Problems & Puzzles: Puzzles

Puzzle 1199 10245 & 10246 Gaydos found (when?) that 10245 and 10246 are the smallest pair of consecutive composite numbers whose prime factors together are pandigital (3*5*683 and 2*47*109). Himself reported that the next such pair is 15734 and 15735. [Gaydos] Q1. Find he largest pair of these. Q2. Redo Q1 but such that both the pair and the prime factors of that pair are pandigitals

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From Dec 1 to Dec 6, contributions came from Adam Stinchcombe, JM Rebert, Emmanuel Vantieghem

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Adam wrote:

As currently formulated the first part of the puzzle has unbounded answers.  If you allow repeat digits (the 3 in 10245=3*5*683 and the 7 in 15734=2*7867), then you can create infinitely many solutions.  By Dirichlet's theorem, there are infinitely many primes of the form   k =  10^11*n+1023456789, then a pair of consecutive composite numbers are 3*k. 3*k+1 and the prime factor k is pandigital, so it doesn't matter what the other digits in the other factors.

 

If you limit the puzzle to using exactly all digits only once in the prime factorization (for instance, square free), I find 8609 is a prime and 8610 = 2*3*5*7*41.  This version of Q1 clearly has an upper bound but it is not immediately clear what that upper bound is.

 

Similarly, for Q2, if you allow repeats of digits.  You just need a seed value a such that both a and 3a are pandigital.  Then, like above, k=10^m*n+a for sufficiently large m, infinitely many different n by Dirichlet, use 3k and 3k+1.  For instance if a = 37074000148185259296123 (which is pandigital) and 3a = 111222000444555777888369 and use m = 25.

 

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Rebert wrote:

Q1:  I found  the following greatest n-digits composite numbers k and k+1.

n k

5 99282

6 999905

7 9999873

8 99999942

9 999999993

10 9999999994

11 99999999992

12 999999999994

13 9999999999998

14 99999999999998

15 999999999999997

16 9999999999999997

17 99999999999999998

18 999999999999999998

19 9999999999999999997

20 99999999999999999998

21 999999999999999999996

22 9999999999999999999998

23 99999999999999999999996

24 999999999999999999999998

25 9999999999999999999999998

26 99999999999999999999999998

27 999999999999999999999999997

28 9999999999999999999999999998

29 99999999999999999999999999998

30 999999999999999999999999999997

31 9999999999999999999999999999998

32 99999999999999999999999999999998

33 999999999999999999999999999999998

34 9999999999999999999999999999999998

35 99999999999999999999999999999999998

36 999999999999999999999999999999999998

37 9999999999999999999999999999999999998

38 99999999999999999999999999999999999997

39 999999999999999999999999999999999999998

40 9999999999999999999999999999999999999998

41 99999999999999999999999999999999999999998

42 999999999999999999999999999999999999999998

43 9999999999999999999999999999999999999999998

44 99999999999999999999999999999999999999999998

45 999999999999999999999999999999999999999999998

46 9999999999999999999999999999999999999999999998

47 99999999999999999999999999999999999999999999998

48 999999999999999999999999999999999999999999999998

49 9999999999999999999999999999999999999999999999998

50 99999999999999999999999999999999999999999999999998

 

The sets of digits in the prime factorizations of k and k+1 are disjoint.

 

88034 = 2 * 44017 and 88035 = 3 * 5 * 5869 with disjoint set of digits [0, 1, 2, 4, 7] and [3, 5, 6, 8, 9] in their prime factorisation, respectively.

998845 = 5 * 107 * 1867 and 998846 = 2 * 499423 with disjoint set of digits [0, 1, 5, 6, 7, 8] and [2, 3, 4, 9] in their prime factorisation, respectively.

9994384 = 2^4 * 624649 and 9994385 = 5 * 17 * 307 * 383 with disjoint set of digits [2, 4, 6, 9] and [0, 1, 3, 5, 7, 8] in their prime factorisation, respectively.

99998068 = 2^2 * 53 * 83 * 5683 and 99998069 = 709 * 141041 with disjoint set of digits [2, 3, 5, 6, 8] and [0, 1, 4, 7, 9] in their prime factorisation, respectively.

999995996 = 2^2 * 249998999 and 999995997 = 3 * 7 * 11 * 13 * 53 * 61 * 103 with disjoint set of digits [2, 4, 8, 9] and [0, 1, 3, 5, 6, 7] in their prime factorisation, respectively.

9999994323 = 3 * 11 * 303030131 and 9999994324 = 2^2 * 47629 * 52489 with disjoint set of digits [0, 1, 3] and [2, 4, 5, 6, 7, 8, 9] in their prime factorisation, respectively.

99999943376 = 2^4 * 6249996461 and 99999943377 = 3 * 883 * 37750073 with disjoint set of digits [1, 2, 4, 6, 9] and [0, 3, 5, 7, 8] in their prime factorisation, respectively.

999999920721 = 3 * 333333306907 and 999999920722 = 2 * 11 * 45454541851 with disjoint set of digits [0, 3, 6, 7, 9] and [1, 2, 4, 5, 8] in their prime factorisation, respectively.

9999999920355 = 3 * 5 * 666666661357 and 9999999920356 = 2^2 * 2499999980089 with disjoint set of digits [1, 3, 5, 6, 7] and [0, 2, 4, 8, 9] in their prime factorisation, respectively.

