A power of 2

Relevant wiki: Proof by Contradiction

Suppose the digits of $2^a$ can be rearranged to give $2^b,$ where $a < b.$

Consider $2^b - 2^a.$

Since the digits of $2^b$ are a rearrangement of the digits of $2^a,$ it follows that $9$ is a factor of $2^b - 2^a.$

As $2^3 < 10 < 2^4,$ there can be at most four powers of 2 with the same number of digits.

Therefore we must have $2^b - 2^a = c \times 2^a,$ where $c = 1, 3,$ or $7.$

The prime factorization of $2^b - 2^a$ does not contain $3^2 = 9.$

Hence the answer is $\color{#D61F06} \boxed{\text{no}}$

Suppose the digits of A can be rearranged to B (A and B are powers of 2 , A>B)

then A≡B (mod 9) , because sum of digits of A is equal to B's

and A and B are powers of 2 so A=2B or 4B or 8B

1) A=2B : 2B≡B (mod 9) so B≡0 (mod 9) : contradiction

2) A=4B 4B≡B (mod 9) so B≡0 (mod 3) : contradiction

3) A=8B 8B≡B (mod 9) so B≡0 (mod 9) : contradiction

By 1), 2), 3) the answer is "No"

Let the two solutions be $2^b$ and $2^a$ and assume $2^a > 2^b$ . For these two numbers to be permutations of each other, we must have $2^a\equiv 2^b \mod(9)$ . As $\phi(9) = 6$ and $2^1\not\equiv 1 \mod(9)$ , $2^2\not\equiv 1 \mod(9)$ , and $2^3\not\equiv 1 , \mod(9)$ , we have $v_2(9) = 6$ . Therefore, we need the difference between $a$ and $b$ to be at least 6 for them to be permutations of each other. However if this were the case, then $2^a = 2^b\cdot 2^6 = 64 \cdot 2^b > 10\cdot 2^b$ and $2^a$ and $2^b$ couldn't have the same number of digits. So they can't be permutations of each other. Therefore, the answer is No.