Powers of 2

For $n \geq 4$ , $2^n$ is a multiple of 16. Also, $10\:000$ is a multiple of 16.

However, since $1111$ is odd, none of its multiples $1111a$ with $a \leq 9$ is a multiple of 16. Therefore $10\:000k + 1111a$ is not a multiple of 16, and cannot be a power of two.

A power of 2 can only end in the digits $2,4,6,8$ .

We note that the only instance where a power of 2 has the last 2 digits the same are with $44$ and $88$ which occur at $2^{18}$ and $2^{19}$ . This can easily be checked by listing out powers of 2 until $2^{20}$ .

Using logical reasoning, this means that the number at hand must end with $4444$ or $8888$ . However if the number ends with $4444$ , the previous power of 2 must be $2222$ or $7222$ $mod$ $10000$ . We know this is not possible as no power of 2 can end with $22$ .

The same logic applies for a power of 2 ending in $8888$ .

Suppose such a number exists, say $N = 10^{4}k + aaaa$ , where $aaaa = a \times 1111$ is the four digit number. Clearly , $a \neq 0$ , because then $5|N$ . Now $16|10^{4}$ , but $v_{2}(aaaa)= v_{2}(a)<4$ as $a$ is a single digit. Thus the maximum power of $2$ in $N$ is $3$ , which makes no sense, as $N$ is at least a four-digit power of $2$ .

Relevant wiki: Divisibility Rules (2,3,5,7,11,13,17,19,...)

The powers of 2 can only end with 2, 4, 6 or 8. We notice that 2222, 4444, 6666 and 8888 are not divisible by 16, then none of them is a power of 2.

All powers of 2 are even numbers, since 2 itself is an even number. Since no power of 2 contains a factor of 5, no power of 2 can end in 0, or ...0000. In order to actually have at least 4 identical digits, n > 10. If 2^n ends in ...2222, then 2^(n-1) must end in ...6111 or ...1111. But no power of 2 is odd, so a =/= 2. If a = 4, it is possible that 2^(n-1) could end in ...7222 or ...2222. Since n>10, n-2 > 8, and 2^(n-2) must be an even number. But 2^(n-2) is odd, an impossibility. Similarly, if a = 8, 2^n = ...8888. 2^(n-1) must equal ...9444 or ...4444. Then 2^(n-2) becomes ...4722 or ...2222. Last, 2^(n-3) is ...2361 or ...1111. Since n-3 > 7, and 2^(n-3) must be even, this is also impossible. Final case, a = 6. Then 2^n = ...6666, and 2^(n-1) can either be ...8333 or ...3333. This, too, is impossible. Therefore, the assumption that some a exists must be false.

Reason this in steps. a can't be 2 or 6 because a number that ends in 22 or 66 is not a multiple of 4, whereas a number with at least 4 digits and an integer solution for n is. Neither can it be 0 because at no point is there a multiplication by a multiple of 5. We then move on to 4 and 8. However, since 2^n is a multiple of 16 if it has at least 4 digits, it can't end with 4444 or 8888 as those are not multiples of 16.

If xnnnn (where x is a string of unknown digits, and n is one digit) is a power of 2 it can be divided by 2 a lot of times. Clearly n must be even. If n is 2 or 6 after dividing by 2 once we have x111 or x333 and are bust, if x is 4 then we are bust after dividing twice when we are on x11, and if x is 8 then after dividing by 2 three times we are on x1. So x must be zero. However if the number ends in zero it is a multiple of ten and cannot be a power of 2. Therefore there is no possible n.