361 (9x2=18 total digits, 2 of each, digit absent = 4) 2) 573206377^4 = 107955321063072794563791236590244641 (9x436 total digits, 4 of each, digit absent = 8) 3) 646582319^5 = 113010584876708763099460467843153845593167599 (9x5=45 total digits, 5 of each, digit absent = 2) If my search was correct, these are the minimal prime p that powered to some exponent k give a certain total length 9m. Q. Find more solutions alike for other m values.

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Contributions came from Fred Schneider, Paul Cleary and Emmanuel Vantieghem

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Fred wrote on Jan 27, 2019:

I checked powers of 3 for any equally-occurring pandigital (EOP) powers.  I did it through 3^108100 which is a 51530-digit number and found no solutions.
 

Factoring in that there are only 9 choices for the lead digit, the number of EOP 10n digits is
(10n -1)!
--------------------
9 * n!^9 * (n-1)!
 

Dividing by 10^10n ...
 

The odds of a 30 digit number being EOP is roughly 1 in 20516106.
 

The odds of a 1000 digit number being EOP is roughly 1 in 112107542996523.
The odds of a 10000-digit number being EOP is roughly 1 in 3518926614958095848.
The odds of a 51530-digit number being EOP is roughly 1 in 5628492465562741668243
 

Clearly, the odds worsen as the number of digits increases.

Assuming even distribution, the odds of an even power of 3 being EOP is approximately the same.  (One obvious difference is that there are only 4 choices for the unit digit.  This difference diminishes as the number of digits increases.. 
 

So, it doesn't seem impossible but just extremely unlikely.

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Paul wrote on Jan 30, 2019

For m = 2 there are 3615 solutions where k also = 2, of all the 3615 solutions there were only 3 different absent digits, these were 1, 4 and 7.  The distribution were 1 = 724, 4 = 1228, 7 = 1663.  Here are the first occurrences of each missing digit and the very last solution.

 

1 , 317912407^2 = 101068298524533649, 2  of each, digit absent = 7

2 , 318374117^2 = 101362078375529689, 2  of each, digit absent = 4

539 , 450661117^2 = 203095442375687689, 2  of each, digit absent = 1

3615 , 999416681^2 = 998833702261055761, 2  of each, digit absent = 4.

 

 

Couldn’t find any with m = 3 and k = 3.

 

these are all the remaining ones I could find.  I searched all k up to 50.  They are in P order.

 

1 , 573206377^4 = 107955321063072794563791236590244641, 4  of each, digit absent = 8

2 , 591733591^4 = 122604071997086238376129447403388961, 4  of each, digit absent = 5

3 , 592540889^4 = 123274513969276740751065045996323041, 4  of each, digit absent = 8

4 , 640691917^4 = 168498864982307046762937031197042321, 4  of each, digit absent = 5

5 , 646582319^5 = 113010584876708763099460467843153845593167599, 5  of each, digit absent = 2

6 , 675377389^5 = 140518201908334268225489915631304536626805949, 5  of each, digit absent = 7

7 , 736221287^4 = 293787613714736930922420880864904161, 4  of each, digit absent = 5

8 , 740888891^4 = 301309157987838504547656967984034161, 4  of each, digit absent = 2

9 , 748055639^4 = 313137878096072468684227394690219041, 4  of each, digit absent = 5

10 , 776860309^5 = 282953236690124480549131866181924053568300549, 5  of each, digit absent = 7

11 , 780009211^4 = 370168044703599576897116358089635441, 4  of each, digit absent = 2

12 , 806577311^4 = 423237367648064719082781930126099841, 4  of each, digit absent = 5

13 , 822665869^5 = 376805642659037238808945979283266725507440349, 5  of each, digit absent = 1

14 , 838035943^4 = 493231247993947012188276886630764001, 4  of each, digit absent = 5

15 , 839105933^4 = 495755073403367041207491566292639121, 4  of each, digit absent = 8

16 , 875064433^5 = 513097810569728611227680695216550372803897393, 5  of each, digit absent = 4

17 , 886546301^4 = 617739949751345650404726103296325201, 4  of each, digit absent = 8

18 , 890016781^4 = 627469731677514302930426510399045521, 4  of each, digit absent = 8

19 , 924094139^4 = 729230564472157536903917396126045041, 4  of each, digit absent = 8

20 , 940278197^4 = 781673634298908269420203730176914481, 4  of each, digit absent = 5

21 , 968135923^5 = 850514408393850596669014214923262836121095843, 5  of each, digit absent = 7

22 , 976940009^4 = 910901848064571768733943958357046561, 4  of each, digit absent = 2

23 , 992047591^5 = 960865353886448602542209346343921101291058951, 5  of each, digit absent = 7

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Emmanuel wrote on Feb 1, 2019:

First of all, it is not impossible that a power of the prime 3  has all digits with the same quantity.

But I could not find such power.

 

To make it easier, I call numbers with exactly one missing digit and all the other digits appearing with the same frequency : pseudo pandigital numbers.
 

Squares

The smallest prime whose square is pseudo pandigital is  13147.

There are lots of others.

The biggest I could find is  316229524664727413.

 

Cubes

The smallest four primes with pseudo pandigital cube are :

667333, 678299, 464207860637, 464159055289103.

 

Fourth powers

The smallest prime with pseudo pandigital fourth power is indeed  30059.

The next sixteen such primes are 

573206377, 591733591, 592540889, 640691917, 736221287, 740888891, 748055639, 780009211, 806577311, 838035943, 839105933, 886546301, 890016781, 924094139, 940278197, 976940009.

 

Fifth powers

The smallest eight primes with pseudo pandigital fifth power are :

646582319 (indeed), 675377389, 776860309, 822665869, 875064433, 968135923, 992047591, 2512711521013.

 

Sixth powers

The smallest five primes with pseudo pandigital sixth power are :

24813047167, 25606402561, 26914871077, 30007974149, 31230364037.

 

I could not find a prime whose seventh power is pseudo pandigital.  If it exist, it mut be bigger than  5183810086813.

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