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Number Theory Level 3

Given that $9<2^n<2^k$ , is the next sentence true or false?

It is possible that $2^n$ and $2^k$ have the same digits, just in different orders.

(For example the numbers $1288, 8128$ have the same digits in an other order, but numbers $1288, 1182$ don't have the same digits.)

True False

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Áron Bán-Szabó
May 14, 2017

We will use indirect proof. Suppose it is possible. Since they have the same digits, $9|2^k-2^n$ , so $9|2^n*(2^{k-n}-1).$ We know that $9$ can't be a divisor of $2^n$ (because $2^n$ have only $2^x$ divisors), so $2^{k-n}-1$ is a multiple of $9$ . $2^{k-n}=2^k/2^n<10$ (because they have the same digits). Then $0<2^{k-n}-1<9$ , so it can't be a multiple of $9$ . We have a contradiction.

So it's not possible.

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