Is 20482048 the highest power of 22 with all even digits (base ten)?

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,,45N1

Number of suffixes with all even digits in cycle: 10,25,60,150,370,925,2310,5780,42.5N1

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 (23789535319=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:


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.)


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 09 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:


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:


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 2a, and the (N+1)th digit is odd, then one of the following will have the (N+1)th digit even:


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 2168 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 kth digit of 2n (with a0 representing the 1’s place.)

ak=2n10k mod 10ak=2nk5k mod 10

We can write
where d is the divisor and r is the remainder of 2nk/5k. So

a_k \equiv \frac{2^{n-k}-r}{5^k} \equiv d \pmod{10}

\Rightarrow a_k \equiv d \pmod{2}
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.


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).

Source : Link , Question Author : Brian Rothstein , Answer Author : Zander

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