n-digit number multiple of 2^n, written with 6s’ and 7s’

 
 
Find the lowest n-digit number multiple of   2^{n},   written only with   6s’   and   7s’

n  \; \leq  \; 76

 

The first few examples,
 

n = 1   …   2^1   …   6
n = 2   …   2^2   …   76
n = 3   …   2^3   …   776
n = 4   …   2^4   …   7776
n = 5   …   2^5   …   67776
n = 6   …   2^6   …   667776
………………………………………..
………………………………………….
………………………………………….

 

6667776766677767667676666776766667777767666677766776777777777777666766667776
= 2^{76} \; \times \; 88247290723637794367680048552939376696792027490086866

is 76-digit multiple of   2^{76}

 

Find a 67-digit multiple of   2^{67}
 

 
 
 
Paul found:

6677767667676666776766667777767666677766776777777777777666766667776
= 2^{67} \; \times \; 45250313829053138281370553831260951302463537892
 

Using the digits 4 and 5:

554445454544445545455455454544555545454444544
= \; 2^{45} \; \times \; 15758287604070504399244767186717

4545554455554445454544445545455455454544555545454444544
= 2^{54} \; \times \; 252328960812200656346451529713240416716

 
 

                      ****************************************                      

Prove, by induction, that for every integer   n   there exists a n-digit multiple of   2^n   – (a multiple formed with digits 6 and 7 only.

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

7 Responses to n-digit number multiple of 2^n, written with 6s’ and 7s’

  1. paul says:

    I think this is the one

    147573952589676412928 x 45250313829053138281370553831260951302463537892 =
    6677767667676666776766667777767666677766776777777777777666766667776

    2^67 = 147573952589676412928.

    Paul.

    • benvitalis says:

      Yes. You may want to prove, by induction, that for every integer n there exists a
      n-digit multiple of 2^n – a multiple formed with digits 6 and 7 only.

  2. paul says:

    Here’s a 45 and 54 digit number consisting of just the digits 4 and 5 that are multiples of 2^45 and 2^54

    554445454544445545455455454544555545454444544 = 35184372088832 x 15758287604070504399244767186717 and

    4545554455554445454544445545455455454544555545454444544 = 36028797018963968 x 126164480406100328173225764856620208358

    P.

  3. paul says:

    Yes it can be done with any pair of digits such that the difference between the two numbers is odd.

    Here is my MMA code that will print all combinations out

    Do[f = s1; s = ToString[s1]; t = ToString[t1]; 
     Do[If[Mod[f, 2^(k + 1)] == 0, f = FromDigits[s <> ToString[f]], 
       f = FromDigits[t <> ToString[f]]]; 
      Print[f, "  ", 2^(k + 1), " x ", f/2^(k + 1), "  ", 
       IntegerLength[f]], {k, 1, 
       Max[(10 s1 + t1) - 1, (10 t1 + s1) - 1]}], {t1, 1, 9, 2}, {s1, 2, 
      8, 2}]

    There is quite a lot when it prints.

    P.

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