Question

Does there exist a (base 10) 67-digit multiple of 2^67, written exclusively with the digits 6 and 7? Does there exist a (base 10) 67-digit multiple of 2^67, written exclusively with the digits 6 and 7? @Mathematics

Answer 1

definitely. You wouldn't ask such a question otherwise :-D

question is, what number is it?

Answer 2

6677767667676666776766667777767666677766776777777777777666766667776

Answer 3

6677767667676666776766667777767666677766776777777777777666766667776 = 2^67×45250313829053138281370553831260951302463537892.

Answer 4

I don't even want to know how you came up with that

Answer 5

Google is your friend :-D

Answer 6

Oh. I thought that maybe you used wolfram alpha

Answer 7

How do you tell wolfram alpha to output 67 digits?

Answer 8

Tell it to 'show more digits'

Answer 9

More generally, there exists a k-digit number, written exclusively with the digits 6 and 7, which is divisible by 2^k. The proof is by mathematical induction on k.

The induction will consist of two steps:

1. Basis: Show the proposition is true for a 1-digit number.
2. Induction step: Prove that if the proposition is true for some arbitrary value, k > 0, then it holds for k+1.

The basis is straightforward: we simply note that 6 is a single-digit multiple of 2.

For the induction step, we assume the inductive hypothesis: there exists a k-digit number, Nk, written exclusively with the digits 6 and 7, divisible by 2^k.

Since nk congruent to 0 (modulo 2k), we have nk congruent to 0 or 2^k (mod 2k+1).

Note that, since 10^k = 2^k×5^k:

6×10^k congruent to 0 (mod 2^k+1)
7×10^k congruent to 2^k (mod 2^k+1)

Hence, we choose:
Nk+1 = 6×10^k + Nk, if Nk congruent to 0 (mod 2^k+1); or
= 7×10^k + Nk, if Nk congruent to 2^k (mod 2^k+1)

It follows that Nk+1 is a ^(k+1)-digit number, written exclusively with the digits 6 and 7, and is divisible by 2^k+1.

This completes the inductive proof. Hence a 67-digit multiple of 2^67, written exclusively with the digits 6 and 7, does exist.

Answer 10

this one should be done analytically; no way am I going to try to brute-force my way to find that number using my calculator/computer program

Answer 11

lol

Answer 12

Gah

Answer 13

whats the gah? o.o