Prove that a power of 2 cannot end with four equal digits.

Question

The questions states that: Prove that a power of 2 cannot end with four equal digits in base 10.

My try:

I tried by assuming there to be such a power of 2.

Since, $2^n≡0$ $(mod$ $16)$

$abbbb≡10^4a+1111b$ $ (mod$ $16)$

And then I can't proceed. Could someone continue the solution or solve it in a better and easier way?

Question didn't state any other bases, so I suppose we have to take base 10. — , Aug 19 '24 at 14:08
Last digits need to be even ; it cannot be 0, neither 2 (numbers finishing with 22 are not multiple of 4) , neither 4, neither 6, neither 8. — Lourrran, Aug 19 '24 at 14:10
Well, $2^{39}$ ends in $888$, so our solution must be something about $4$ equal digits, and not $3$. — GEdgar, Aug 19 '24 at 14:27
Hint: $10^4a+1111b\equiv 7b \bmod 16$. Can $7b\equiv 0 \bmod 16$? — Keith Backman, Aug 19 '24 at 15:59

Bill Dubuque · Answer 1 · 2024-08-19T18:37:42.363

2

The last $4$ digits of $\,2^n$ can't be equal (i.e. of form $\,b\cdot \color{#0a0}{1111},\, b\le9),$ since, more generally, they can't have the form $\,b\:\!\:\!\color{#0b0}c\,$ for $\,b<2^4<\color{#0a0}{c\ \rm odd}.\,$ Proof: $ $ if $\,2^n = 10^4 a \!+\! bc\,$ then $\,\color{#c00}{b\neq 0}\,$ (else $5\mid 2^n)\,$ so $\,2^n\ge \color{#c00}bc\ge \color{#c00}1\cdot c>2^4\,$ so $\,n > 4\,$ so $\,2^4\mid bc,\ c\,$ odd, so $\,2^4\mid b,\,$ contra $\,\color{#c00}{0< b} < 2^4$.

edited Aug 19 '24 at 18:37

answered Aug 19 '24 at 18:30

Bill Dubuque

282,220

Same for any composite radix $,de,,(d,e)!=!1,,$ by $,(d,c)!=!1,\ d^k!\mid bc,\Rightarrow, d^k!\mid b,$ by Euclid's Lemma. $\ \ $ – Bill Dubuque Aug 19 '24 at 19:22

user1176409 · Accepted Answer · 2024-08-21T12:51:37.773

1

Let’s consider the value of $1111b$ can’t be bigger than $ 10000$ (yes, you can bigger than $10000$ but the last four digits will still from a range of $0000-9999$, so it will be easy if we think $1111b$ less than $10000$ as if $b=1-9$, all the value from $0000-9999$ will be included.)

First, b can’t be an odd integer, or else $1111b$ also will be an odd integer.

Second, $b$ can’t be $0$. Or else the number will end as 0 ( divided by $5$).

Test yourself. If $b=2,4,6,8,$ what will happen?

edited Aug 21 '24 at 12:51

answered Aug 19 '24 at 14:10

user1176409

1

Here. I guess the OP asked for base $10$. – user1176409 Aug 19 '24 at 14:11
How did you discard $a$ from the expression? – Aug 19 '24 at 14:11
Is it because divisibility of 16 is dependent on last four digits? – Aug 19 '24 at 14:13
You use “mod $16$” in the expression, so you must only focus in the last four digits in the expression. – user1176409 Aug 19 '24 at 14:13
Oh, so none of the values of b will satisfy because we get $≡7b (mod16)$ – Aug 19 '24 at 14:14
@MathBOT00101 Yes, you are right! – user1176409 Aug 19 '24 at 14:16
I'm confused...is it $1111b=(1111) \times b$ ? Or is it $1111b$ a 5-digit number where the first $4$ digits [most significant] are $1$ and the last digit is $b$? – Mike Aug 19 '24 at 17:18
1

@Mike The key property that makes this work is that the last $4$ digits can't have form $,b\cdot\color{#0a0}{1111},\ b< 9,$ because they cannot have form $,b\cdot \color{#0a0}c,$ for for $,b<2^4<\color{#0a0}{c\ \rm odd}.,$ This is not clear in the above answer so I posted an answer highlighting it explicitly (which immediately generalizes to any composite radix). – Bill Dubuque Aug 19 '24 at 19:32
I still don't get why 1111b can't be greater than 10000. Is there any way to prove that the digits won't be equal? – Aug 20 '24 at 14:44
@MathBOT00101 added in the solution. – user1176409 Aug 21 '24 at 12:47

score 1 · Answer 3 · answered Aug 19 '24 at 17:28

A power of $2$ other than $2^0=1$ must be even, so that immediately rules out $1111, 3333, 5555, 7777,$ and $9999$.

Also, no power of $2$ ends with a zero, so that rules out $0000$.

It remains to consider $2222, 4444, 6666,$ and $8888$. None of them are divisible by $16$, so they cannot be the last $4$ digits of any power of $2$ (as $10000$ is divisible by $16$ and so are all powers of $2$ greater than or equal to $16$).

Prove that a power of 2 cannot end with four equal digits.

3 Answers3