Is $2048$ the highest power of $2$ with all even digits (base ten)?

Question

I have a friend who turned $32$ recently. She has an obsessive compulsive disdain for odd numbers, so I pointed out that being $32$ was pretty good since not only is it even, it also has no odd factors. That made me realize that $64$ would be an even better age for her, because it's even, has no odd factors, and has no odd digits. I then wondered how many other powers of $2$ have this property. The only higher power of $2$ with all even digits that I could find was $2048.$

So is there a larger power of $2$ with all even digits? If not, how would you go about proving it?

I tried examining the last $N$ digits of powers of $2$ to look for a cycle in which there was always at least one odd digit in the last $N$ digits of the consecutive powers. Unfortunately, there were always a very small percentage of powers of $2$ whose last $N$ digits were even.

Edit: Here's a little more info on some things I found while investigating the $N$ digit cycles.

$N$: $2,3,4,5,6,7,8,9$

Cycle length: $20,100,500,2500,12500,62520,312500,1562500,\dotsc, 4\cdot 5^{N-1}$

Number of suffixes with all even digits in cycle: $10, 25, 60, 150, 370, 925, 2310,5780,\sim4\cdot2.5^{N-1}$

It seems there are some interesting regularities there. Unfortunately, one of the regularities is those occurrences of all even numbers! In fact, I was able to find a power of $2$ in which the last $33$ digits were even $(2^{3789535319} = \dots 468088628828226888000862880268288)$.

Yes it's true that it took a power of $2$ with over a billion digits to even get the last $33$ to be even, so it would seem any further powers of $2$ with all even digits are extremely unlikely. But I'm still curious as to how you might prove it.

Edit 2: Here's another interesting property I noticed. The next digit to the left of the last $N$ digits will take on every value of its parity as the $N$ digits cycle each time. Let me illustrate.

The last $2$ digits cycle every $20$ powers. Now examine the following:

$2^7 = 128$
$2^{27} = \dots 728$
$2^{47} = \dots 328$
$2^{67} = \dots 928$
$2^{87} = \dots 528$
$2^{107} = \dots 128$

Notice that the hundreds place starts out odd and then proceeds to take on every odd digit as the final 2 digits cycle.

As another example, let's look at the fourth digit (knowing that the last 3 digits cycle every 100 powers.)

$2^{18} = 262144$, $2^{118} = \dots 6144$, $2^{218} = \dots 0144$, $2^{318} = \dots 4144$, $2^{418} = \dots 8144$, $2^{518} = \dots 2144$

This explains the power of 5 in the cycle length as each digit must take on all five digits of its parity.

EDIT 3: It looks like the $(N+1)$^st digit takes on all the values $0-9$ as the last $N$ digits complete half a cycle. For instance, the last $2$ digits cycle every $20$ powers, so look at the third digit every $10$ powers:

$2^{8} = 256$, $2^{18} = \dots 144$, $2^{28} = \dots 456$, $2^{38} = \dots 944$, $2^{48} = \dots 656$, $2^{58} = \dots 744$, $2^{68} = \dots 856$, $2^{78} = \dots 544$, $2^{88} = \dots 056$, $2^{98} = \dots 344$

Not only does the third digit take on every value 0-9, but it also alternates between odd and even every time (as the Edit 2 note would require.) Also, the N digits cycle between two values, and each of the N digits besides the last one alternates between odd and even. I'll make this more clear with one more example which looks at the fifth digit:

$2^{20} = \dots 48576$, $2^{270} = \dots 11424$, $2^{520} = \dots 28576$, $2^{770} = \dots 31424$, $2^{1020} = \dots 08576$, $2^{1270} = \dots 51424$, $2^{1520} = \dots 88576$, $2^{1770} = \dots 71424$, $2^{2020} = \dots 68576$, $2^{2270} = \dots 91424$

EDIT 4: Here's my next non-rigorous observation. It appears that as the final N digits cycle 5 times, the $(N+2)$^th digit is either odd twice and even three times, or it's odd three times and even twice. This gives a method for extending an all even suffix.

If you have an all even N digit suffix of $2^a$, and the (N+1)^th digit is odd, then one of the following will have the (N+1)^th digit even:

$2^{(a+1*4*5^{N-2})}$, $2^{(a+2*4*5^{N-2})}$, $2^{(a+3*4*5^{N-2})}$

Edit 5: It's looking like there's no way to prove this conjecture solely by examining the last N digits since we can always find an arbitrarily long, all even, N digit sequence. However, all of the digits are distributed so uniformly through each power of 2 that I would wager that not only does every power of 2 over 2048 have an odd digit, but also, every power of 2 larger than $2^{168}$ has every digit represented in it somewhere.

But for now, let's just focus on the parity of each digit. Consider the value of the $k^{th}$ digit of $2^n$ (with $a_0$ representing the 1's place.)

$$ a_k = \left\lfloor\frac{2^n}{10^k}\right\rfloor \text{ mod 10}\Rightarrow a_k = \left\lfloor\frac{2^{n-k}}{5^k}\right\rfloor \text{ mod 10} $$

We can write $$2^{n-k} = d\cdot5^k + r$$ where $d$ is the divisor and $r$ is the remainder of $2^{n-k}/5^k$. So $$ a_k \equiv \frac{2^{n-k}-r}{5^k} \equiv d \pmod{10} $$ $$\Rightarrow a_k \equiv d \pmod{2}$$ And $$d\cdot5^k = 2^{n-k} - r \Rightarrow d \equiv r \pmod{2}$$ Remember that $r$ is the remainder of $2^{n-k} \text{ div } {5^k}$ so

$$\text{The parity of $a_k$ is the same as the parity of $2^{n-k}$ mod $5^k$.}$$

Now we just want to show that for any $2^n > 2048$ we can always find a $k$ such that $2^{n-k} \text{ mod }5^k$ is odd.