99999999925593 = 3 * 33333333308531 and 99999999925594 = 2 * 49999999962797 with disjoint set of digits [0, 1, 3, 5, 8] and [2, 4, 6, 7, 9] in their prime factorisation, respectively.

999999999569979 = 3^2 * 111111111063331 and 999999999569980 = 2^2 * 5 * 49999999978499 with disjoint set of digits [0, 1, 3, 6] and [2, 4, 5, 7, 8, 9] in their prime factorisation, respectively.

9999999999917181 = 3^3 * 370370370367303 and 9999999999917182 = 2 * 4999999999958591 with disjoint set of digits [0, 3, 6, 7] and [1, 2, 4, 5, 8, 9] in their prime factorisation, respectively.

99999999999998577 = 3^3 * 3703703703703651 and 99999999999998578 = 2 * 49999999999999289 with disjoint set of digits [0, 1, 3, 5, 6, 7] and [2, 4, 8, 9] in their prime factorisation, respectively.

999999999999885875 = 5^3 * 167 * 47904191616761 and 999999999999885876 = 2^2 * 3 * 83333333333323823 with disjoint set of digits [0, 1, 4, 5, 6, 7, 9] and [2, 3, 8] in their prime factorisation, respectively.

9999999999999930992 = 2^4 * 624999999999995687 and 9999999999999930993 = 3 * 3333333333333310331 with disjoint set of digits [2, 4, 5, 6, 7, 8, 9] and [0, 1, 3] in their prime factorisation, respectively.

9999999999999930992 = 2^4 * 624999999999995687 and 9999999999999930993 = 3 * 3333333333333310331 with disjoint set of digits [2, 4, 5, 6, 7, 8, 9] and [0, 1, 3] in their prime factorisation, respectively.

99999999999999804095 = 5 * 19999999999999960819 and 99999999999999804096 = 2^6 * 3^3 * 7 * 37 * 223437723437723 with disjoint set of digits [0, 1, 5, 6, 8, 9] and [2, 3, 4, 7] in their prime factorisation, respectively.

 

Q2; 

 

I found these greatest n-digit composite numbers k and k+1, for which their concatenation is pandigital.

 

n k

 

9 867410592

 

10 9982104356

 

11 99912643507

 

12 999830172845

 

13 9999243716084

 

14 99996123074805

 

15 999987610341351

 

16 9999995116270863

 

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Emmanuel wrote:

Q1.
I found  4769  numbers  m < 10^6  such that the prime factors of  m  and  m+1  together are pandigital.
So, It seems clear that there will not be a biggest  m.
If  "pandigital expression" means  "every digit appears once and only once in the expression", then  Gaydos' solutions do not satisfy (the digit  3  is repeated in the prime factors of  10245, 10246 ; the digit  7  appears twice in the prime factors of  15734).
In that case, the smallest  m  would be  69040 (prime factors  2, 5, 683 ; prime factors of  69041 : 7, 1409)..
The next pair then would be : (4055552 (with prime factors 2, 89), 4055553 (with prime factors  3, 450617)).  Is it the biggest  ?  

Q2.
When repetition of digits is not allowed, there cannot exist a pandigital pair  (m,m+1).
When we allow repetition of integers, the smallest  m  is  12678349 ;
  Prime factors of  m : 31, 408979
  Prime factors of  m + 1 : 2, 5, 253567.
Also, there I presume there will be an unlimited number of solutions.

Later he added

If we do not allow repeated digits and if we do not require  m, n  consecutive then there are 86 "solutions" :
{{10928,34765},{12860,57439},{16,45823079},{16784,23905},{172,5046389},{175,9426803},

{185,2407396},{2045,819376},{2164,589037},{25,16087934},{25,46310789},{25,47619038},

{25,73896104},{25,90871634},{25,96718034},{287,5613904},{301,8974256},{31049,76528},

{32081,97456},{34208,96175},{35,68971024},{35,76821904},{35,79420816},{365,1879204},

{365,8297104},{4,127809365},{4,170826935},{4,173629085},{4,193862705},{4,238679105},

{4,239718605},{4,269701385},{4,278190365},{4,315867209},{4,370169825},{4,381679205},

{4,862903715},{5063,784912},{5873,902416},{6904,173825},{80,21463579},{80,34216579},

{80,54273691},{8,126930547},{8,145209367},{8,175346029},{8,175694203},{8,243970561},

{8,264975013},{8,294057613},{8,320941567},{8,321054697},{8,329051467},{8,360472591},

{8,365412907},{841,3796205},{8,417206953},{8,419062573},{841,9762305},{8,431205967},

{845,3072196},{8,461320957},{8,470235169},{8,495102763},{8,496235017},{8,524176903},

{8,570126493},{8,573614029},{8,601294537},{8,603215479},{8,605249371},{8,614792053},

{8,635291047},{8,670531429},{9025,681374},{9025,768134},{9025,813674},{9025,813746},

{9104,562387},{94,36821705},{9475,136082},{95,63184027},{976,4235801},{976,5308241},

{98,31640725},{98,34601725}}
The smallest  m + n  is  40689 :
   m =16784  with prime factors  2, 1049  
   n = 23905  with prime factors  5, 7, 683
The biggest  m + n  is  862903719 :
   m = 4  with prime factor  2
   n = 862903715  with prime factors  5, 19, 478063
Of course, there are other ways to declare what is "smallest" and "biggest".
Observe that in none of the  86  solutions  m, n  are both squarefree.

 

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