Problems & Puzzles: Puzzles

Puzzle 210.  Jeff's numbers

Jeff Heleen sent the following nice puzzle:

Let H(k) be the earliest number such that the digits in its k>1 prime factors be the same as in the number itself with no extras.

For example, the smallest number for k=2 to have this property is 1255=5*251.

The earliest H(k) for k=2 to 5 are:

k H(n) Factorization
2 1255 5*251
3 163797 3*71*769
4 11937639 3*7*61*9319
5 1037715385 5*7*7*83*51031

Question 1: Can you get the solutions for k=6 to 10?

I would like to add the following extensions to the Jeff's puzzle:

Q.2 Find a titanic H(k) (disregarding the "earliest" condition)

Q.3 Find the earliest Jeff's number H(2) expressible in two different ways as a product of two (not prime) factors

Solution:

Contributions came from J. C. Rosa, John Warf, J. K. Andersen & Johan Wiesenbauer.

J. C. Rosa wrote for question 3:

About the Puzzle 210 , question 3 , I have found the following results :

1260=6*210=21*60
1530=3*510=30*51
6880=8*860=80*86
12060=6*2010=60*201
and so on...

I asked him to avoid solutions ending in zero and this is what he replied in return:

if H(2) is not divisible by 10 : H(2)=105264=51*2064=204*516
If H(2) has not the digit 0 : H(2)=258795=9*28755=27*9585
and also the nice following solution : 268398=39*6882=93*2886

Maybe we have a follow-up: What is the next H(2) such that:
H(2)=C1*C2=R1*R2 ( R1 is the reverse of C1 and R2 the reverse of C2)

Later he added:

Two results for question 3 and for H(3)

127680=8*21*760=8*76*210=21*76*80
2669436=6*462*963=6*642*693=42*66*963

***

John Warf wrote for question 2:

Question 2: (while I'm thinking about the other questions) (59*10^1199+21) * 5 = 295*10^1199+105 is a titanic Jeff's number; well, probably (PFGW).

As an easier provable prime I'm working on (295*10^n+1) * 41. Of course we're unlikely to beat JK Andersen's excellent square vampire number solutions to Puzzle 199.

***

J. K. Andersen wrote for questions 1, 2 & 3:

Jeff's numbers are a type of prime vampire numbers. Puzzle 199 are about these.

Question 1:

H(6) = 117295838975 = 5*5*7*89*821*9173
H(7) = 11099654778737 = 7*7*7*7*89*541*96013
H(8) = 1091778783077899 = 7*7*7*7*7*809*8191*9803
H(9) = 1023976197718878397 = 7*7*7*7*61*89*971*983*82301

I have only tested cases with largest prime factor below 10^8 for H(9). There could be a smaller solution with a larger prime factor but it seems unlikely.

Question 2:

I used PrimeForm/GW to find many titanic solutions on the form prime^2 in puzzle 199. The largest was for a prime with 45429 digits: (94892254795*10^45418+1)^2 = 9004540020079200492025*10^90836+189784509590*10^45418+1.

Later (27/3/2003) he added:

I looked for usable base 10 primes among the 5000 largest known primes and found 193*10^69004+1, discovered by Milton L. Brown with PrimeForm/GW. I used PrimeForm/GW to find probable primes on the form ((10^24+9)/193)*10^x+(10^60+3)/193. Marcel Martin's Primo proved the prime with x=1444. This leads to a solution with 70473 digits:

(193*10^69004+1) * (((10^24+9)/193)*10^1444+(10^60+3)/193)

There are similar solutions with variations in the second prime. The form gives more non-zero digits than previous big solutions.

 

Question 3:

A trivial way to construct two-way solutions is by adding a zero in one of the factors for a one-way solution. 126 = 6*21 leads to the minimal two-way solution: 1260 = 6*210 = 21*60 The minimal solution not constructed this way is: 105264 = 51*2064 = 204*516 Only the digit 6 changes place in the two factorizations.

***

Johan Wiesenbauer wrote for question 1:

As for question 1 of puzzle 210 some results:

H(6) = 117295838975= 5^2*7*89*821*9173
H(7)= 11099654778737 = 7^4*89*541*96013
H(8) = 1091778783077899 = 7^5*809*8191*9803

***

J. C. Rosa continued his search for the question 3. Here are summarized his results (4/4/03):

  k  |  earliest H(k)                        earliest  H(k)
          | divisible by  10               not divisible by 10
--------------------------------------------------
      2  | 1260=6*210                    105264=51*2064
          |        =21*60                             =204*516
----------------------------------------------------------------------------------------
      3  | 13950=5*9*310                 2669436=6*462*963
          |          =5*31*90                           =6*642*693
          |          =9*31*50                           =42*66*963
----------------------------------------------------------------------------------------
       4 | 1169370=3*9*61*710      168478464=4*66*784*814
          |             =3*9*71*610                    =7*66*88*4144 
          |             =3*61*71*90                    =7*66*448*814
          |             =9*30*61*71                    =7*88*444*616

Later he added:

Two news:
1°) The earliest H(5) expressible in five different ways
divisible by 10 is :137979450=5*7*9*93*4710
                                         =5*7*9*471*930
                                         =5*7*90*93*471
                                         =5*9*70*93*471
                                         =7*9*50*93*471
 
 (About H(5) not divisible by 10 , I have found this number: 18738929664 which is expressible in 8 different ways of 5 factors !  but maybe it's not the earliest )

***

One week ahead he added:

In order to complete my contribution to the puzzle 210,
please , can you add the following result :
 
  " The earliest H(5) not divisible by 10 is : 1949688832
           1949688832=4*8*8*91*83692
                            =4*8*91*98*6832
                            =4*61*98*98*832
                            =8*8*49*91*6832
                            =8*49*61*98*832         

 

***

J. C. Rosa continues obtaining new interesting results related to this issue. On May 2003, he wrote:

Here are three new palindromic Jeff's numbers:
 

100199991001=11*9109090091   ( 9109090091 is prime )

       and the following pattern :
         1001*10^(2*k+6)+9999*10^(k+3)+1001
          =11*(91*10^(2*k+6)+909*10^(k+3)+91)  for k=>1
        The second factor is prime or probable prime for :
             k=1 ; 7 ; 143 ; 243 ; 439 ; 1903 ;...
             (I used PrimeForm up to k=2025 )

 

 
 10120799702101=11*920072700191   ( 920072700191 is prime )
 
and the  pattern :
 
1012*10^(2*k+6)+7997*10^(k+3)+2101
=11*(92*10^(2*k+6)+727*10^(k+3)+191)  for k=>1
 
The second factor is prime or probable prime for :
 k=2 ; 6 ; 423 ;...
(I used PrimeForm up to k=1718 )
 
 101972279101=11*9270207191   ( 9270207191 is not prime )
 
but we have  the following pattern :
 
10197*10^(2*k+5)+22*10^(k+4)+79101
=11*(927*10^(2*k+5)+2*10^(k+4)+7191)  for k=>1
 
The second factor is prime or probable prime for :
 k=23 ; 365 ; 1037 ;...
(I used PrimeForm up to k=2430 )

***

 
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