I'm not sure if this actually helps or if I've just sort of paraphrased the problem.

EDIT 6: Thinking about $2^{n-k}$ mod $5^k$, I realized there's a way to predict some odd digits.

$$2^a \pmod{5^k} \text{ is even for } 1\le a< log_2 5^k$$

The period of $2^a \pmod{5^k}$ is $4\cdot5^{k-1}$ since 2 is a primitive root mod $5^k$. Also

$$2^{2\cdot5^{k-1}} \equiv -1 \pmod{5^k}$$

So multiplying any $2^a$ by $2^{2\cdot5^{k-1}}$ flips its parity mod $5^k$. Therefore $2^a \pmod{5^k}\text{ }$ is odd for

$$1 + 2\cdot5^{k-1} \le a< 2\cdot5^{k-1} + log_2 5^k$$

Or taking the period into account, $2^a \pmod{5^k} \text{ }$ is odd for any integer $b\ge0$ such that

$$1 + 2\cdot5^{k-1} (1 + 2b) \le a< 2\cdot5^{k-1} (1 + 2b) + log_2 5^k$$

Now for the $k^{th}$ digit of $2^n$ ($ k=0 \text{ } $ being the 1's digit), we're interested in the parity of $2^{n-k}$ mod $5^k$. Setting $ a =n-k \text{ } $ we see that the $k^{th}$ digit of $2^n$ is odd for integer $b\ge0$ such that

$$1 + 2\cdot5^{k-1} (1 + 2b) \le n - k < 2\cdot5^{k-1} (1 + 2b) + log_2 5^k$$

To illustrate, here are some guaranteed odd digits for different $2^n$:

(k=1 digit): $ 2\cdot5^0 + 2 = 4 \le n \le 5 $
(k=2 digit): $ 2\cdot5^1 + 3 = 13 \le n \le 16 $
(k=3 digit): $ 2\cdot5^2 + 4 = 54 \le n \le 59 $
(k=4 digit): $ 2\cdot5^3 + 5 = 255 \le n \le 263 $

Also note that these would repeat every $4\cdot5^{k-1}$ powers.

These guaranteed odd digits are not dense enough to cover all of the powers, but might this approach be extended somehow to find more odd digits?

Edit 7: The two papers that Zander mentions below make me think that this is probably a pretty hard problem.

Answer 1

This seems to be similar to (I'd venture to say as hard as) a problem of Erdős open since 1979, that the base-3 representation of $2^n$ contains a 2 for all $n>8$.

Here is a paper by Lagarias that addresses the ternary problem, and for the most part I think would generalize to the question at hand (we're also looking for the intersection of iterates of $x\rightarrow 2x$ with a Cantor set). Unfortunately it does not resolve the problem.

But Conjecture 2' (from Furstenberg 1970) in the linked paper suggests a stronger result, that every $2^n$ for $n$ large enough will have a 1 in the decimal representation. Though it doesn't quantify "large enough" (so even if proved wouldn't promise that 2048 is the largest all-even decimal), it looks like it might be true for all $n>91$ (I checked up to $n=10^6$).

Answer 2

This sequence is known to the OEIS.

Here are the notes, which give no explicit answer but suppose that your conjecture is correct:

Are there any more terms in this sequence?

Evidence that the sequence may be finite, from Rick L. Shepherd (rshepherd2(AT)hotmail.com), Jun 23 2002:

1) The sequence of last two digits of $2^n$, A000855 of period $20$, makes clear that $2^n > 4$ must have $n = 3, 6, 10, 11,$ or $19 (\text{mod }20)$ for $2^n$ to be a member of this sequence. Otherwise, either the tens digit (in $10$ cases), as seen directly, or the hundreds digit, in the $5$ cases receiving a carry from the previous power's tens digit $\geq 5$, must be odd.

2) No additional term has been found for n up to $50000$.

3) Furthermore, again for each n up to $50000$, examining $2^n$'s digits leftward from the rightmost but only until an odd digit was found, it was only once necessary to search even to the 18th digit. This occurred for $2^{12106}$ whose last digits are $\ldots 3833483966860466862424064$. Note that $2^{12106}$ has $3645$ digits. (The clear runner-up, $2^{34966}$, a $10526$-digit number, required searching only to the $15$th digit. Exponents for which only the $14$th digit was reached were only $590, 3490, 8426, 16223, 27771, 48966$ and $49519$ - representing each congruence above.)

Answer 3

Here's a heuristic justification that none exist.

The odds of a uniformly-randomly chosen $n$-digit number has all of its digits even is $$ p(n) = \frac{5}{9} \frac{1}{2}^{n-1} $$

(note that your last digit data supports this method of analysis: roughly $1/2^n$ of the last $n$-digits are all-even)

On average, there are $\log_2(10)$ $n$-digit numbers in the sequence of powers of 2.

So, at random, the odds that there does not exist a power of 2 of $k$ or more digits that are all even is $$ \prod_{n=k}^{\infty} \left(1 - \frac{5}{9} \frac{1}{2^{n-1}}\right)^{\log_2 10}$$ For $k=10$, wolfram alpha computes this to be $0.992814$. So there's a less than 1% chance of there being another power of 2 whose digits are all even.

Of course, after you exhaust the small ones to see one doesn't exist, you can plug in much larger values into $k$....