## 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.

math grad - Interest: Number theory
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### 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.

• benvitalis says:

Yeah, digits don’t cooperate when the difference is even. I’ve tried to find few examples
e.g. combos (2,8), (6,4)